PNP-based linear regulator  - Page 1
Forum section: electronics / beginners

author: Recyclojunk64
author_rank: Contributor
timestamp: July 07, 2014, 05:28:16 pm
content: Hi allI'm trying to build a simple linear regulator that will output about 9V from an input of just above 9V up to maybe 70V. So that rules out using an LM317 or a 7809 as they only take an input of up to 40 volts. The regulator only needs to power a switchmode PWM controller (most likely a TL494) driving several nF of gate-capacitance (some paralleled IRFP250s) at maybe 30kHz. So while not much power is needed, it's still too much for a zener shunt regulator I think.I have do have a large assortment of scavanged parts, but while I've searched dozens of industrial UPS and server PSU boards, I can't find a single P-channel mosfet. And I'm in too much of a hurry to get one off the net (and too much of a cheapskate to get one locally) so I opted to use a PNP darlington transistor, aTIP126(first one I could find), driven with aIRFD210mosfet, and of course a TL431 for feedback.While it does output a voltage around the correct value, the voltage increases significantly as the input increases (output will be about one volt higher with 15 volts in than with 12 volts in). I'll have to do some testing with the scope tomorrow but for now if anyone can see any obvious "traps for young players" I may have fallen into with the design I might just be able to get my bike dynamo battery charger (this is part of it) done before the bike camp I'm going on.

author: homebrew
author_rank: Frequent Contributor
timestamp: July 07, 2014, 06:16:20 pm
content: Hi,I'm just guessing ...According to the data sheet of the TL431 (http://www.fairchildsemi.com/ds/TL/TL431.pdf) the component's performance is specified around a current (IKA) of 10mA. You won't reach that with your 39k-R4 resistor. "Minimum Cathode Current for Regulation" is specified as 0.45mA(Typ) and 1.0mA(Max) (Page 3). You will miss that one, too.According to Fig. 2 (Page 5) your reference voltage is dependent to the cathode current in that range. That would support your measured behaviour ...But hey, just a quick guess ...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 08, 2014, 12:00:04 am
content: Mind a couple things when using a circuit like this:1. Don't hard saturate any of the transistors. (Exercise for the student: can you tell why?) Use current limited or current mirror type structures.2. Because you're using current mirror type structures, your circuit is a current source. Feeding a filter cap, which acts as an integrator. So in addition to the high transconductance of the loop (and there's going to be phase shifts in there already), you've got a free 90 degree phase shift and more potential gain. In short, as shown, you have an oscillator, not a regulator! You must use compensation networks to stabilize it.3. For such a wide operating range, you'll need something fancier than a resistor to provide bias. A simple yet effective method: use a larger resistor to bias a pair of diodes in series. This makes a sort-of-regulated 1.2V. Use this to power other transistors, using an emitter resistor to set collector current.Example use case.Tim

author: TerminalJack505
author_rank: Super Contributor
timestamp: July 08, 2014, 02:23:27 am
content: There is one obvious error I see. You are expecting 9V output but have programmed the TL431 for about 12.25V. Change R2 to 27K (nominal value 26K) and you'll get roughly 9.3V output.I'd also change R4 to 10K to ensure the TL431 gets enough current like homebrew mentioned. I'd also change R3 to 10K since you'll have plenty of gain as-is with the Darlington.If you see oscillation then it might be due to the gate capacitance putting the TL431 into its "unstable region." (TL431s can't drive a particular range of capacitance values. Consult the datasheet regarding this.) If this is the case then you will have to add extra capacitance across the TL431 to move it out of that region.Another potential problem you might have is that the TL431's cathode can only go down to about 1V so it may not be able to completely shutoff the MOSFET. If that's the case then you may have to put a diode (or two) between the MOSFET's source and ground.I did a simulation of your circuit with the changes I suggested and it looks pretty good. I didn't have the exact components you are using but they're likely to be probably pretty close. I couldn't simulate past 60V since the MOSFET in my simulation is a 60V part.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: July 08, 2014, 02:31:22 am
content: Hi,I would not use the TL431 in this application. I would use a 6.2V Zener in this circuit:Here is the dropout behaviour:To check for stability I did an AC analysis:The AC analysis result shows the power supply is stable.Regards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: July 08, 2014, 03:13:44 am
content: Hi,If I replace the pass transistor with a TIP127 Darlington transistor, this increases the loop gain. To preserve the loop stability, I have added C2.The results:Regards,Jay_Diddy_B

author: LvW
author_rank: Frequent Contributor
timestamp: July 08, 2014, 06:49:34 am
content: Hi Jay_Diddy_B !Only in the interest of accuracy I like to comment your ac analyses.I think, it is not correct to use for the loop gain the ration Va/Vb. This is allowed only if at the opening a large load is connected to a small source resistance.Otherwise loading errors occur (as in your case).That means: Either you must use* the MIddlebrook two-injection method, or* a large ac decoupling inductor together with a large ac coupling capacitor, or* another node for opening the loop without heavy loading errors.

