Pass Transistor Driver Problems - Page 1
Forum section: electronics / projects

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 25, 2016, 09:57:33 pm
content: Hi,I'm designing a psu and I'm having problems with my pass transistors driver.the problem is that if I for example set the current to be limited to 3A, it does limit to 3A, but when I change to load to higher resistence that is less then the set current (for example 10mA load while the current is limited to 3A) the output voltage is higher then set.It does regulate the voltage correctly for set currents less then 1A, but for anything 1A the output voltage is not what I set.Schematic here:

author: TLengr
author_rank: Contributor
timestamp: March 26, 2016, 02:43:59 am
content: Hello there,Did you design this regulator circuit yourself? One possible answer is that the series pass transistors may not be operating in their linear region with load current above 1A. You might want to measure the series pass C-E voltage while varying the load current. What is the expected output voltage and at which point does it appear? I think maintaining voltage regulation over a 0.01A - 3A (300:1) range is probably asking a lot of any regulator. However, at low load currents, it looks like U3/Q3 should provide some shunt loading to help hold the output voltage down. Hope this helps.Regards

author: mij59
author_rank: Frequent Contributor
timestamp: March 26, 2016, 05:06:35 am
content: The output voltage is reference to Q2, the voltage_set is reference to ground, therefore the voltage drop across R3 is subtracted from the output voltage.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 26, 2016, 11:47:36 am
content: Quote from: mij59 on March 26, 2016, 05:06:35 amI think I fixed it, I'm trying to use summing amplifier that will take the set voltage + the voltage drop across the sense resistor, and add them together - so voltage set = voltage output.now it seems like it fixed.. but I'm still playing with it.

author: TLengr
author_rank: Contributor
timestamp: March 26, 2016, 01:46:51 pm
content: "I think I fixed it ..."Congratulations! If I may offer a word of caution; be careful, you have not modeled a source impedance for V3. In your SPICE model this allows the collector voltage of the series pass to remain constant which is usually not the case in the "real world". Good luck.

author: mij59
author_rank: Frequent Contributor
timestamp: March 26, 2016, 02:14:21 pm
content: Quote from: ZeTeX on March 26, 2016, 11:47:36 amOr disconnect the ground from V3 and R3, connect the ground to Q2 and connect the lead from R3 to the negative terminal of V3, and use a inverting current sense amplifier.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 26, 2016, 03:56:49 pm
content: Quote from: mij59 on March 26, 2016, 02:14:21 pmDoesnt work, the loop osciliates..what have I done wrong?

author: Kleinstein
author_rank: Super Contributor
timestamp: March 26, 2016, 04:48:47 pm
content: I would expect the down-programmer part to oscillate. There is another mistake in the part, as there is no well defined bias between the two possible outputs. So any offset in the OPs changes things a lot. So feedback for the sink circuit should have an extra small resistor to the output and feedback from there instead of the output.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 26, 2016, 04:53:30 pm
content: Quote from: Kleinstein on March 26, 2016, 04:48:47 pmDown-programmer unfourmently has nothing do with it, if I remove it completely it doesnt change anything.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 26, 2016, 05:17:45 pm
content: The output stage still has a lot of large power transistors but only a simple resistor providing base current. This might be rather slow and would not give much current anyway. So at least something like a Darlington configuration is needed, thus one more NPN transistor to drive all the TIP35 in parallel. Only 0.1 Ohms emitter resistors is rather low - this could also make stability difficult.The capacitor C10 might need to be much larger or a combination of several.The combination of output capacitors usually needs to be a a combination of low ESR capacitor and one with an ESR in the 0.1-1 Ohms range. The capacitor with some ESR (comparable with emitter resistors) might be needed. Only 1 µF is also very low - so may be difficult to get stable under all conditions, especially with slow transistors like the TIP35.Depending on the performance of the output stage, the compensation may need adjustment. Currently there is only a single cap in feedback - this might be Ok for low capacitive load, but could be a problem for higher capacitance.The capacitor directly to ground after the transistors is also not very helpful for stability.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 26, 2016, 08:34:46 pm
content: Quote from: Kleinstein on March 26, 2016, 05:17:45 pmThanks for your suggestions, really helping me alot.After all of your suggestion the loop still osciliates.here is a picture of the osciliation, blue waveform is the voltage at V5, green waveform is the curent across R9.And here is the schematic now:

author: TLengr
author_rank: Contributor
timestamp: March 27, 2016, 01:43:12 am
content: I think the next thing is to re-establish a ground return for source V3. As drawn, V3- is floating. One could ground “O2” and remove the connection of R29 from O2 and move it to the junction of R3 and V3-. Regards.

author: amspire
author_rank: Super Contributor
timestamp: March 27, 2016, 03:36:07 am
content: The oscillations are almost 100% guaranteed in the initial stages of a power supply design like this.Ideally, the opamp has a 90 degree phase shift, you have a Darlington pass stage that often adds significant phase shift, R2 will add a little phase shift, R12/C7 adds a bit, and the 22uF on the output is adding significant phase shift. You can almost bet on getting 180 degrees total phase shift somewhere and ending up with an oscillator.You are at the point you want to start to use LTSpice to look at loop gain/phase plots. There is a great tutorial here:Code:https://www.youtube.com/watch?v=YYWlPFBebfcThe two key concepts are (1) that by adding a voltage source set to 0V DC into the loop, you can use it to inject a signal for AC analysis, and by plotting the ratio of this injected AC to the output AC, you get the open loop response. Concept (2) is that you can use the STEP option to run the curve with multiple component values or multiple loads together, so you can easily see the effect of any changes.Make sure you try different loads - particularly capacitive with different ESRs. Don't get too worried about some instability with a 10,000uf capacitor with no ESR - most supplies are not fully stable for every load, and instability with a huge capacitance is not too serious as the big capacitance is flattening the voltage anyway. In fact, I would add an extra resistor to simulate the ESR of the 22uF capacitor as it does affect the phase margin with low loads and resistive loads.Pay particular attention at the phase shift at the 0dB loop gain point with the different loads.Once you get the hang of analysing loop gain, you will find you can analyse the gain and phase shifts of individual sections - like the Darlington stage. Adding a resistor from the emitter of Q4 to the junction point of all the 1 ohm emitter resistors will make the current in Q2/Q4 more constant over the different loads and may make their behaviour more predictable as far as AC analysis goes.Once you get the voltage loop right, you have to do it again for the current limit loop.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 27, 2016, 04:43:43 am
content: Quote from: TLengr on March 27, 2016, 01:43:12 amAnd then where would the opamp non inverting input would go? because if it will stay ground, the op amp inputs would be shorted, if I change the non inverting input to the negative terminal of V3, it doesnt work (the current doesnt regulate, the voltage is kinda weird but regulates.amspire - thanks your very much for helping, I will come back later and watch, hopefully I could understand all of this.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 27, 2016, 05:30:53 am
content: Gah, why are you people always recreating this circuit, and why is it always so bad? Is there really truly no reference design out there to start from?The "30V, 3A power supply" must be the most abused circuit in the history of this forum.Tim

