# Current-voltage converter exercise

Discussion in 'Homework Help' started by Burzuzuzum, Sep 23, 2016.

1. ### Burzuzuzum Thread Starter New Member

Sep 23, 2016
1
0
I've a question for you.
In the circuit shown below I've to find the values of Rin and Ro in order to have the maximum (20mA) and the minimum (4ma) output current on the load, with Vin swinging from 0V to 3V3. The OP-AMP has to be considered as an ideal OP AMP, and the BJT too. Can someone help me finding the solution? My problem is that I can't figure out how to consider the feedback ring. I've tried to put Vin to 0V. So I get 0V on V-, 0V on V+. Is it correct? The current Io on Ro is Vcc/Ro. The current on the "feedback ring" is Ifb= Io-IR2. And this is where I'm stuck. I've to put this current Ifb equal to what to solve the equation and get Ro?

In circuit like this one is it wrong to use the virtual ground aproximation? Is it beacause I have no negative feedback?
Thanks!!

2. ### ci139 Member

Jul 11, 2016
341
39
according to your schematic you need to consider your control from Q1 U.base2emitter point of view
. . . if the ▽ is the "Virtual ground" = (Ucc+Uee)/2 then --
**** the op amp is turned below (U.base2emitter + U.Rlim) (> ▽ or virtual- or signal- ground) quite abruptly when --
I.R2 by VL.12V forms there a negative voltage drop that cancels the voltage drop on R1 (e.g. U.R1 – U.▽ = 0)

. . . so the Q1 "Jumps" "close" and again --
the U.(R1+R2) - U.▽ becomes greater the 0 and the Q1 is abruptly set to "Jump" "open"
return ****

. . . you likely will do better if you reference your Op Amp inputs to (Q1,R.lim)'s threshold point
e.g. a voltage at OA1 OUTP where the Q1 starts conducting and add some negative feedback to make the operation smooth