I have a DC to DC converter that outputs 28v and charges a very large cap. (10mF).

Cap acts as an energy source for a load that consumes lots of energy for a very short time. (A few amps for micro seconds).

I currently have a small resistor between cap and DC DC converter and this solves the inrush current and ensures we don't burn DC DC converter. However I like to improve this circuit and do a more sophisticated current limiter which can pump up to 100mA to capacitor regardless of its voltage level. (Current solution pumps current based on the voltage level of the cap. If voltage is very low, resistor allows a larger current. But say if the delta is 1v, resistor allows a smaller current to pass through. )

Ideally I like a solution where I can pump up to 100mA even when the voltage on the cap is slightly below the 28v.

  • \$\begingroup\$ A current source, then? \$\endgroup\$ Commented May 7, 2016 at 12:19
  • 2
    \$\begingroup\$ You're looking for a constant current source. (1) Do you want it on the high side or the low side of the capacitor? (2) How much voltage drop on the full charge can you tolerate? \$\endgroup\$
    – Transistor
    Commented May 7, 2016 at 12:20
  • \$\begingroup\$ 10 mF isn't very large, it's just large. \$\endgroup\$
    – pipe
    Commented May 7, 2016 at 12:40
  • \$\begingroup\$ Yes you can make such system. How complex design do you want: is OP AMP ok for you? \$\endgroup\$
    – Master
    Commented May 7, 2016 at 19:26
  • \$\begingroup\$ @transistor I am not so sure I look for a constant current source. I am looking for a variable current source capped at a value. Since I want current flow to stop when the capacitor voltage is 28v. After a next consumption event, I want current to start flowing again. On other questions, high side and as little as possible but even 0.5v drop should be ok. \$\endgroup\$
    – TGG
    Commented May 8, 2016 at 0:24

1 Answer 1


The simplest way to create a constant current is to use the inherent current regulating characteristic of a bipolar transistor. If you feed a fixed current into the Base, current at the Collector will be multiplied by the transistor's current gain (β or HFE), which is relatively constant over a wide range of Collector-Emitter voltages. The circuit is very simple:-


simulate this circuit – Schematic created using CircuitLab

R1 sets Q1's Base current to 1mA. Q1 has HFE of about 100, so ~100mA flows from Emitter to Collector to charge the capacitor.

Unfortunately there is a catch:- current gain is a difficult parameter to control in manufacture. So while 100 is the 'typical' figure, an individual transistor might be <50 or >150 - and affected quite strongly by temperature. You can adjust R1 to suit an individual transistor, and use a good heatsink to keep the temperature stable, but you can't expect the current to stay at exactly 100mA.

If you don't need an exact charging current and are willing to 'tune' it to suit the individual transistor then this simple circuit may be enough for you.

But if you want better accuracy and stability then you need a circuit which measures the current and compares it to a stable reference, turning the transistor on more or less to keep the current constant. This technique does not rely on HFE so it can use a bipolar transistor or a FET. Here's an example circuit:-


simulate this circuit

R1 and R2 provide a reference voltage of 100mV (relative to 28V) when the potentiometer is centered (variable from 0 to 200mV to adjust the output current). R5 senses the output current by dropping a voltage of 1V per Amp. Op amp OA1 compares these two voltages, and varies Q1's Base current (via R4) until they match.

Because the sense resistor must drop some voltage to measure the output current, this circuit has a slightly lower maximum output voltage than the simple transistor regulator. However it is much more accurate, and not sensitive to the characteristics of the transistor.

  • \$\begingroup\$ The opamp solution is nice. There is one key question. Once the cap is charged to its maximum voltage, the current flow must stop. In this case system will still try to push the constant current, at least that's my interpretation of the circuit. \$\endgroup\$
    – TGG
    Commented May 8, 2016 at 0:22
  • \$\begingroup\$ Charging current drops to zero when the capacitor charges up to the power supply voltage, because it cannot charge any further. However if the capacitor is discharged then charging current will start to flow again. If you don't want this then you will have to turn the current regulator off some time after the capacitor is charged, and turn it on again when you want to recharge. You could put a switch or transistor between R1 and GND. Turn the switch/transistor on for charging and off for discharging. \$\endgroup\$ Commented May 8, 2016 at 0:47
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    \$\begingroup\$ BTW this circuit can also tell you when the capacitor is fully charged. Normally the op amp's output voltage will be quite high (>20V) because the transistor doesn't need much Base current. However at full charge the transistor becomes saturated, and the op amp pulls its output down to GND in an attempt to keep the current flowing. This signal could be used to trigger a flop/flop that turns the charging current off automatically when the capacitor is fully charged. \$\endgroup\$ Commented May 8, 2016 at 1:01
  • \$\begingroup\$ I suggest using p-channel MOSFET for the similar circuit. The advantages are: the voltage drop is almost 0; this solution does not require R4. \$\endgroup\$
    – Master
    Commented May 10, 2016 at 10:13

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