Voltage to Current Converter

Thread Starter

jason_bourne

Joined May 4, 2010
7
Hi,

Need some urgent help, this is from an exam question. The question requires
us to work out the expression for Io/Vi

This is what i have tried so far,
oamp 1 e+ = (Vi/2)
using e+ = e- The base of transistor is now Vi/2,
i assumed Ve = Vi/2 as well (or should it be Vi/2 - 0.7)
Now ie = Vi/2/R3 and ie = ic thus Vc = Vi/2 + 0.2 (say saturated)

Now e+ of opamp2 is (Vi/2 + 0.2) say Vi/2 again thus e- = Vi/2
Thus the voltage drop across the resistors is V-Vi/2.
Not sure after this, i understand some aspects of the PMOS, when VGS is < Vt then current flows.

I have attached the diagram for reference.
 

Attachments

hgmjr

Joined Jan 28, 2005
9,029
HINT: An ideal opamp in the presense of negative feedback tries to adjust its output to make its negative and positive input terminals equal.

hgmjr
 

Adjuster

Joined Dec 26, 2010
2,148
Do you need to concern yourself with the detail of the MOSFET characteristics, or indeed those of the transistor? For this circuit to be functioning in a predictable way, it seems likely that we should assume that both the amplifiers will generate suitable output voltages so that their differential input voltages become quite small.

I see that hgmjr has said the essence of this in fewer words.
 

hgmjr

Joined Jan 28, 2005
9,029
Adjuster,

I think your rephrasing of the circuit behaviour is an excellent supplement to my comments. It can often be helpful to have the same thing stated in slightly different terms.

Let's hope that together we have imparted sufficient knowledge to jason that he can begin to understand the problem better.

hgmjr
 

Efron

Joined Oct 10, 2010
81
using e+ = e- The base of transistor is now Vi/2,
Be careful, your strategy (using e+ = e-) is correct. However, e- is connected to the emitter of transistor, not to the base.

As a consequence, Ve is forced to Vi/2.

i assumed Ve = Vi/2 as well (or should it be Vi/2 - 0.7)
Now you should be able to estimate Vout of first OA = Vb (although this information is not relevant for the exercise).

Now ie = Vi/2/R3 and ie = ic thus Vc = Vi/2 + 0.2 (say saturated)
This assumption is dangerous at this stage. If you think so, you will have to demonstrate it afterwards.

Now e+ of opamp2 is (Vi/2 + 0.2) say Vi/2 again thus e- = Vi/2
Thus the voltage drop across the resistors is V-Vi/2.
Not sure after this, i understand some aspects of the PMOS, when VGS is < Vt then current flows.
What could be important to state here is:
* The current over the transistor is proportional to Vi because Ve = Vi/2 and R3 is constant

If V is also constant, the same for Voltage_R2, and therefore Ve+(2) is function of Vi, so for Ve-(2) (assumed negative feedback through Mosfet).

I let you find out the equation Io/Vi.
 
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