author: amspire
author_rank: Super Contributor
timestamp: March 27, 2016, 06:32:31 am
content: Quote from: T3sl4co1l on March 27, 2016, 05:30:53 amThere are reasons. First, everyone needs a power supply, so it is usually the first major learning project people tackle.Also every power supply design is imperfect in some way - stability, transient behaviour, drift, noise. If you look at any of the HP supply ranges where they make a 15V model, a 30V model, a 60V model and a 100V model, all use the same basic circuit, but everyone has a fair number of different parts. The compensation parts are all different. The nature of power supplies is that you change a few things and you have to re-optimise the design.If someone can design a single power supply design that is easy to customise, is not over complicated, is not expensive, does not require a specific transformer made from unobtainium, never fails, can work with other supplies in parallel or tracking mode and does absolutely everything almost perfectly would be great. Please, I want to see this circuit!Richard

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 27, 2016, 06:36:35 am
content: I'd love to design one, but there's no incentive to do so...And who cares about HP anything: the only people who are interested in making/buying these things are cheap, or beginners, for which one of those shitty nameless Chinese units is fine*.So I'm left yelling at the screen whenever these things pop up...(*Fine in the sense of, "it'll blow up on you after a few weeks/months of use, but it was only $30, so who cares.")Tim

author: Kevin.D
author_rank: Frequent Contributor
timestamp: March 27, 2016, 02:54:59 pm
content: Zetex :- As you have it at the moment You have left V3 floating .You have to put a GND connection at top of R3 (O2). Then invert U4 connections ( so leave noninv to GND as it is , then connect R4 to low side of R3) .Regards

author: Kleinstein
author_rank: Super Contributor
timestamp: March 27, 2016, 04:06:21 pm
content: The circuit is trying to work with a kind of integral regulator only - this seldom works.The first step would be to test an optimize the output stage alone, usually there is a resistor bypassing the base emter junction of the final transistors. This helps to turn of the transistors a little faster and improves operation at low output currents. For protection there usually needs to be a kind of diode to prevent the base of the Darlingtion to go to negative - as a consequence the current through the PNP also needs some current limiting. This looks like small details, but its better to have them there from the start, not to get surprised later from possible negative effects.For later stability the output current into a high capacity load as a function of input voltage is important. This should give a relatively fast (high) frequency response and not too much phase shifts (preferably less than 90 deg. for most of the range).A second point for the output stage it too look at the output impedance of the transistor stage and the output caps alone. In the simulation a current source (as a spice element) as a load is very handy. This combination should be well behaved (e.g. less than 90 deg. phase shift - so representing a passive circuit). In this step the output caps, especially the damping part may need adjustment. So you see if you need 100 µF or can get away with 1 µF and suitable ESR of output capacitance for stability. Later one might want to add more low ESR capacitance to get better regulation / pulse response.The third step is than to match the voltage control loop to the output stage. This usually needs at least a proportional and integral term, that is a resistor in series with C4. One might also need a kind of phase boost across R8 to get a really fast response, but it should also work without. One way to to this is by looking at the AC output impedance with feedback. Here again like in the step before the phase shift must be less than 90 degree at all frequencies.Usually one needs to start with little feedback to start from a stable point and than adjust the compensation so that more feedback is possible. As an addition degree of freedom one could also add additional output capacitance in parallel to the one from the step before. Usually one get an output impedance to start of very low at DC or low frequencies than increase about proportional to frequency than has a maximum somewhere in the 10-1000 kHz region and than fall down to the ESR of the output capacitor. Something like a peak output capacitance of about 1-10 Ohms at 100 kHz would be a reasonable value.The final test is the pulse response, especially with pulses down to rather low currents (e.g. 1 A to 10 mA) and different extra output capacitance. Even if the simulation is stable in the small signal regime of AC simulation, the pulse response could show trouble and may require to adjust the output stage and start over. If you are lucky just add enough low ESR ouput capacitance to get good pulse response.Adjusting the current regulation is similar but usually not that critical. The cross over between CC and CV mode may still be tricky, to avoid overshoot here.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 27, 2016, 04:16:52 pm
content: Quote from: Kleinstein on March 26, 2016, 04:48:47 pmAlright, I finally got it working but then as you said the down programmer started to osciliate because of the offset between the comprator inputs." have an extra small resistor to the output and feedback from there instead of the output."so you are suggesting resistor from O1 (output) and then take the feedback for the comporator from there?if yes, then how would that work? what would the resistor do?

author: diyaudio
author_rank: Frequent Contributor
timestamp: March 27, 2016, 05:16:07 pm
content: Quote from: T3sl4co1l on March 27, 2016, 05:30:53 amI agree with Tim on that.If you really want to build a functional working circuit why not use a proven reference design ? like the e3610a ? instead you at the mercy of guesstimation.http://sites.fas.harvard.edu/~phys191r/Bench_Notes/A1/agilent_e3610a.pdf

