Design Requirements:

The purpose is a linear DC power supply for 12-volt aircraft, marine, and amateur radio equipment

  • power source 20 volts @ +/-10%. (output 85W = 6A*14.2V)
  • Rser on V1 = 0.4 or just say Output Voltage = 14 volts as shown
  • adjustable 10 to 15V via R5/R6 with pot.
  • Tested and works Output Current = 6-amps.
    • Tested and works Current Limit not shown. Have a working model.
  • Line/Load Regulation: 14 VDC @ 0 to 5-Amps = <0.1% ( Model exceeds spec.)
  • Ripple and Noise <200 uVpp
  • Load Step Response 0 to 4 to 0-amps: better than +/- 200mv recovering in 10 us. This is the one I am having trouble with.

I need a mentor to help me put the finishing touches on a 14-volt 5-amp DC power supply. I am an EE, RF engineer, ham radio operator, and first-time LTSpice user. It has been 30 years since I last worked on discrete components.

I did some homework, looked at some designs I liked, and came up with the circuit below. It worked great on LTSpice until I added the second test circuit on the right. I am sure the circuit design needs some Compensation somewhere, or perhaps a different op-amp, but I am lost.

Brief description of the circuit. It is a Floating Ground voltage regulator. U1, C1, C2, R1, and R2 create a soft-start -5-volt voltage reference applied to U2 inverting input. R5 and R6 provide feedback to the op-amp non-inverting input.

D1 is an OR gate on the op-amp output to allow a current control circuit (not shown) to integrate into the control loop. As load current increases, the regulator drive voltage increases to compensate. At the same time, the current control voltage falls lower as the load current increases to a point where it takes control of the loop. R4 is a 10K pull-up resistor that guarantees M1 will start and force U2 to sink current discharging gate capacitance on M1. R3 is a 100-ohm gate resistor, which helps with compensation according to manufacturing data.

M1 is the pass element capable of supplying 10-amps of current. Or can be reconfigured to drive multiple parallel NPN transistors for higher output currents. C3 is a 4700ufd filter connected across the output. R7 is a bleeder resistor that doubles as a load resistor in LTSpice and output voltage monitor.

M2, V2, R8, and R9 are test circuits turning on/off a 4-amp load to test the output response and why I am here. Not happy with what I am seeing. The trouble starts when I connect the test circuit and configure V2 as a PULSE circuit—configured as a Sine Wave generator with 7-volt DC bias, 4 volt 1000 Hz signal performs as expected, and the output voltage is a flat line. The output signal from the op-amp is a pretty sinewave, as is the output current seen in R8 with a magnitude of 4 amps...

Configure V2 as a Pulse Generator as shown, and things get ugly depending on how fast/slow I set the rise and fall times. At 1us and faster rise/fall times, note the voltage spikes across the output represented by the current in the IR7 trace. It dips to 13.88 volts and peaks at 14.1 volts.

The second trace is the op-amp's output voltage measured at the gate of M1. I expect it to look like a nice squarewave but see that nasty falling edge spike. I suspect a result of gate capacitance but am clueless about how to fix it. Increase rise/fall times, and things improve. At 10us, results look acceptable. At 100us looks excellent.

Can someone help me out? Appreciate all comments, suggestions, and answers.

Voltage Regulator IR1 Output Voltage Gate Input

  • \$\begingroup\$ Why not use a proper switching regulator chip, instead of trying to make your own? \$\endgroup\$
    – Hearth
    Jun 2, 2022 at 1:46
  • \$\begingroup\$ @Hearth thx for the comment. I want to design an ole fashion boat anchor linear power supply with updated circuitry.. I do not ant a switcher and the noise associated with them. Technically this is an LDO linear design. In practice, V1 is replaced with 17-volt AC transformer, rectifiers, and a massive filter capacitor. Designed to replace voltage regulators in Astron and other like 12-volt DC high current power supplies ham radio operators use. \$\endgroup\$
    – Dereck
    Jun 2, 2022 at 2:07
  • \$\begingroup\$ Always start with expectations for all variables: Vin range, Voutput range error. { also min/maxLoad & load reg error Dc and step) your zener is in the wrong place and you are measuring zener current not as a stable voltage. Define Vo/Io=Zo ... / load = load regulation error (%) . Add 500 ohm + C in series across R5 for Kd gain with RC to match load rise time or slightly slower eg 10us , C= 10u/500=20nF. Add current sensing for better prediction of boost. Otherwise, the control gain is asymmetrical and many other issues. \$\endgroup\$ Jun 2, 2022 at 2:31
  • 2
    \$\begingroup\$ @Hearth, I read "ole fashion boat anchor" as a euphemism for "big, heavy", and not necessarily anything to do with a boat. \$\endgroup\$
    – brhans
    Jun 2, 2022 at 8:34
  • 1
    \$\begingroup\$ @Hearth The boat anchor comment was tongue-n-cheek humor in reference to older equipment like a linear DC power supply with a great big heavy transformer or radios and TV's with tubes. All can be used as boat anchors. \$\endgroup\$
    – Dereck
    Jun 2, 2022 at 18:25

2 Answers 2


OK, it would be better with a switcher, everyone said it.

With 20V input, 10V output, 6A, the MOSFET will dissipate 60W. If there is a 6A current limit and the output is shorted, it will dissipate 120W.

The first problem is IRF530. It has RthJC of 1.7°C/W, which means it's smoking, and we haven't added the thermal resistance of the interface compound or grease to the heat sink.

