I wish to use this circuit

0–300V variable supply without precise current limiting

to achieve a variable 0–300V DC supply. However, I'm interested in keeping the current below 50 mA, with as little variance from that value as I can achieve without doing anything incredibly expensive.

According to the circuit's designer, maximum current is determined by the resistor R2 and the transistor Q2:

The transistor Q2-BC337 and the shunt resistor R2 – 3.3ohm are added for as a current limiter. When the current output is too increased, the Q2 will stop the gate pin of the Q1 immediately, which will be guard the higher current output

But it was made clear to me in a previous question that this does not produce an exact cap on the maximum current, instead constraining it to more of a "ball park".

My question, then, is (adding a hard value): how to precisely limit the current output of this supply, within a 5mA maximum deviation from the desired maximum value, for as cheaply as possible.

I would want the circuit to cease supplying power if the current fluctuates above the maximum tolerance.

  • \$\begingroup\$ The first step is to define the behaviour you want upon reaching current limit. Remember that if you want a constant output voltage, the output current is determined by the load. Current limiting circuits can transition into constant current mode when the maximum permissible current is reached, by varying the output voltage. On the other hand, given the dangerous voltages present here, you may decide that if the current limit is exceeded, the supply should be disconnected. Which behaviour do you want? \$\endgroup\$
    – user133493
    Commented Apr 23, 2017 at 3:50
  • \$\begingroup\$ You specified 3 parameters with no tolerances, making this unsolvable. Pls define now, 50mA tolerance, cost tolerance and power dissipation design assumption. \$\endgroup\$ Commented Apr 23, 2017 at 4:16
  • \$\begingroup\$ @replete I'd definitely want to cut off above the maximum \$\endgroup\$
    – Jack Lynch
    Commented Apr 23, 2017 at 4:26
  • \$\begingroup\$ @TonyStewart.EEsince'75 Let's go with 5mA tolerance, cost below $50 (unless that makes no sense, in which case, what does seem feasible?) and power dissipation at whatever seems best to you. Can I handle larger power dissipation by adding something like a heat sink? If so, anything's fine—I'm not space-constrained. \$\endgroup\$
    – Jack Lynch
    Commented Apr 23, 2017 at 4:36
  • \$\begingroup\$ How tightly regulated must the output voltage be? What are the expected loads? The wider the variety of loads connected to a power supply, the more difficult is the design of its control loop. If you will only use specific loads, such as tubes, this simplifies the design. I realize your question is about current limiting but these subcircuits will interact. \$\endgroup\$
    – user133493
    Commented Apr 23, 2017 at 7:43

3 Answers 3


Q2 limits current by pulling the Gate voltage down when Vbe gets to ~0.6V. This corresponds to 50mA when R2 = 12Ω. However due to the logarithmic response of the Base-Emitter junction the cutoff is quite 'soft', starting to limit at about 40mA. Another potential problem that the junction voltage has a temperature coefficient of about -2mV/ºC, so the current limit will reduce as the transistor's temperature rises.

You could replace Q2 with a TL431. This device has a precision 2.5V reference and comparator connected to a bipolar transistor. It should precisely and sharply limit current to 49mA with With R2 = 51Ω (51Ω*49mA = 2.5V).

Here's the circuit:-


simulate this circuit – Schematic created using CircuitLab

The TL431 is very cheap and readily available. If you have old switch-mode mains power supply that you can rob for parts then you might even get it for free!

  • \$\begingroup\$ In similar designs one can find an RC filter after the bridge rectifier for further smoothing and something like 1uF from the wiper to ground. How would you recommend for the OP to assess whether these would help in his application? \$\endgroup\$
    – user133493
    Commented Apr 25, 2017 at 0:20
  • 1
    \$\begingroup\$ At 50mA the supply ripple would be ~2.7V or 0.9%. A capacitor on the pot wiper would reduce this ripple, but the question was about how to 'precisely limit the current... within a 5mA maximum deviation from the desired maximum value', not how to improve voltage stability. However the OP now says that voltage regulation is most important (when the circuit he supplied has none) and is asking for a way to cut the power (like a fuse) if current goes over the limit - a quite different function. So it seems all the time and effort I put into answering his question was wasted... \$\endgroup\$ Commented Apr 25, 2017 at 1:27
  • \$\begingroup\$ I believe we discussed the cutoff issue in the comments to the question before you posted. You may have missed it. At any rate, your excellent and instructive answer will surely be useful to others - it was to me - so I don't think it was wasted effort. \$\endgroup\$
    – user133493
    Commented Apr 25, 2017 at 2:19

Your bridge Cap will reach 365Vdc and with 0.05 mA any linear regulator will dump the V difference to output * I =Pd = 18W say for an output of 3V @50mA of 150mA meaning you need a small CPU heatsink and fan to keep it cool.

