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I have used the above simple circuit for a constant source where \$V_{\text{in}} = -5\$V, \$R_{\text{in}} = 390\Omega\$, \$R_f\$ is the load which ranges from \$110\$ to \$170\Omega\$....would using this circuit for a constant current source for the load be recommended?

As current $$i = -\frac{V_{\text{in}}}{R_{\text{in}}}$$ $$V_{\text{out}} = -V_{\text{in}}\frac{Rf}{R_{\text{in}}}$$ so it should serve as a constant current source, but while browsing the Internet for a constant current source I typically found these sorts of circuits:

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Is there a compulsion to include the transistor in my circuit? If yes, why wouldn't my circuit serve the purpose for producing a constant current?

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    \$\begingroup\$ Are you aware that you have created a POSITIVE feedback loop? \$\endgroup\$
    – LvW
    Commented Jun 24, 2015 at 11:07
  • \$\begingroup\$ Didn't notice it while posting I just selected an example circuit, anyways thank you for telling it :) \$\endgroup\$
    – Night_Fury
    Commented Jun 24, 2015 at 17:20

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If you don't need a true ground on the load and the load is not remote then the first circuit is preferable (an example would be an integrator where the load is a capacitor). It would not be great if the load was a temperature sensor with 20m leads.

The transistor circuit has the load grounded, will work even if the load is below the V- of the op-amp. It has less compliance in the positive direction (only works up to V+ -Vref + 100mV or so), and somewhat lower output resistance, but in practical applications it's most often a much better choice (at least when implemented correctly, as LvW points out!).

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    \$\begingroup\$ What about current limit for the first circuit? Wouldn't there be a ~20mA current limit (op-amp maximum output current) that can pass thru the load? The second circuit can source alot more current since the transistors usually have a high current gain. \$\endgroup\$
    – Golaž
    Commented Jun 24, 2015 at 11:34
  • \$\begingroup\$ @Golaž good point. The op-amp circuit can only source voltage and current the op-amp itself can handle. A 1A or 300V current sink/source is easy with a transistor or Darlington, and does not require a special $300 Apex op-amp. At the low end <<1uA the transistor circuit may be limited by leakage, but it's easy to make a 1nA source with a low Ib op-amp. \$\endgroup\$ Commented Jun 24, 2015 at 11:43

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