# Constant current source math - Using AD8276 differential Op-Amp

So I'm trying to implement a constant current source to conduct a specific measurement using a AD8276 differential amplifier. I'm trying got derive the equations my self for sanity check, and to give a little me weight when I present my idea. This is what I have work out but my math doesn't checkout, here is a [circuit diagram][2] for my calculations

simulate this circuit – Schematic created using CircuitLab

things to know Rf1=Rf2=Rg1=Rg2 and R1=R2

The equivalent resistance of the parallel circuit consisting of the load resistance and the negative feedback lines $$R_{eq} = {({R_L}^{-1}+(R_1+R_2+R_{f1}+R_{g1})^{-1}})^{-1}$$

The voltage at the positive terminal, using the voltage divider. This is mainly where I thin I have gone wrong. $$V_1 = V_{ref} * ({R_{f2}+R_{eq}})/({R_{g2}+R_{f2}+R_{eq}})$$

Next it is understood that the negative terminal of the op amp will have the same voltage as the positive terminal, 0 potential diffrence $$V_1 = V_2$$

V2 will be propped completely across Vg1 and therefore the current through R1,R2 and Rf1 can be calculated by calculating the current through Rg1 $$I_{Rg1} = V_2 /R_{g1}$$

The total current supplied by Vref is $$I_{total} = (V_{ref} - V_1)/R_{g2}$$

And therefore the current through the load is $$I_L =I_{total} - I_{g1}$$

This does not give me the same results as the equation provided from the sources of www.analog.com/library/analogDialogue/archives/43-09/current_source.html figure 5 without the transitor. $$I_o = I_L = V_{ref}((1/40K)+(1/R_2)$$

Any thoughts?

EDIT: I fixed my diagram error

• Analog's circuit is different. The op-amp is "upside down" compared to your diagram. Plus other details, like the transistor, and the other op-amp doing the sensing feedback. In that circuit, the current across $R_l$ has no other path to go by out across $R_{load}$. (Since the + input on the AD8603 op-amp in the feedback has a high impedance) So if the circuit regulates the voltage drop across $R_l$, it thereby programs the current across the load.
– Kaz
Apr 16, 2013 at 1:20