A smarter way is a Buck SMPS that will be much more efficient. I suggest you buy one.

Do you need galvanic isolation?

What ambient temp range?


This circuit will operate in the following manner. As the current increases at some point the Q2 transistor will start to turn on. However the exact point at which it turns on will vary with temperature and will be nonlinear because it will turn of Q1 so quickly it that Q2 won't have a chance to get to the linear region. In the case of a large overload this will not be a problem because Q2 will be fully bias very quickly.

By add some resistance (try 10K) between the collector of Q2 and the gate of Q1 it will require a higher current to flow through Q2 to turn Q1 off. This will produce a cleaner current reduction point. However this current limiting method will start limiting at some point but won't shut off completely until a higher current is present. Off course when Q1 shuts off the current will fall and Q1 will turn back on thus creating an oscillation. This oscillation can be dampened by putting a capacitor between the source and gate of Q1. This may very well suit your needs as it has a more predictable current limit. If you are really implementing a 0 to 300 V and not just a 300 V power supply you will need a voltage regulator to go with this as this circuit is really just variable resistance power source.

Also keep in mind the voltages that components will have to withstand. VR1 has to withstand 300+ volts. Most reasonable sized pots will not handle this. This can be remedied but putting a 1/2W 50-100K resistor between the pot and ground. It is likely that the lower part of the range for the pot is useless anyway. Additionally R1 will be subject to nearly the full voltage if the load is great enough to trigger the current limiting. Using a 1 watt or 2 1/2W resistors will help to ensure this resistor can handle the voltage.

A fuse on the hot side of the primary is extremely important. I would also recommend a 150K 1 W resistor across C1 will with discharge the capacitor when it is unpowered which will make the circuit safer. Using a 10 OHM 3W resistor between one output winding of the transform and the bridge diodes will limit the surge current into C1 to safe level for the diodes and allow the fuse to be rated closer to your maximum current rating without blowing the fuses when applying power with C1 fully discharged. A MOV across C1 can also save Q1 from damaged due power spikes from the power line. The 10 OHM resistor will help keep the spikes contained and improve the life of the MOV. The reason I am making these suggestions is that high voltages need to be handled with extreme care. Resistors are cheap but they must be operated without their capabilities. Read datasheets for everything you use to be sure you understand all the limits of the parts.

If you are trying to maintain full output voltage and turn off quickly when the current limit is exceeded then you need a circuit with a non-linear but well defined switching point. The easiest way to do this is to use a comparator circuit. The tricky part is how do you want it to behave once the current limit has tripped. As with the circuit above most current limiters will turn back on once the current limiting kicks in. So if you want the circuit to stay off until power is removed you need a circuit that latches it's state. Another possibility is to have the power cut off and then turn back on after a delay.

High voltages complicate things and thus provide a challenge to keeping things simple and cheap, but this chip (STFC01) is design for use to around 50V but can be easily used for much high voltages using the techniques shown in the data sheet. This solution is very flexible but as I could not find a source for the chip so great idea but not happening. I did a little looking around for another device but did not find one that worked with high voltages.

If I knew what you really need this circuit to do for you I could give you some better information. If you needs are simply enough the circuit above with the tweaks I suggested may do what you need. If you needs are more complicated then you may want to consider entirely different solutions such as a DC to DC convert type circuit similar to what is used in computer but with a high voltage output instead of a low voltage output. The advantage of this type of power supply is it is efficient and there are several chips that do most of the work on the line voltage side and provide both regulation and current limiting secondary side. It does require finding an appropriate transform for the job but since the transforming with be small the cost should be reasonable.

  • \$\begingroup\$ The pot will act as a bleeder resistor to C1. \$\endgroup\$
    – user133493
    Commented Apr 25, 2017 at 1:01

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