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 01, 2016, 07:51:35 pm
content: I've made some changes with the driver and some small things, but I have this weird problem where the output current would not limit to 3A (while the voltage is set to 25V and the current to 3A) with 8.1ohm load connected, what happens is that the current just stays at 532mA. the resistor should try to draw 3.086A.I cant figure out why it keeps happeing.Schematic:

author: Kleinstein
author_rank: Super Contributor
timestamp: April 02, 2016, 07:42:10 am
content: A problem could the limited base drive current. The TIP31 don't have a very high amplification, and the 100 Ohms in series limits the current at higher output current. One might need a 3 rd transistor, one with higher gain for the first stage or a slightly lower resistance at the OP.The supply of the driving OP might limit the output voltage anyway to something like 5 V lower than it's supply.Such problems are usually easy to spot in the simulation. It's one reason to do the simulation. Though the more important purpose of the simulation is to check stability.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 02, 2016, 11:16:00 am
content: Quote from: Kleinstein on April 02, 2016, 07:42:10 amUnfortunately its not the base drive current, I've changed the transistors to FZT849, Tried reducuing R6 to lower resistence, tried replacing the op amp with LT1886 (http://www.linear.com/product/LT1886) and still the same issue.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 02, 2016, 11:36:49 am
content: My guess would be the diodes D3 and D4 - but the simulation will show for sure.

author: amspire
author_rank: Super Contributor
timestamp: April 02, 2016, 11:43:07 am
content: I agree with Kleinstein. The positive supply for the opamps is too low for the current circuit. The opamp output has to be above 28.7 volts to even turn on the output transistors with a 25V output. Opamps can easily have at least a half volt drop or more on the output, so the maximum base current the opamp can supply will be under 10mA. Maybe much less. If you try a opamp positive voltage of 35 or 40, it will probably work.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 02, 2016, 11:49:13 am
content: Quote from: Kleinstein on April 02, 2016, 11:36:49 amYes, it is those 2 diodes, The circuit is working correctly now, Thanks.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 16, 2016, 08:25:15 pm
content: I think I finished designing the regulator, it seems to be stable in simulation, no osciliations.Is there anything that stands out as "Not good"?Any improvement that could be made, or any feedback about the circuit?Thanks.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 16, 2016, 09:16:45 pm
content: The LT1037 is a good choice for U2, but not for the other OPs. It's not unity gain stable and could thus cause trouble in real life, even if it work in the simulation.C5 makes the voltage regulation relatively slow - so likely a smaller Value with a resistor (e.g. 1 K) in series is a better choice.The output stage should have a relatively small resistor (e.g. 20 Ohms range) from the emitter of Q2 to the anode of D5. This resistor helps to speed up the transistor stage.There definitely should be a diode that prevents a to large base to emitter voltage for the output stage. A fast transient could damage Q7 otherwise. This also helps a little to turn down the output - not much, but a little.With C1 and C2 one should check the type. Usually there should be one with very low ESR (e.g. 100-500 nF ceramic / film type) and one with some ESR (e.g. low ESR electrolytic). Something like 0.1 Ohms or 1 Ohms of ESR can make quite a difference here. I would guess C1 should be somewhat larger (more like 100 µF) to get a better pulse response and to reach the 0.1-0.5 Ohms range for the ESR.For simulation I would replace V4 and the switch with current source element - that's easier and allows direct changes in load current and run AC simulation to get the output impedance. The curve for the output impedance gives a hint if the supply is stable with any "passive" load.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 16, 2016, 09:44:31 pm
content: Quote from: Kleinstein on April 16, 2016, 09:16:45 pmThanks for commenting and helping, you are helping a lot and I appreciate it.the OPs are not going to be LT1037, the OPs are going to be "OP275" (DS:http://www.analog.com/media/en/technical-documentation/data-sheets/OP275.pdf).I've added base emitter diodes + emitter to D5 anode resistor, that does not seems to make things faster but not worse actually, but I have nott tested it much.I've played with C2 & C1 ESR and capacitance a lot, I figured out that with 4.7uF 0.5Ohm capacitor and 100nF at 0.01Ohm I get the supply most stable. reason for the low capacitance is because with high capacitance the current limiting takes a lot of time to start limiting. lower capacitance - faster current limiting, too low makes oscillations.One thing I cant understand is using current source as a load, its a current source, it does not sink current, how would that even work? is there any tutorial about this that I can read?I've tried plotting the output impedance by this tutorial:http://www.rocklinger.se/Elektronik/Simulering/output_imp/output_impedance_ltspice_en.htmlIt seems about 0-1ohm depending on the load. (simulation syntax is .ac oct 100 1 1000000)

author: Kleinstein
author_rank: Super Contributor
timestamp: April 16, 2016, 10:27:34 pm
content: The current source is also working as a current sink - it just set the current, not matter what. So if the circuit can't provide the current you might see negative voltages.If the circuit only works with a small output capacitor this is a bad sign. An external load might add capacitance and the supply should work in this case too. It's also hard to find a capacitor 5 µF with a fitting ESR. The speed of current limiting is more limited by the output transistors/regulator part - not by the capacitor. As the regulator is rather slow, a large capacitor might be needed to prevent large transients on load changes. So even 100 µF might be to little.The OP275 is a low noise JFET OP. Such a simple supply does not need this. So I would suggest the OP37/LT1037 (essentially the same) for U2 as this OP should be fast and low drift / offset. The other OPs are less critical, except for a high supply voltage. Sp I guess there a simple types.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 16, 2016, 10:49:24 pm
content: Quote from: Kleinstein on April 16, 2016, 10:27:34 pmDidnt know about the current soruce as current sink, learned something new- going to test it.the circuit can work with high output capacitance, it is just that the current limiting then becomes slow as the capacitor has alot of charge left, for example, 10,000µF capacitor, with C1 set to 100u, the current limiting takes about 200uS to get to the set current, while with C1 set to 4.7µF, it takes about 20uS to get to the set current.I have already OP275 from a different project - better to use them instead of ordering new OPs.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 21, 2016, 08:50:42 pm
content: I'm just posting the same schematic just with the fixed cc indicator + the things "Kleinstein" pointed out.As I said previously I dont want to make C1 any larger because it will slow the current limting.Any suggestion will be appreciated

author: StillTrying
author_rank: Super Contributor
timestamp: April 21, 2016, 10:41:26 pm
content: The inputs to the U7 and U8 LED indicators can't be right.Surely each op-amps inputs should go to both 22ks, with +/- reversed for one, so that it's output is inverted

