1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

The above circuit delivers a predetermined current to the load: $$i_L=-\frac{u_s}{2R}$$

Is it possible to extend it, so that it does the same thing, but for two diffrent loads, while using only one opamp (and resistors)?

Basically, I am looking for a way to create two independent current sources.

\$\endgroup\$
3
  • \$\begingroup\$ No, it can't do that. \$\endgroup\$
    – Andy aka
    Commented Jul 22, 2020 at 18:09
  • \$\begingroup\$ You can make dependent current sources that work in parallel, but they'll all have the same current. \$\endgroup\$
    – Aaron
    Commented Jul 22, 2020 at 18:47
  • \$\begingroup\$ I don't think your circuit does what you think it does. Just intuitively, take the extreme cases from Load resistance is very small (close to zero), and then very large (close to infinity) and see if you think the same Us/2R will flow through the load in both cases. \$\endgroup\$
    – td127
    Commented Jul 22, 2020 at 23:26

2 Answers 2

0
\$\begingroup\$

As Andy states you will need an op-amp per current source. One can simply not direct a specific current in more than one direction.

\$\endgroup\$
1
  • \$\begingroup\$ Please take a look at Voltage Spike’s proposed solution \$\endgroup\$ Commented Jul 22, 2020 at 20:36
3
\$\begingroup\$

A current mirror might work:

You can use a current mirror with 2 NPN's in the configuration shown to mirror a current. This ensures that the Load R4 has the same current as the current running through R2.

schematic

simulate this circuit – Schematic created using CircuitLab

This circuit shows what is possible (and is unfinished because I have not sized the Vcc's or the resistor). But you could make an adjustable current mirror with this circuit. The idea is the opamp adjusts M1 to create the same voltage on the negative terminal as the positive terminal, so if you have 1V on the positive terminal there will be 1V on R2 (and the current \$ i = \frac{v}{2R}\$). If you use a mosfet for M1, make sure the gate voltage is high enough to turn on M1, this means you'd need to have a higher Vcc on the op amp to turn on the gate than the Vcc of M1 or use an NPN with an appropriately sized resistor). These circuits are hard to compensate to avoid ringing, so this circuit will need some filtering on the feedback path.

schematic

simulate this circuit

\$\endgroup\$
2
  • \$\begingroup\$ Yeah, I forgot to mention matching, a few ways around this are get a current matching IC or get matched transistors (2N3904's probably aren't that great for matching but I haven't checked the datasheet). \$\endgroup\$
    – Voltage Spike
    Commented Jul 22, 2020 at 22:05
  • 1
    \$\begingroup\$ It's not really a true current mirror unless the emitters are tied together too (to force Vbe to be the same on both transistors). \$\endgroup\$
    – td127
    Commented Jul 23, 2020 at 3:02

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.