0
\$\begingroup\$

I've been studying DC electronic loads and almost all of them used mosfet as main transistor, while researching I find out that mosfets that are designed to work in their linear region are expensive and also it's not easy paralleling them because their threshold voltage can vary and we need one op amp for each mosfet.

Long story short; I decided to use BJTs instead of mosfet, below circuit is the raw idea that I end up with:

schematic

simulate this circuit – Schematic created using CircuitLab

I haven't built this yet but on the simulator even when the load is not connected, the circuit is drawing some current from the source:

  • How can I compensate for that extra current or can I somehow null it out?
\$\endgroup\$
5
  • \$\begingroup\$ You must be turning on the leftmost op-amp slightly. What's its output voltage? What model is the simulator using for the op-amps? Are they ideal, or are they specific part numbers -- and if they're specific part numbers, what are their power supply rails connected to? \$\endgroup\$
    – TimWescott
    Nov 9 '20 at 16:44
  • \$\begingroup\$ According to the current state the output of the rightmost non-inverting amplifier is almost zero (it is actually amplifying the input offset). The leftmost comparator will logically ramp up its output. Isn't that the expected behavior? \$\endgroup\$
    – vtolentino
    Nov 9 '20 at 16:50
  • \$\begingroup\$ @TimWescott When the load is not connected, op amp output is 12V, it's just a simple java based simulator and the op amp is an "ideal" op amp, op amp supply is ±12V. \$\endgroup\$ Nov 9 '20 at 17:11
  • \$\begingroup\$ @vtolentino Yes the non-inverting input is higher and the output will go to positive rail, but in order to null the current, the op amp output should be 0V. and that's where I stuck. \$\endgroup\$ Nov 9 '20 at 17:15
  • 1
    \$\begingroup\$ @vir, you should post your answer as an answer. But another solution would be to move the on/off switch into the setpoint circuit (where there's currently just a +1 V supply). \$\endgroup\$
    – The Photon
    Nov 9 '20 at 18:13
1
\$\begingroup\$

That's expected base current. When the switch is open, the feedback loop is disconnected so the 10x amplifier is putting 0V or close to 0V to the comparator, which is driving the 2N3904 as hard as it can to try to reach the setpoint. The 2N3904 is letting base current from the MJ2955s through which comes from your source.

With this configuration, the base current through the MJ2955s cannot be nulled out since base current is, so to speak, what causes current to flow through them and is therefore unavoidable. You could switch to an NPN transistor on the low side of your load and shunt resistor so that the base current can be provided by a different source. You'd probably have an unacceptably high common mode voltage across whatever amplifier is connected to the shunt resistor so a Hall-effect or flux gate current sensor might make sense.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.