author: Kleinstein
author_rank: Super Contributor
timestamp: April 22, 2016, 07:14:30 am
content: If you really insist in such a small output capacitor, it gets a little more tricky: In this case the regulator would need to be really fast unless you can accept a considerable drop / overshoot in voltage on fast load changes.A faster regulator should also use faster output transistors. At least the middle one can be rather easy changed to a faster one, as it does not need that much power. For getting the right compensation you would essentially start over for this - a faster compensation also needs a more accurate choice for C5,C6, R16 and so on.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 22, 2016, 09:46:53 am
content: Quote from: Kleinstein on April 22, 2016, 07:14:30 amI never got the fascination with huge caps. The leads you put on outside add more than enough to ruin it.Big caps are only ever a cheap hack of a fix for a slow regulator design.My bench supply is little more than a 20V, 10A audio amplifier; it can deliver surges up to 40A, as fast as the cables can handle it. During such a surge, the waveform is stable at both ends of the cable -- it's not dropping out at all.A faster regulator also has less windup between voltage and current modes. I still wouldn't suggest shorting it out into LEDs, but at least the LEDs won't be discharging a big stupid cap, when you do this accidentally.Tim

author: Kleinstein
author_rank: Super Contributor
timestamp: April 22, 2016, 10:47:31 am
content: There is not only the possibly current pulse from the capacitor, but there is also a delay from a slow regulator reacting to a dropping voltage / short. This is especially by this type of regulator shown here. So even it there is only a 5 µF cap at the output, one a short, a regulator might still behave like a 1000 µF capacitor. So it is not only the capacitor that is actually at the output. Current regulation in principle behaves like a capacitive source - the faster the smaller the "simulated" capacitance. Windup of the regulator can add an additional current peak - so for fast reaction a different type of regulation (e.g. shunt on the high side and floating current regulator).Usually an output capacitance in the 100 µF range is not bad. Many commercial supplies have more - sometimes in the 1000 µF range. Getting really fast, so you can use a rather small capacitance also makes the circuit layout sensitive.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 22, 2016, 12:32:51 pm
content: Yup. Also, in case it wasn't clear: my bench supply doesn't have current limiting, so it just keeps belting out the amps until something else gives. (I think it's actually hFE or PS limited around 10A.)The limited response time of a CV supply looks like inductance (i.e., the error amp can't regulate fast enough and the output dips a little, then recovers); while, that of a CC supply looks capacitive (voltage doesn't change immediately in response to current, but takes some catching up).The interaction of the two (CV to CC or vice versa) typically looks like diode reverse recovery: upon passing the threshold condition, it stays in whatever mode it was, until after that time has passed.The mechanism for this, in most circuits, is integrator windup: you have two independent integrators (normally, an integrator is an inverting amp with a C across it; in this case, it's the op-amp's intrinsic integrating behavior (dominant pole compensation) acting as a smaller C, in parallel with the R+C network explicitly used), and while one of them is inactive, it saturates to the opposite rail. The time taken to go from rail to setpoint is the recovery time, which varies with output level and load, but is on the order of the loop time constant (~ms for a switching converter -- awful!).Unfortunately, as transconductance amps are rarely used, it's very difficult to avoid this in a practical circuit.Some solutions include:https://www.eevblog.com/forum/projects/limiting-op-amp-output/msg450476/#msg450476andhttps://www.eevblog.com/forum/projects/limiting-op-amp-output/msg732595/#msg732595but these are more applicable tovoltage --> current --> Gm amp --> outputloops (the loops are stacked), not merging two loops into one.Tim

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 22, 2016, 06:57:57 pm
content: Quote from: Kleinstein on April 22, 2016, 10:47:31 amthanks, I got the psu to float but the current limiting does not work with capacitive load at the output, but does work the resistive load.C3 get charged to 3V and thats it. it starts with 12A charging current and drops slowly.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 22, 2016, 07:34:24 pm
content: Just moving the shunt to the high side is making things only worse - if you move the shunt high side, you would need to change the current regulator too, e.g. make the current regulator floating, possibly with a second isolated supply, a separate reference and so one - so not a small change but more like a new start.The downside with the old regulator type is, that the current regulator needs to bring the voltage down from a possibly high value. This takes quite some time and thus makes response slow, especially if the current loop is also running into windup, that is the OP going all the way to the positive limit and charge the capacitor all the way. At first the current can rise quite far above the set point.The old circuit uses the diodes to get the output voltage from the voltage or current regulator which ever is lower. This kind of works, but not very well with critical loads and is slow for current regulation. The alternative way is having a current setting output stage and use the diodes to use the lower current level from both regulator parts. This type of regulation is faster with current limiting, but needs a fast tuning to make the voltage regulation work well. If done right is can give good voltage regulation too. Usually it needs a slightly larger capacitor at the output, but has less virtual capacitance coming from the regulator part - so the overall current pulse on a short is usually smaller.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 22, 2016, 08:25:34 pm
content: Quote from: Kleinstein on April 22, 2016, 07:34:24 pmAlright, new start is no good. I'm back to the the previous design..I'm aware of the this downside, but I could not think of a way to stop it. "The alternative way is having a current setting output stage and use the diodes to use the lower current level from both regulator parts."could you explain about the "current setting output stage"? how does it work?I've seen this schematic :https://www.circuitspecialists.com/pdf/1802X_schematic.pdfit seems pretty simple psu, similar to mine but floating, is it good?

author: StillTrying
author_rank: Super Contributor
timestamp: April 22, 2016, 10:56:44 pm
content: Quote from: EmmanuelFaure on April 22, 2016, 09:39:46 pmThe .asc of his last but one version is on a post above.This is near enough it, but with the op-amps dragged into better positions - so I can understand it!

author: Kleinstein
author_rank: Super Contributor
timestamp: April 23, 2016, 08:53:59 am
content: The Circuit from the linkhttps://www.circuitspecialists.com/pdf/1802X_schematic.pdflooks good. This is the alternative design with current setting output stage and floating regulator circuit, very much like the older HP power supplies. So if the caps / resistors for compensation are chosen right it provides good performance. Though usually there is one more cap and 2 extra diodes.The way it is drawn is a little strange - so it may even more difficult to follow.The 3rd transformer winding is just for the relay to switch between the transformer taps. For a high current version this might to be modified a little (e.g. use 4 taps or use transistors instead of relay).

