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

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 07, 2016, 08:18:17 pm
content: Hi,The pass transistor driver that works by the opamp need at least 45V supply to work propaly, as the voltage at the base=voltage at the emitter, the supply must be rated at least higher then the output voltage and The problem is that the max output voltage is 30V while the driver supply (V6) need to be at least 45V, so my only opinion is to use a voltage doubler (MAX1044 for example).My question is if there is any way to make it so you would not need that high voltage for the driver?Schematic:

author: Kleinstein
author_rank: Super Contributor
timestamp: March 07, 2016, 09:38:48 pm
content: In this type of circuit the driver needs something like 5 V more than maximum output voltage. As there are not many cheap OPs for much more than 40 V, the output is effectively limited to about 30 V.There are several options to have a output voltage higher than the supply of the OPs. One is having a transistor stage to amplify the OPs output. Another solution is using an output stage that is giving a output current for a given control voltage - e.g. using a PNP transistor in emitter circuit. There are also other circuits to get this kind of behavior.A common solution is using a floating regulator using a second independent supply for the regulator. This is how many of the HP supplies were build. Effectively it's also using the power stage in in an common emitter circuit. The nice thing about this circuit is, that one could essentially use the same principle circuit and even board to work for low voltages (e.g. 5 V max) and high voltages (e.g. 600 V) with just changing a few resistors, caps and transistors. Its also well suited for digital control, as the auxiliary supply can also power a µC.

author: uncle_bob
author_rank: Unknown
timestamp: March 08, 2016, 02:20:26 am
content: HiAnother very common solution is to simply accept that the bulk supply will always be at least 5V higher than the output voltage. Linear supplies like this one are well known for being inefficient.Bob

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 08, 2016, 02:41:23 pm
content: Quote from: Kleinstein on March 07, 2016, 09:38:48 pmhow exacly do I float the supply?I've seen how hp did it but it doesnt make sense to me, it seems they grounded the positive output so can it even regulate propaply?Also I would need to move the shunt to high side and not low side, as it would work better?

author: Kleinstein
author_rank: Super Contributor
timestamp: March 08, 2016, 05:15:53 pm
content: The HP supplies often use the floating regulator. So the reference point ("ground") for the regulator circuit itself is the positive output. So effectively they are controlling the negative output. This is not a problem for the regulator, it's just a little confusing to read such a plan, as ground is not the same as the negative side of the input or output side.Than having the shunt at the high side means having the shunt with one side at ground (for the regulator part), so this is the logical position for the shunt in this case. The nice thing on having the shunt at the emitter side of the transistor is that the regulator does not need much gain for current regulation and thus can be quite fast. The downside is that you need the extra floating supply (e.g. transformer) for the regulator.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 08, 2016, 05:47:13 pm
content: Quote from: Kleinstein on March 08, 2016, 05:15:53 pmAlright, thanks for the comment, anyways, I disconnected the current error amplifier and now I'm left with the voltage regulation that I'm trying to get to work, unfourmently, it doesnt seem to work at all.the voltage at the base of the pass transistor is exacly the output voltage but the load sees 30V (max voltage)all of this floating thing doesnt still get in my head..

author: Kleinstein
author_rank: Super Contributor
timestamp: March 08, 2016, 08:16:42 pm
content: The voltage regulator still needs the feedback. With the floating regulator the feedback is from the negative output terminal with a resistor to the non inverting input of the regulator OP. The reference voltage also needs a resistor to the non inverting input.This type of regulator usually needs a slightly larger capacitor at the output - something in the 100 µF range is more suitable.The current from Q2 looks like a little to much for the OP - so you either need less current there and than likely a darlington for Q1, of alternatively transistors in stead of the diodes.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 08, 2016, 08:54:16 pm
content: Quote from: Kleinstein on March 08, 2016, 08:16:42 pmso I took the inverting input to the negative output teminal with 22k, It didnt work.why do you need resistor to the non inverting input of the op?

author: Kleinstein
author_rank: Super Contributor
timestamp: March 08, 2016, 09:15:57 pm
content: The output stage is inverting the signal. So you have to use the feedback for the global voltage regulation to thenoninverting input. The typical way is having the inverting input to ground (= positive output terminal) and have two resistors from the non inverting input to the negative output and a positive reference / set-point voltage. This is similar to a inverting amplifier, just with the inversion after the OP and thus making the + input a virtual ground.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 08, 2016, 09:40:47 pm
content: Quote from: Kleinstein on March 08, 2016, 09:15:57 pmI'm still having problems with that, is there any way you can draw quickly a schematic with the connections? as I'm sure I will quickly understand it better.here is the connection I currently got, and it still doesnt work:

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 10, 2016, 07:19:40 pm
content: I havent got floating the power supply and make it working, how can I amplifiy the voltage after the opamp without amplifing the input voltage?or what other solution I can use that will not reduce performance?

author: Kleinstein
author_rank: Super Contributor
timestamp: March 10, 2016, 07:54:50 pm
content: The circuit above for the floating regulator is still missing a resistor (e.g. 10 K) in series with V4. Then is should work - at least in principle with a ohmic load.One alternative to the floating regulator is an output stage that controls the output current. It helps to have resistors to have a well defined change in current from a change in input voltage. This is about how low dropout regulators are build. This usually means the output transistor with the load at the collector side.