author: mij59
author_rank: Frequent Contributor
timestamp: July 08, 2014, 07:13:12 am
content: It is easier to use a special regulator like the one below.http://www.ti.com/product/tl783

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 08, 2014, 09:39:43 am
content: Jay: what does it do with a low-ESR load capacitance?Tim

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: July 08, 2014, 11:19:16 am
content: Quote from: T3sl4co1l on July 08, 2014, 09:39:43 amIf I add a reasonable amount of capacitance, 100uF with 50m Ohm ESR and a 10uF Ceramic Capacitor in parallel on the output the loop gain is indicating stability:Quote from: LvW on July 08, 2014, 06:49:34 amI am not sure about this point. The technique I am using is Middlebrook's method from 1975.If I add buffers E1 and E2, I have a low impedance on one side (0 Ohms) and a high impedance (Infinite) on the other, but the result is identical.Can you show me the improvements?Regards,Jay_Diddy_B

author: LvW
author_rank: Frequent Contributor
timestamp: July 08, 2014, 12:23:54 pm
content: @Jay-Diddy_B:It´s not easy to say how large the error is but both methods suffer from loading errors:*Error without buffers: The ac source is causing a voltage drop (error) across the resistances connected to the upper (positive) node - in addition to the voltage caused by the current coming out of the path transistor.*Error with buffers: This configuration does not reflect the real operating conditions because R11 and the following feedback path does not load the collector side of the path transistor. A similar error would occur using a large inductor for ac opening the loop.______________Middlebrooks method from 1975 is valid for very low output impedances only (opamp output, for example). If this cannot be met a second injection (current) is to be made for correcting the error. This is described in Middlebrooks "General Feedback Theorem (GFT)".

author: Recyclojunk64
author_rank: Contributor
timestamp: July 08, 2014, 02:07:25 pm
content: Thanks for all the replies guys I didn't expect so many suggestions. Been a little busy with another o-scope I bought off gumtree today (BWD 521 with differential inputs, I'll upload some pics inside of it as I can't find anything about it on the net) so only just had a chance to implement your suggestions.QuoteThanks I must have overlooked that, I chose the value based on not passing more than 100ma at full voltage.QuoteThat would explain why it emitted an audible tone when under load, and the voltage on the mosfet's gate was oscillating quite a bit.QuoteQuite a good idea I'll have to remember that if I redesign it or something simmilar.QuoteHmm, that would probably be because I forgot to update the values on the schematic after I decided to use 9v instead of 12.QuoteLooks like a pretty useful regulator I might get some of them for future use.QuoteThanks for that diagram, I built it up and tested it. With a 1uF ouftput cap it outputs a very clean 11.5V (20mv ripple) with an input from just over that up to 80 volts (didn't test further least I have to find another TIP126 if it blows). I did use a 7V zener that's why it's not 9V exactly. I think I will probably stick with this arrangement.

author: Recyclojunk64
author_rank: Contributor
timestamp: July 08, 2014, 02:33:23 pm
content: Looks like I spoke too soon. Just as I hit post I heard a loud squeal comming from the ZVS driven transformer (that I was using to generate the 75VDC needed to test it) and one of the transistors (Q2) exploded. Q3 has a short between EB and the zener diode is shorted aswell, and R5 has increased to 2k. However the main transistor is unharmed. I guess since I have quite a few more 2N5551 transistors and zener diodes I'll reproduce this mishap and try and work out what went wrong.Edit: Just before anyone points it out I know that R5 was dissipating over half a watt but I wouldn't think the lack of a load if it burned out first would cause everything else to go too?