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 23, 2016, 11:59:34 am
content: Quote from: Kleinstein on April 23, 2016, 08:53:59 amwhat does this part do?also they are not using current sense amplifier, I guess because they just use low voltage to adjust. (Isense point goes directly to the current error amplifier.) would not that cause problem because the low voltage adjustment (mV range)?

author: Kleinstein
author_rank: Super Contributor
timestamp: April 24, 2016, 11:00:44 am
content: The circuit part looks like compensation for the current needed for the output divider. So the meter for the current does not include that little current used by the regulator itself.They do not use an extra amplifier for the current signal, as they don't need to. The output of the regulating OPs should be somewhere in the 0.5-1.5 V range, so not such a big change. Looking only at the current loop, this very much looks like the classical constant current sink.The circuit with the output stage as an emitter follower needs the OPs to go all the way from near zero to a little more than full voltage. So the current regulating OP may need to work with a much higher gain. In this case the extra amplification of the current signal helps to make the regulator faster - at least if the amplification is fast.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 24, 2016, 12:34:43 pm
content: Quote from: Kleinstein on April 24, 2016, 11:00:44 amThey have not specified resistor values, as for now I copied the circuit to ltspice but it doesn't work, I'm guessing because the resistors values are not chosen correctly.Whats the problem ?original circuit:https://www.circuitspecialists.com/pdf/1802X_schematic.pdf

author: Kleinstein
author_rank: Super Contributor
timestamp: April 24, 2016, 01:24:16 pm
content: The circuit has resistor value - just the strange coding as used with SMD resistors. So a resistor marked 103 should be 10*10^3 or 10 K.One also has to be careful abut where ground is connected. I think there is a small mistake in that pin 1 of the connector CON2 -> CN2/CN4 should not go to ground.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 24, 2016, 02:08:09 pm
content: Quote from: Kleinstein on April 24, 2016, 01:24:16 pmok, I will put the correct resistor values and see if it works,I noticed the mistake - didnt include it in the schematic.

author: StillTrying
author_rank: Super Contributor
timestamp: April 24, 2016, 05:01:25 pm
content: I did a LT version of that pdf. Here's a 0-1A-0 load.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 24, 2016, 05:13:39 pm
content: The circuit might need to have some ESR (e.g. 0.1 - 1 Ohms range) for C28 (the output capacitor). Normally electrolytic caps have this, but a simulation be default assumes 0 ESR.Usually there is also a capacitor in parallel with parts of R54-R56. This might be needed for extreme loads, like a perfect current sink often used in simulation.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 24, 2016, 05:22:57 pm
content: Quote from: StillTrying on April 24, 2016, 05:01:25 pmyou did what and what pdf?

author: StillTrying
author_rank: Super Contributor
timestamp: April 24, 2016, 06:12:40 pm
content: The same pdf you were trying to get working.There might be mistakes, LT1006s and LM358s showed the same 1.8k osc.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 24, 2016, 06:23:50 pm
content: Quote from: StillTrying on April 24, 2016, 06:12:40 pmno problem for me, pretty stable but I'm still working on the schematic, in the current state it doesn't oscillate. notice I used 33pF caps with 1k resistor, I'm surprised it doesn't oscillate.EDIT: I noticed in your schematic that you didnt provide any ESR to any of the caps, change the ESR of the output cap to 0.5ohm (didnt test any value yet expect 0.5?) and watch the magic.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 24, 2016, 07:49:59 pm
content: The circuit is rather stable, at least with ideal parts. The FTZ849 transistor is relatively fast which allows a fast response. 33 pF is a rather small capacitance, but something like 100 pF would not be that unusual for this type of regulator.In real life inductance of the shunt and not so perfect raw voltage (e.g. ESR/ESL of capacitor) might require a slightly slower circuit.

author: StillTrying
author_rank: Super Contributor
timestamp: April 24, 2016, 09:00:30 pm
content: I hate 33p being written as 330.Yep, 0.5R esr on the output cap is/was the only solution, - after spending an hour carefully comparing the 3 schematics, - all 3 are the same!Only did a quick test going into and out of 2A current limit using a triangular wave on the voltage control and 3R on the output - no oscillation at least.

author: timsu
author_rank: Contributor
timestamp: April 24, 2016, 10:05:47 pm
content: What would be the right way to control the output voltage with a voltage?I tried applying a voltage to the inverting input of U2, that kind of worked, but that was not linear.

author: StillTrying
author_rank: Super Contributor
timestamp: April 24, 2016, 10:55:04 pm
content: On ZeTeX's image above you connect the voltage source across R15. Yes it's non-linear and don't forget the output voltage is sort of inverted.If varying, you need 10R across the output to discharge the output cap, to see the output.Here's a 1V to 2V square voltage source.Edit here's a 10K pot to fiddle with.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 25, 2016, 09:27:57 am
content: Quote from: StillTrying on April 24, 2016, 10:55:04 pmbut it is linear, it just took me about 15 minutes to get it right.(the load was 1k)

author: Kleinstein
author_rank: Super Contributor
timestamp: April 25, 2016, 04:48:42 pm
content: The more normal way would be having the resistor at inverting input of the OP going to GND, and have the variable reference voltage at resistor R1 (in the last schematics) instead of the fixed reference.The down side is, that this will not give you an constant current load, but this is relatively easy to add. If separate it could be before the shunt, so that one might get away without the current compensation any more.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 25, 2016, 05:17:09 pm
content: Quote from: Kleinstein on April 25, 2016, 04:48:42 pm"that this will not give you an constant current load, but this is relatively easy to add. If separate it could be before the shunt, so that one might get away without the current compensation"Can you explain more? what constant current load?I did the way you said and it works great.

