Voltage to Current Converter

Discussion in 'Homework Help' started by jason_bourne, Oct 24, 2011.

  1. jason_bourne

    Thread Starter New Member

    May 4, 2010
    7
    0
    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.
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    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
     
  3. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    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.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    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
     
  5. Efron

    Member

    Oct 10, 2010
    81
    15
    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.

    Now you should be able to estimate Vout of first OA = Vb (although this information is not relevant for the exercise).

    This assumption is dangerous at this stage. If you think so, you will have to demonstrate it afterwards.

    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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