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 10, 2016, 08:50:08 pm
content: Quote from: Kleinstein on March 10, 2016, 07:54:50 pmI tried putting a resistor in series with V4, the output just shoots up to the maximum input voltage, unless I change R1 to 2k and then it kinda works on 1V output, but higher then that? nope.I rather go with a diffrent design then floating the supply, unless floating the supply is the most recommend and worth the hassle.could you explain more \ give an example for an "output stage that controls the output current"?

author: uncle_bob
author_rank: Unknown
timestamp: March 10, 2016, 11:26:04 pm
content: Quote from: ZeTeX on March 10, 2016, 08:50:08 pmHiOnly two real choices:1) Floating supply with another winding on the power transformer2) More headroom in front of the output transistor.Both work and work pretty well.Bob

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 11, 2016, 11:39:42 am
content: Quote from: uncle_bob on March 10, 2016, 11:26:04 pmI see, then I will have to go with option 2, As option 1 doesnt work for me, and get extrenal transformer for the high voltages (my transformer is rated exacly at 1A, and I dont want to draw more then 1A at full load because of the opamp \ bias supply.)

author: Kleinstein
author_rank: Super Contributor
timestamp: March 11, 2016, 12:18:58 pm
content: With the current rating of the transformers you have to be careful anyway. With the normal bridge rectifier and filtering cap, for an 1 A DC current the transformer sees an AC RMS current somewhere in the 1.5 - 2 A range, depending on the size of the filtering caps, resistance of the winding / fuses and stray inductance.

author: uncle_bob
author_rank: Unknown
timestamp: March 11, 2016, 12:34:08 pm
content: Quote from: ZeTeX on March 11, 2016, 11:39:42 amHiWhile you are buying a transformer, consider that for various odd reasons a 1A AC transformer does not produce 1A DC and stay at a reasonable temperature on a linear regulator. (The current is pulled in bursts and P= I^2 * R). Figure that you will need a 1.5A transformer for 1A out. The same issue gets you on rating diodes. There the factor may be a bit higher.One simple approach is to use a dual secondary transformer. They go in parallel for the low voltages and in series for the high voltages. That way you get twice the current on the "low range" of the supply. Yes, you need to rate all the "other stuff" for the higher current, but that usually is pretty easy.Bob

author: ZeTeX
author_rank: Frequent Contributor
timestamp: March 11, 2016, 01:03:52 pm
content: Quote from: uncle_bob on March 11, 2016, 12:34:08 pmHi bob,first I'm not buying any transformer for the opamp supply\bias supply, I already got 10VA 12V "P.M.A" brand transformer desolderd from something, I might be able to multiplie the voltage for such low currents. (about 80mA max)As far as I know linear regulators doesnt take current from the input in bursts, thats a switching regulator, isnt it?My main 1A transformer is like this:(As you said, duel secondery transformer)so I can wire it to have only 15VAC output but at 2A, but then there is still a problem - I have already created an AC-DC Rectifier + Filtering with a choke (LC FILTER) and I really dont want to use only half of the filter (the middle winding where supposd to be shorted to get 30VAC, a then I would get 15V 0V 15V, and each 15V was filtered sepretly), and if I would use only 1 winding then half of the filter would not work, and thats a waste of money and componets..Also if it matters I have other 25V 2x12V duel secondery transformer..

author: Kleinstein
author_rank: Super Contributor
timestamp: March 11, 2016, 02:01:21 pm
content: The linear regulator take a continues current in, so there ideally 1 A in for 1 A out. But the simple rectifier with filter cap takes a pulsed input current around the voltage peaks. So something like 1.6 A RMS in for 1 A out for a 20 VA transformer. With a filter that uses a choke, the current is a little more smooth, so you might get something like 1.4 A in for 1 A out. However using a filter with choke is rather uncommon for such a small power. Depending on the design the choke could be on the primary side - so it would work independent of the transformer tap used. One downside of the choke is, that the voltage under load will show quite a drop. So you might not get the full voltage at the full current.For a linear regulator one can switch the taps or windings. The conventional way is using a relay to switch on the AC side. Or alternatively the rectifier could deliver 0 - 15 V - 30 V and the power stages uses the 15 V or the 30 V rail as needed.For a transformer with dual secondary the rating might be a little more complicated, as the loss is from both the primary and secondary.. This could be something like 1 A when using both windings or 1.5 A when using only one winding.

author: uncle_bob
author_rank: Unknown
timestamp: March 12, 2016, 01:10:53 am
content: Quote from: ZeTeX on March 11, 2016, 01:03:52 pmHiLong before the current gets to your linear regulator, it hits a diode. That diode connects to a capacitor. When the voltage on the transformer secondary is higher than the voltage on the capacitor, the diode begins to pass current. As soon as the voltage on the secondary goes below the voltage on the capacitor, the diode stops conducting current. The result is a capacitor that charges quickly to the peak voltage of the transformer secondary and then decays until the next peak comes along. You get a sawtooth (fast charge / slow decay) waveform on your bulk storage capacitor. The current in the diode is only moving during a small percentage of the waveform.The net result is that you have 1A out of the bulk capacitor. The capacitor charges over (maybe) 10% of the waveform. You now have current (on average) moving at 10A rather than 1A. Since the voltage is rising quickly, the actual current is a bit "interesting". If P = I * R none of this would matter. Since P = I^2 * R it does matter. 10A * 10 A = 100 times 10% gives you 10 whatever. 1A full time gives you 1 whatever.You *do* need to watch the secondary ratings.Bob