0
\$\begingroup\$

The circuit is a constant current source:

schematic

simulate this circuit – Schematic created using CircuitLab

When SW1 is open and no load is connected output of integrator is clipped to positive rail, this forces the op amp to swing its output all the way from positive rail to negative rail when a load is connected. obviously it makes sparks and overshoot on the RL which I'm trying to avoid (no matter how fast is the op amp slew rate).

To solve the issue I added a comparator to raise the inverting input of OA1 above the set voltage V1:

schematic

simulate this circuit

Even though it does work in a crude simulator, but because I'm not an EE I believe there's a better way to this without an op amp or maybe an improvement is needed to the current circuit that I don't see it!

  • How do I avoid sparks at RL and make the load to turn on slowly?
\$\endgroup\$
0

1 Answer 1

0
\$\begingroup\$

The simplest solution to prevent current overshoot when connecting the load is to disconnect the load by short circuiting it instead of opening the circuit.

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ You mean that I should draw a little current to keep the op amp output at a relatively low voltage? but then I'll have some unnecessary current flow! \$\endgroup\$ Commented Jun 8, 2021 at 6:31
  • \$\begingroup\$ Yes, though it's just 10 mA at 100 V, so 1 W of power. \$\endgroup\$
    – jms
    Commented Jun 8, 2021 at 6:55
  • \$\begingroup\$ Which is wasted and makes heat... \$\endgroup\$ Commented Jun 8, 2021 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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