1
\$\begingroup\$

enter image description here

I have this circuit here and need some help understanding it. The values for it is as follows:

  • VDD = 8 V
  • k'*W/2L=10mA/V^2
  • R=200 ohms
  • Laserdiod has U=2.2 volts

Now, I think the current through the diode and the current through the resistor should be the same, since no current goes in to the OP-AMp. Is this correct? Then the current through the diode I is given by I=Uin/R. I also want to know how to find the maximum current through the diode and calculate it.

\$\endgroup\$
2
  • \$\begingroup\$ Welcome to EE.SE. Please embed the photo properly in the post so it's still there when the link dies. You'll get better answers as well if you save the readers having to follow links. \$\endgroup\$
    – Transistor
    Sep 27, 2017 at 23:10
  • \$\begingroup\$ It's generally better to use an N-channel as a low side switch (source at ground) if only conceptually because V_gs (and therefore I_ds) depends only on V_g, and not on the load current. In this topology V_s increases with load current. Any reason for this topology? \$\endgroup\$
    – Cuadue
    Sep 27, 2017 at 23:25

1 Answer 1

1
\$\begingroup\$

I think the current through the diode and the current through the resistor should be the same, since no current goes in to the OP-AMp. Is this correct?

Yes.

I also want to know how to find the maximum current through the diode and calculate it.

It depends on information you haven't provided.

First, you need to know the maximum output voltage of the op-amp, which could be nearly equal to Vdd, or it could be 2 or even 3 V less.

Then you need to figure out what the LED current will be when the op-amp is producing that output. That depends on information about the MOSFET that you haven't included in your questions.

\$\endgroup\$

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.