While browsing the lab power supplies market and watching the price ranges, I've decided that I'd be better off with building my own, since I've got most of the needed parts anyways.

I've searched for schematics and stumbled across the one included below. It is self-explanatory in many ways, but there are two questions that keep bothering me.

The lower part of the schematic with the LM377L regulator handles the negative power rail for the TL082. I completely understand why is it needed when using the TL082 opamp, as it is not a rail-to-rail opamp.

  • The first question is: Why not just use a LM324 rail-to-rail opamp, for example? Will there be any drawbacks?

  • The second question is: Why is the BD139 transistor on the right part of the schematic needed at all? If it is there to increase the current gain - aren't there any high-power transistors with higher hFE that could be used instead of suggested 2N3055?

The schematic

  • 1
    \$\begingroup\$ Yeah, it's there for current gain since it is wired up as a part of the Darlington including that BJT on the heatsink. There are some high power BJTs with high beta (the NPN D44H11 and the PNP D45H11 come to mind.) But even then probably not enough. I haven't looked over the rest, but I'm guessing that this lab supply wants to be able to actually get down to a TRUE 0 V at the output. Not even a rail-to-rail will do that. You simply need the extra headroom. \$\endgroup\$
    – jonk
    Commented Oct 8, 2017 at 1:16

1 Answer 1


One specification common to all power supplies is Load Regulation ( usually 1~2%) This means at (max V/max I)=R load the V drops 1~2%. From Ohms Law this means the output impedance of the regulator is equal to the Load Regulation (%)*R(rated)

How do you compute your load Regulation?

  • Assuming sufficient input power and voltage, examine TL082 schematic and see R out=100+200=300 Ohms.
  • We know emitter followers reduce the source impedance by hFE.
  • We know that closed Loop negative feedback also reduces Zout by the feedback gain H(s) but if saturated, there is no gain and just Rout of 300 Ohms.
  • We also expect hFE to drop rapidly near rated currents but we can examine specs to get an estimate.

    • Let's look at worst case saturated Op Amp drive output high, then for a 30V/3A supply the rated load is 10 Ohms.
    • We expect Zout to be <=1% or 0.1 Ohms thus you need a total current gain of 300/0.1=3000=hFE for 1% load regulation at max load.
    • The 2N3055 has a hFE of 37 at 3A @2V ( from interpolating ON Semi's figure 5) , so you need only 3000/37=81=hFE for BD139.

hFE=3000. That's why you need a Darlington!


  • Then you guys probably want to know how I can interpolate hFE=38 @ 3A, 2Vce so easily from figure 5. but that's a different question.
    • Of course there will be a tolerance on this. enter image description here

enter image description here


This design has a "nominal" load regulation <1% at full load with saturated Op Amp Output. (Worst case)

bonus info

The initial assumption was that you can make DIY PSU's cheaper than buying them since you have all the parts. Consider that Laptop chargers are $0.25~$0.50/W and your supply has a rating of 90W and that you have < 50% efficiency at 15V 3A or dumping >50W into the 2N3055. Also consider that it is possible to easily modify Universal 19V 85W universal laptop chargers to go up to 30V by varying the feedback voltage and checking the Cap rated voltages. I have done this before, to make an 80W LED dimmer.

You can make it low noise by using a variable SMPS with a good Pi (CLC) filter after it. enter image description here

  • \$\begingroup\$ Thank you very much for your reply! Just a couple of another questions. 1) How did you get Rout = 100+200=300Ohms? 2) Am I understanding it right that increasing the maximum power output will be problematic since I'll have to increase overall hfe even further? \$\endgroup\$
    – sx107
    Commented Oct 8, 2017 at 13:46
  • \$\begingroup\$ By the way, using a laptop charger is not an option for me. I need a well-stabilized low-noise power output. The high-power situation is just "in case of". Since this is a lab supply and not a power supply in a consumer product, the power waste doesn't matter that much for me. \$\endgroup\$
    – sx107
    Commented Oct 8, 2017 at 13:50
  • \$\begingroup\$ 300 Ohms comes from knowing where to look in datasheets and decades of experience. ( I'll post it) Yes increasing current out may require more current gain only IF you want low ripple from load regulation.at max V and I , for lower V load regulation improves from negative feedback , \$\endgroup\$ Commented Oct 8, 2017 at 14:26
  • \$\begingroup\$ What if I just use an opamp with a lower output resistance? Like the NE5532, for example. ti.com/lit/ds/symlink/sa5532.pdf It seems to have about 15 ohms of output resistance. \$\endgroup\$
    – sx107
    Commented Oct 8, 2017 at 14:47
  • \$\begingroup\$ That'll work better at <30V since Vdropout is 2~3V from Vcc and then cut off at >45mA. est. but maybe not at 30V, ..depends on V drop. You have 33V input. so 3V drop at 30V out. Then 1.5V dropout from Darlington. so NG. A better choice would be an Half bridge FET rated for 50W, then you get 0~33V \$\endgroup\$ Commented Oct 8, 2017 at 15:01

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