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I would like to implement the current source circuit above to output a 4-20mA. Vin coming in from non-inverting input on IC1 is 0-5V. I am also adding 1nF capacitor and 10k resistor as shown to prevent any occsilation. The question is I dont understand the equation that calculate Iout. Also choosing value for R1,R2 and R3. I know R3 should be a small value to have low voltage drop. but what about R2 and R1. I dont want to plug and play with number. I would like to have full understanding of this concept.

Your input is very much appreciate. Thank you,

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  • \$\begingroup\$ You don't need the added parts on IC1 for reasonable values of R1/R2 and most op-amps. On IC2 it may be necessary, especially with micropower op-amps. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Aug 2, 2021 at 3:43

2 Answers 2

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IC2 will regulate FET to achieve the same voltage on Vin-=Vin+ of IC2 by negative feedback.

Filters should not be necessary as no voltage gain is added and very little capacitance. But if there is any layout error-inducing crosstalk to positive feedback nodes, it will reduce gain margin and could oscillate.

  • thus Iout= (Vcc-Vin+)/R3 when Q1 is conducting a current controlled by the input voltage on the left

  • with \$I_{R1}=I_{R2}\$ and \$V_S * I_{R3}= V_C * I_{R2} , I_{R3}=I_{OUT}\$

  • Thus due to the matching input voltages with any linear output on both OpAmps, Iout is just determined by R Ratios to scale the input voltage.

For these low currents, R3 is in the < 100 Ohm range <0.4V drop to <2 V drop or less if desired. as long as Vce is not saturated where hFE drops to 10% of its max value. So Vce must be > 0.7V or pref.=2V to have lower differences from Ic = hFE * Ib = Ie-Ib in order to maximize hFE.

enter image description here

You could consider this as a voltage-controlled current-mirror where the input current in R1 is shared by R2 and amplified to an equal drop in voltage (Vc=Vs) V = I * R current ratios.

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  • \$\begingroup\$ is there a typo on red under image? Vc+Ic/R2 = +Vcc = Vs +Iout*R3. where Ic/R2, should that be * instead of /. \$\endgroup\$
    – RC-95
    Commented Jun 3, 2021 at 18:29
  • \$\begingroup\$ Yes it was “a typo “ \$\endgroup\$
    – D.A.S.
    Commented Jun 3, 2021 at 18:35
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As a good approximation, when negative feedback is present, the op amps drive their outputs to voltages which make their inputs equal to each other.

So, current through R1 is equal to Vin/R1

This same current (approx) flows through R2 and so the voltage drop across R2 is equal to (Vin/R1)*R2

The voltage drop across R3 is equal to the voltage drop across R2 and so the current through R3 is equal to ((Vin/R1)*R2)/R3 = (Vin * R2)/(R1 * R3)

The current through R3 is equal to Iout

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  • \$\begingroup\$ Can you explain why is the voltage drop across R3 is equal to R2? \$\endgroup\$
    – RC-95
    Commented Jun 3, 2021 at 18:46
  • \$\begingroup\$ @RC-95 Read the first sentence of my answer again. The two inputs of IC2 are virtually at the same voltage as each other and the input bias current is so small that the voltage drop across your added 10k resistor is very small. \$\endgroup\$
    – user173271
    Commented Jun 3, 2021 at 18:55

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