author: StillTrying
author_rank: Super Contributor
timestamp: April 27, 2016, 02:01:52 am
content: This is where I ended up with this experiment. It's doesn't seem three bad!Max V and I is around 3.3A at 26.6V, - you can have a bit more of one for a bit less of the other.The graph shows 2R being put on the output while the output is set at 20V and 1A, the slopes are mostly the output cap discharging and recharging when the 2R is removed.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 27, 2016, 10:43:47 am
content: Quote from: StillTrying on April 27, 2016, 02:01:52 amToo slow! ~3ms for the current limit to hit is pretty slow, maybe down programmer will fix it! because the regulator is fast, but the caps at the output have some energy left. with only 100nF caps at the output the supply is fast in the nS range, but the op amp oscillate and that's not a good sign.have not had much time to play with it in the last days but I'm going to play it with today and see.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 27, 2016, 01:11:24 pm
content: This type of circuit usually needs more that a 100 nF cap at the output. Depending on the output transistors used and the compensation , the minimum capacitance is something like 10 µF with an ESR in the 0.1-0.5 Ohms range and about 100 nF with very low ESR in parallel. Though you can get the supply stable with such a small capacitance, one might want a little more (like 100 µF) to reduce the drop and overshoot on load changes. This is still a rather good value.It's usually not that good to set the voltage by adjusting the feedback divider, as this also makes the loop gain dependent on the setting. So it's better to adjust the ref. voltage instead.If one wants it even faster, I would chose a true two quadrant output stage, which allows the output transistors to work at a minimum bias level under all conditions. But this type needs some extra effort and has higher standby power consumption / heat production. A fast regulator also gets sensitive to the layout - not a surprise if you go for an output impedance (mainly inductance in the 100 nH range) comparable to a 10 cm piece of thick (e.g. 100mm²) wire.

author: timsu
author_rank: Contributor
timestamp: April 27, 2016, 01:20:21 pm
content: Quote from: ZeTeX on April 25, 2016, 05:17:09 pmI'm not quite sure, but if I understood it right, but the current regulations seems a bit off.The current through the 0.1R shunt resistor is not exactly the same as through the load (it seems to be around 1-2mA off).I'm not quite sure where it comes from and it is also just a very small error.Also the forward voltage drop in diodes can quite differ if I read the datasheets correctly.Will this become an accuracy problem?

author: StillTrying
author_rank: Super Contributor
timestamp: April 27, 2016, 01:47:41 pm
content: Quote from: ZeTeX on April 27, 2016, 10:43:47 amI agree, could do it 30 times faster in firmware!I got it down to 0.6ms by reducing the output caps to 47u+4.7u+0.1u all with 0.5 ESR, but then the output rises by 0.5V for 20+ms when the load is removed.I've got it down to 1ms ATM with 110uf output cap for a 1A load, and 0.2V output rise. I don't think that's too bad considering the rubbish components I'm using - LM324 & 2N3055.Quote from: Kleinstein on April 27, 2016, 01:11:24 pmI agree, even on the current control op-amp where the reference is simply applied to the +ve input, there's changes in the bias currents especially while it's being used as a comparator, I'm surprised it's simulating as good as it is.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 27, 2016, 02:33:54 pm
content: Quote from: StillTrying on April 27, 2016, 01:47:41 pmI'm trying to implement a down programmer that will sink current, that way the output caps can get discharged fast, it seems to work, but the ESR of the output caps play a big rule here, The down programmer works very good with the output cap at ESR of 0.01ohm, but the circuit osciliate when the down programmer is off, but with 0.5ohm ESR the down programmer is a little bit slower but the circuit is stable.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 27, 2016, 03:41:05 pm
content: The current through the shunt is not exactly the same as the output current. There is also the base current for the output stage and the current for the voltage divider flowing through the shunt. The current for the voltage divider was compensated in the original circuit with the one extra OP.If you need an accurate current, one should use a separate shunt on the other side of the GND connection and than have a separate emitter resistor. You need emitter resistors anyway, if more than one transistor in parallel is to be used.Just as a note: essentially the same circuit also works with a MOSFET instead of the darlington transistor. This might be attractive if high speed at relatively low output voltage is needed. However MOSFETs don't work that well at relatively low current.If you want a down-programmer to make it a limited 2 (or more limited even 4) Quadrant supply, one could use a PNP transistor at the + 5 V auxiliary supply. Just to give an Idea I add an LTspice file. There are also a few modifications for slightly faster crossover from CV to CC mode. Accurate current regulation and measurement would need an extra shunt. V5 could be used to adjust the standing current or choose class B operation.

author: Kalvin
author_rank: Super Contributor
timestamp: April 27, 2016, 04:08:54 pm
content: This has been quite educating discussion thread. Designing a stable control loop is not easy. But it can be made easier using a simulator and analysing the loop stability with different loads and conditions, thus reducing the guess work.Edit: You may find the following LTSpice tutorial quite useful for analysing the loop stability:http://www.linear.com/solutions/4449For someone wanting to take a bit closer look at the basics of the feedback systems, here is an excellent set of tutorials on feedback systems and stability, breaking the loop etc. :http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-1-general-structure-and-essential-concepts/http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-3-improving-noise-linearity-and-impedance/http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-4-introduction-to-stability/http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-5-gain-margin-and-phase-margin/http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-6-new-and-improved-stability-analysis/http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-7-frequency-dependent-feedback/http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-9-breaking-the-loop/http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-10-stability-in-the-time-domain/For those who like to see videos instead, here is a nice introduction to feedback systems by Prof. J. K. Roberge:A lecture and a demo for a feedback system compensation by Prof. J. K. Roberge:

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 28, 2016, 03:36:17 pm
content: Quote from: Kleinstein on April 27, 2016, 03:41:05 pmI have not found a difference in speed between MOSFET and NPN transistor, Could you explain about the opreation of the down programmer?I've downloaded your file, but its a little bit messy and I'm having hard time understanding how it works, because copying it directly to the schematic doesn't work.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 28, 2016, 04:50:25 pm
content: The down programmer / second quadrant part might be a little difficult to under stand. Sorry for the confusing order, the negative output is on the upper side of the current source/sink and the positive side is at the bottom, where the ground symbol is. So a little explanation:The main idea is to use the positive auxillary supply to pull up the negative output side. So using this as a way to allow slightly negative output voltages and have room for the transistors to work even at zero output voltage. The very basic circuit would be just Q3 and a resistor from the +5 to the emitter of Q3 - this would give a constant standing current.V5 and Q8 check if the controlling voltage for the normal output stage is going to low so that the current is set to zero or very low. At low control voltage more current is flowing through Q8. From V5 one could adjust the point where the negative side sets in.Q9 with R16 and R15 works like a crude kind of current mirror and amplification of the current. The voltage amplification from Q8 compensates for the higher value of R15 compared to the R1 (0.1 Ohms shunt). Q10 is a crude, fixed current limiting for the negative side current.Q3 is than just there to reduce the power for Q8 and thus allow a small and fast type there. As a second effect it prevents output voltages below about -0.7 V.At high currents MOSFETs are rather fast compared to power BJTs, especially if it comes to turning off. This is at least true if the circuit is made to have a reasonable strong gate drive. However the downside is that they behave different at low currents, so not so much speed advantage unless a considerable standing current is used. Normally I see not big advantage in using MOSFETs - for this circuit the current reading would be a little more accurate compared to the BJT version.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 29, 2016, 11:11:19 am
content: Quote from: Kleinstein on April 28, 2016, 04:50:25 pmI have a problem where, if I set V5, to higher voltage, then at the output the current get limited to higher value then I set in op amp that limits the current.Are there any advantages to this circuit then just shorting the output if the current that flowing is higher then the set current like I did before?