A chunkier MOSFET would be a better choice. There aren't many options, due to Spirito effect in modern MOSFETs, so it has to be the usual suspects: IRFP140N will do the job. Thanks to TO-247 package, it has RthJC of 1.1°C/W, and the much larger area at the back of the package reduces the thermal resistance to heat sink in proportion. With an uninsulated mount (grease) or Keratherm red, it should survive a shorted output for long enough until the temperature probe on the heat sink shuts it down.

Now, the problem is your specs...

  • Ripple and Noise <200 uVpp

That's 74dB PSRR, considering 1V ripple on the input caps. It's doable.

However, if your load draws a dynamic current, for example because it's a radio with an audio amp and a RF amp whose supply current depends on signal... then, supposing its current has say 1A ripple, then your desire for 200µV ripple on the output of the regulator implies an output impedance of 200µOhm, wires included. I've done much worse than that, but there were no wires. With wires, ain't gonna happen.

That's a bit like the audiophiles who get a 1µV ultra low noise power supply, and use that to power a raspberry pi, whose supply current is a mess of spikes and transient. So, in real use, the output voltage has a lot more noise, due to load current. It's a waste of money to optimize a paper characteristic that won't exist in real use.

  • Load Step Response 0 to 4 to 0-amps: better than +/- 200mv recovering in 10 us. This is the one I am having trouble with.

Now it's normal that this one is giving you problems. You have to spec it better, and add another parameter: current rise time.

If your current rises from 0 to 4A in 1µs, that's di/dt=4A/µs. Now, say you put the load drawing that current step at the end of 30cm of zipcord, that's about 150nH. e=Ldi/dt, so that's a whooping 0.6V, just due to the inductance of the wire.

So if you plan to connect your power supply to the load with wires, there's no point to design a power supply with an output impedance that is way lower than the impedance of the wires. Especially considering the load will have capacitors inside, so the high frequency performance part of the transient response of your regulator is only relevant through the way it influences loop gain and behavior at lower frequencies.

Basically the kind of transient response you're asking is absolutely doable at the load, but... that will require a power and ground plane with a large quantity of capacitors right on it, to get very low output inductance. And you don't need that, because your load is not a desktop CPU, it's a radio.

One good starting point would be to stick a 10 mOhm resistor in series with the power supply you're currently using, connect it to your radio, probe it with a scope, and check what kind of current the radio is actually drawing. If it doesn't look like a square wave (and it won't) then it is pointless to optimize transient response for a load that draws square wave current. If the load current is civilized, then it would be more productive to optimize output impedance, make sure it's low enough, without bumps or peaks, some of which may be due to the interaction of the power supply, its output cap, wire inductance, and capacitors/filters/inductance inside the load.

Note the current measuring resistor must be non-inductive, and especially not wirewound. A bunch of 0R1 SMD resistors soldered on a spare bit of pcb works wonders.

If I had to build such a PSU, I'd use a common drain configuration. I don't like the extra pole from common source. That requires a boosted voltage for the driver, which can easily be generated with a voltage doubler from the rectifier.

enter image description here

I've opted for discrete, not just for aesthetic reasons, but also because that allows to drive the FET gate with the output current from the long tailed pair. This means the FET capacitance determines the dominant pole, and unlike an opamp based design, we don't have an extra annoying pole from the opamp itself. So, compensation is pretty simple. The drawback is that variations in the FET capacitance will influence ULGF, so C5 may or may not be necessary, or it may need a bit of adjustment for optimum stability over the whole range of voltages and current.

Transient response with 4A/1µs:

enter image description here

I1 adjusts the overcurrent protection. It should depend on heat sink temperature. Current sense resistor R1 must be low inductance.

  • \$\begingroup\$ thank you, sir. I can work with you and your answers. About the smoking IFR530, I have a CC and Fold Back current limiter not show. CC= 6-amps,FB=2-amps. Note diode OR gate. I understand the di/dt delta step response and why it is overkill but humor me, please. Where do I add a Cf or Snubber, and how do I calculate? FWIW a switcher is not applicable. I am designing a replacement voltage regulator using linear power supplies. The board drives a parallel bank of NPN transistors. In my case it will drive a single Mitsibushi QM100HY-H transistor. VCEO 600, IC = 100-amps, PD = 620 \$\endgroup\$
    – Dereck
    Jun 3, 2022 at 17:32


simulate this circuit – Schematic created using CircuitLab

I think there are too many problems with your present config.


  1. Battery charger design 14.2V + motorcycle battery, stable, most power
  2. LM317 with bypass
  3. DCDC regulator to 1V above a linear power regulator, More efficient, less heat rise.
  4. DCDC regulator to meet or exceed all above requirements. Most efficient, quiet like PC power supplies.

modified Option 2. enter image description here

Won't work
Input is too low and needs 20V with 1 ohm drop @ 6A for 14.2 out

Option 2 needs a rugged FET to handle linear failure modes using LM317 (search site for details)

Best bang for the buck

Option 1 using Option 2 with thermal protection (both FET and LM317 mounted to same heatsink. Car batteries have very low ESR thus can be regulated well at high step currents.


Future work

enter image description here https://www.analog.com/en/design-center/design-tools-and-calculators/power-management-tools.html LTPowerCAD

  • \$\begingroup\$ There is no current ir derivative feedback and poor voltage reference and high source impedance . Do you understand the stability and regulation solutions yet? Questions? \$\endgroup\$ Jun 3, 2022 at 16:59
  • \$\begingroup\$ I give you solutions to your problem. What Do you not understand , so I may explain better. \$\endgroup\$ Jun 3, 2022 at 17:01
  • \$\begingroup\$ thank you very much for drawing the circuits out and looking at them. Just one comment. I am using a -5 volt reference to the Inverting input. \$\endgroup\$
    – Dereck
    Jun 3, 2022 at 17:45

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