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: July 08, 2014, 05:49:30 pm
content: Hi,R5 is a 50mA test load. It is not be needed in the circuit. I should have indicated thatRegards,Jay_Diddy_B

author: LvW
author_rank: Frequent Contributor
timestamp: July 08, 2014, 07:13:14 pm
content: Jay_Diddy_B : The situation (measuring loop gain) is even more complicated than indicated in my last post.Your simulation results cannot be correct because forlow frequencies(including dc) the phase shift of the loop gain function must be-180deg.Otherwise, the whole circuit has no stable operating point.The fundamental error is as follows: For each system with feedback we always havetwo loop gains: Current loop gain and voltage loop gain.And Middlebrook´s two-injection method takes this into consideration.However, in most cases, one of both is dominating.Example: In opamp circuits we have a voltage loop gain.However, in the example under consideration this is NOT the case.Hence, we have to use Middlebrook`s extended method.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: July 08, 2014, 08:19:35 pm
content: Hi,I have had another look at the ac analysis. I did find an issue, but it is not the technique.The issue is that there is another feedback path in the circuit from the output. The path is a divider R6 and the slope resistance of D1. If I replace the Zener diode with a voltage source, this path is broken. I can also break this path by moving the connection for R6 to the other side of the disturbance source.The Zener is biased from the output to reduce the effects of input voltage changes. Thismaybe a bad idea, because it introduces positive feedback.Revised ModelRevised ResultsThis has the desired characteristics. The low frequency phase response is 180 degrees, it is 90 degrees with the dominant pole.Op-Amp ModelIf I build a simple model using an op-amp and an emitter follower, I have circuit that is easy to check.The results for the op-amp model are:The Middlebrook (1975) technique is valid here, especially if the LDO has an output capacitor, the impedance of the output is very low compared to the impedance of the divider.In this thread I compared the model versus measurement of an electronic load.https://www.eevblog.com/forum/projects/dynamic-load-bode-plot-using-hp-35665a-dsa/msg309192/#msg309192There is a good match between the model and the measurements.Regards,Jay_Diddy_B

author: LvW
author_rank: Frequent Contributor
timestamp: July 09, 2014, 08:28:12 am
content: Hi Jay_Diddy_B,interesting discussion, I think.And I agree - your last BODE plot looks much better (means: much more realistic).It would be interesting to know if you are able to get approximately also the same phase margin withcurrentinjection.I also did perform some simulations - however (due to non-avalability) with some other but similar parts.I have found out that the phase margin is about the same for (a)voltageinjection (as in your last simulation) and for (b)currentinjection as described next.For this purpose, I have placed an ac current source between the common node of the collector of Q2 and the base of Q4 (of course, without any opening).Then the loop gain isI(coll. Q2)/I(base Q4).It would be interesting if you could confirm these findings.EDIT: More than that, after modifying the load conditions the observed phase margin was altered, of course. And this margin change was - in principle - confirmed byclosed-loopsimulations. For example, for a reduction of the margin the closed-loop magnitude shows a corresponding gain peaking.

author: LvW
author_rank: Frequent Contributor
timestamp: July 16, 2014, 08:06:30 am
content: I believe that it would be most unfortunate if this discussion would stop at this point.According to my experience, the problem of loop gain/stability margin detrmination really deserves some additional considerations.And - in this respect - the shown regulator circuit based on transistors only is a very interesting circuit because it contains more than one loop and it does not contain elements which can be treated as ideal (like opamps, for example).Nobody interested?

author: Recyclojunk64
author_rank: Contributor
timestamp: July 16, 2014, 05:06:03 pm
content: Well I can do some practical testing of the circuit. I have already built one of them on protoboard that's happily powering aSG3526Npwm controller but I do have another TIP126 transistor so I can put another one together on a breadboard for testing. I might try it with some other ordinary PNP transistors aswell. And perhaps even reverse the entire circuit with an NPN to make a negative supply regulator.I do have a sweep function generator (2MHz), though I have no idea of how to measure the phase offset between Va and Vb. Maybe my old tektronix DPO can do it (I'll have to research that) or perhaps it's possible somehow with a (analog) differential-input scope? Would there be any useful measurements to gain from testing it in real life or is the simulation result good enough?I could of course conduct some more destructive testing if it would be useful.Another thing, there is a 20mV output ripple, I'm not sure what's causing it, perhaps noise from the zener diode?I do think this is quite a useful circuit as a (relatively) high voltage linear regulator chip isn't quite as cheap as these discreet components. Well at least not to someone with large quantities of such transistors.

author: LvW
author_rank: Frequent Contributor
timestamp: July 17, 2014, 11:01:53 am
content: Recyclojunk64,thank you for replying.However, my primary concern is loop stability and the different methods available for phase margin determination - and this is certainly primarily a simulation task rather than a hardware test.As mentioned, I did some simulation runs with interesting results - but with other parts as shown in the circuit diagrams.However, there is one test which could be performed with hardware: Measuring the output impedance as a function of frequency in the frequency region where the loop gain approaches unity. There is a published article which describes how such a measurement can give reliable information on the phase margin (non-invasive stability testing):http://powerelectronics.com/power-electronics-systems/new-technique-non-invasive-testing-regulator-stabilityIt would be interesting to know if this method works in the case under discussion.