author: Kleinstein
author_rank: Super Contributor
timestamp: April 29, 2016, 01:30:12 pm
content: The down programmer should operate in a linear mode too, so allowing for 2 quadrant operation and support regulation at very low current. This way you could even do load steps like +1 A to -10 mA and they behave very much like a 1 A to 100 mA step. Just switching in a resistor will not work well: to much at high voltage and to little at low voltage plus some extra glitches - so I don't think this would be an option at all.If you set the voltage limit to high (e.g. more than about 2 V for V5 in the circuit) there can be too much standing current. But there is a reasonable large range (e.g. 1.2 - 1.9 V) that should work OK. Even without a standing current and thus in class B mode the dead zone is not that large. One could get away without Q3 and the extra 10 µF capacitor for speed up if you want is simpler. In a real circuit the voltage should be something like 3 diode drops or a VBE multiplier to compensate for temperature dependence. It is rather similar to setting the bias current in an audio amplifier, though it is less critical here as we are not concerned with crossover distortion.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 29, 2016, 02:03:34 pm
content: Quote from: Kleinstein on April 29, 2016, 01:30:12 pmthen, maybe not switch just a resistor, what about turning on constant current sink? therefore the drawn current will always be the same, but even if I will just short the output with 0ohm resistor (direct short) it will still work great even at low voltages.It doesn't matter if I set V5 to 1.2V or 1.9V, the output waveform is the same thing.I'm also trying to a find a "true" 2 quadrant bench power supply schematics on google and I did had much luck with that.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 29, 2016, 05:07:01 pm
content: One can use a constant current sink. It's possible with just the PNP transistor with the base to GND and a resistor from the emitter to +5 V.However this gets a little wasteful if a higher negative current is desired. So something like 50 or 100 mA might be ok, though already adding some waste heat all the time, even if a large positive current is needed. The maximum current is reduced by the maximum negative current set. Having a variable negative current might also allow for a slightly faster setting of the regulation as the range with very low current (and thus usually the lowest speed) of the power transistor can be avoided. However this only works if the negative side is faster than the NPN at low currents.

author: StillTrying
author_rank: Super Contributor
timestamp: April 30, 2016, 12:31:07 am
content: This 2 op-amp circuit is always going to have a delay/gap in swapping control from V to I. When the V op-amp output is controlling at around 0.5V, the I op-amp output has to come down from 3.6V to take over, not only is there a gap/delay before it does, but during this time the V op-amp(in linear mode) moves in thewrongdirection and overshoots before the I amp gets down to 0.5V.Here's a messy graph showing the pass tr (grey) overshooting by 1.5A before the I amp takes over to bring it down to the 0.5A set limit.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 30, 2016, 07:20:55 am
content: Quote from: StillTrying on April 30, 2016, 12:31:07 amThis problem can be reduced by having the local capacitive feedback at the OPs mainly from behind the diodes. Besides the down programmer this was the second change I had in the circuit. The plan is a little messy in that area and there is still some direct feedback capacitor - some OPs might need that. No need for the series resistors with these caps.This will not fully eliminate the problem, but it can be reduced by something like a factor of 10.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: April 30, 2016, 10:54:04 am
content: Quote from: Kleinstein on April 30, 2016, 07:20:55 amIs moving the output caps behind the current sense resistor an option? that way the caps get charged faster when the current limit is at low value, and the voltage ramps up quickly, but I wonder if it can effect stability?EDIT: it those, got it to osciliate with just 0.01ohm load when the caps are behind the sense resistor.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 30, 2016, 03:07:59 pm
content: Moving the output caps to the other side of the shunt does not help at all. Capacitance there is not helpful but more a problem.To reduce the current overshoot one can limit the voltage at the output of the current regulation to a value just higher than needed worst case. Also making the regulation part relatively fast helps. It's the principle problem of having the current limiting first inactive not to interfere with current regulation and than having an integrating regulator to start.

author: StillTrying
author_rank: Super Contributor
timestamp: April 30, 2016, 07:55:28 pm
content: Well this is where I'm up to in StillTrying to improve the 2 amp circuit's V to I crossover, don't.The 1st graph shows the two amps/comp outputs swapping control from V to I, and the pass tr's current during this time. The swap over only takes 880ns, but then 5uS to stabilize to the set current, - and the load still has the charge in the (now 3.2uF) output cap to deal with.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 30, 2016, 08:28:00 pm
content: The LM339 is a comparator - this might cause trouble in a real circuit. Also 3 meg resistors are rather large, not practical for a fast real circuit. This circuit might very well oscillate.It can help to have the current regulation considerably faster than the voltage regulation. So the current won't go up that much until current regulation sets in. For the cross over from current regulation to voltage regulation the capacitor in the divider and the limiting diodes can help - at least in may cases. Cross over will be slow, but at least usually coming from below.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: May 04, 2016, 10:30:05 am
content: Quote from: StillTrying on April 30, 2016, 07:55:28 pmCould you upload your .asc file please?

author: StillTrying
author_rank: Super Contributor
timestamp: May 04, 2016, 05:23:54 pm
content: Quote from: ZeTeX on May 04, 2016, 10:30:05 amI don't think it's very good.https://github.com/evenator/LTSpice-Libraries/blob/master/sub/LM339.sub

author: ZeTeX
author_rank: Frequent Contributor
timestamp: May 07, 2016, 01:12:21 pm
content: Just wondering, for performance, how about using IGBT as a pass element?As "Kleinstein" pointed out that MOSFET can be faster, but they don't work well at low currents, and they need protection and things like that, but NPN are pretty robust but slower compare to some MOSFETS, then how about combine both (IGBT) and get fast speed + robustness? do people using IGBT as a pass element? if I would be able to find IGBT model to LTspice I might try it out.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 07, 2016, 03:30:53 pm
content: IGBTs are usually not that fast. There are a few audio amplifiers that use IGBTs, but that is more an exception. Most IGBTs are made for switching applications, so they may have similar SOA problems as modern MOSFETs. Also most IGBTs are made for high voltage (e.g. > 600 v) as this is the area where they might make sense in switching applications. In Principle one could combine a large NPN with a N-MOSEFT to get similar properties.For most cases my choice would be BJTs (e.g. 2SD1047) made for audio applications: they are reasonably fast and have good SOA curves, that are reliable, as they are made for analog operation. One problem might be that good audio transistors are also candidates to find fakes, but I don't think one needs so fast ones (e.g. SD4700 or similar) unless one gets a really good layout. Fast circuits can also get sensitive to parasitic inductance and coupling.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: May 07, 2016, 05:06:55 pm
content: Quote from: Kleinstein on May 07, 2016, 03:30:53 pmOK then, if IGBTs are not that fast that they are wroth over an NPN in my case then its good.Is there any transistor that you will recommend from here?:http://www.taydaelectronics.com/t-transistors/mj-series.htmlhttp://www.taydaelectronics.com/t-transistors/tip-series.htmlhttp://www.taydaelectronics.com/t-transistors/other-transistors.htmlWhat about 2SC5200? DS:https://www.fairchildsemi.com/datasheets/2S/2SC5200.pdfIt cost about 2.7$ which is nice.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 07, 2016, 05:54:45 pm
content: TO247 / TO218 and similar cases are good, as you can have them an the board, so no loose cables. TO220 is not that good for higher power - so OK for something like 40 W at most. So it might work for an 1 A supply. TO3 can be difficult to mount and often is more expensive. I personally don't like them.The 2SC5200 is a very good audio transistor. So it is a viable choice, if from a credible source, but it's a type where you can find fakes if you get them from EBAY or a discount seller. $2,70 is not too bad, but still not cheap.TIP3055 (cheap), TIP35 (more robust) and TIP140 (Darlington) are good for a low cost version, though not really fast. If you can live with something like 100 µF (per transistor) at the output, they should be fast enough. In the simulation I even got 5 µF output cap with a 2N3055 stable, though a faster transistor might make it easier.With these lower cost types you are less likely to find fakes.It depends on the source, which transistors you might get for a good price.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: May 07, 2016, 08:35:37 pm
content: Quote from: Kleinstein on May 07, 2016, 05:54:45 pmI will get the 2SC2500, not from ebay of course.In the meanwhile, should I be concerned about high peak currents when the current limiting kicks in?Here is 1ohm resistor with voltage set to 25V and current to 0.3A.here it is only 6A~ but the lower the resistance the higher the peak current (can get to 400A+).expect increasing the capacitance (or lowering to about 1uF when its nice a fast, so the peaks exist but they are for much shorter time), and the speed of the regulator, what can I do about it if I even should?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 07, 2016, 08:41:25 pm
content: IGBTs have higher current density, and therefore higher power density, than BJTs. BJTs have to be specially designed to dissipate high power levels at large voltage drops (high power density). Otherwise you get second breakdown.IGBTs aren't rated for more than ~100s of microseconds of linear operation. They are much more prone to 2nd breakdown for the above reason.MOSFETs are what you need for linear.Tim

author: Kleinstein
author_rank: Super Contributor
timestamp: May 07, 2016, 09:14:58 pm
content: MOSFETs also need to be designed for linear operation, and modern ones are usually made for switching. It's rather hard and expensive to get MOSFETs designed for linear operation. They don't have 2nd breakdown, but a similar effect just with an other name. The trouble with MOSFETs is that some datasheets have wrong SOA curves - so even if there is a DC SOA curve, it might not be reliable.BJTs for linear operation are readily available - not every type is well suitable, but many types are.I have tested a way to limit the output of the OP that is not actively controlling the output. At least in the simulation it works reasonable well. The circuit is using a separate shunt resistor an the other side of the "GND" point to make current regulation also work with the 2 quadrant output stage (one transistor less here - still fast enough). The new part is the transistor to enable extra direct feedback when the other OP is active. Shown here only for one OP, but works for the other half too if needed.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: May 10, 2016, 03:40:52 pm
content: Quote from: Kleinstein on May 07, 2016, 09:14:58 pmWhat is the purpose of R12 and C5? Also the set_current voltage is negative, is there a way to adjust it using positive voltage?Schematic now:

author: Kleinstein
author_rank: Super Contributor
timestamp: May 10, 2016, 08:05:07 pm
content: It is possible to change the circuit to use a positive voltage to set the current. Just add a divider at the non inverting input to mix the positive set signal with the negative measured shunt voltage. It might interact a little with the measured current, but not much. So not a problem unless one wants really high resolution ( <0.1 mA range) current readings.C12 and R5 are there to improve stability with large capacitive loads (e.g. 10 mF). They also help to reduce the bandwidth requirements for the OP. You find that part in many similar circuits, sometimes without R5. Without this cap the output impedance in the 10 Hz - 1 kHz range gets very close to a perfect inductance and thus could give very long ringing with large low ESR caps.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: May 13, 2016, 07:37:42 pm
content: Quote from: Kleinstein on May 10, 2016, 08:05:07 pmIs the connection supposed to be like this?Like you said, I'm mixing the positive set voltage and the negative measured shunt voltage.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 14, 2016, 08:27:08 am
content: There still needs to be the resistor from the inverting input to round. The set voltage is than positive.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: May 14, 2016, 01:31:16 pm
content: Quote from: Kleinstein on May 14, 2016, 08:27:08 amSuch an obvious thing and I forgot about it, after you get the answer suddenly everything makes sense.Anyways, I'm going to use LT1074 as a per-regulator because I'm unable to find large heat sinks cheaply.I got the pre regulator to work:https://i.gyazo.com/d103e9bd25098711af27f600869e5c85.pngbut I need to find the correct values for the output caps & inductor.