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Date Posted: 7/21/2008 11:50:59 PM  Status: Live
Power Converters
Course Textbook Chapter Problem
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Question Details:
The figures below show simple: (a) current-to-voltage converter (a.k.a. transimpedance
amplifier) and (b) voltage-to-current converter (a.k.a. transconductance amplifier).
(a) (b)
a) Show that the voltage Vout in Fig. (a) is proportional to the current generated by
the cadmium sulfide (CdS) solar cell, Is. Also show that the trans-impedance of
the circuit, Vout/Is = R.
b) Show that in Fig. (b), the current iD through the light-emitting diode (LED), and
therefore its brightness, is proportional to the voltage source, Vs (as long as Vs > 0.
Also find the trans-conductance of the circuit, iD/Vs.

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Date Posted: 7/22/2008 4:28:02 AM  Status: Live
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Response:
(a)
Given circuit diagram can be represented as below:
 
   
 
This circuit is operating in inverting configuration.
Assume that op-amp is ideal one.
If op-amp is ideal one then there is no current flow in inverting and non-inverting terminals.
So and
Total current in cadmium sulfide cell will flow through feed back path (R).
By applying KVL to the  loop as shown in figure with red arrow we can get
From this equation we can say that output voltage is proportional to the current generated in the cadmium sulfide cell.
Trans-impedance of the above circuit  =
 
(b)
Given circuit diagram can be represented as below:
 
 
 
This circuit is operating in non-inverting configuration.
Assume that op-amp is ideal one.
If op-amp is ideal one then there is no current flow in inverting and non-inverting terminals.
So
................(1)
Voltage across resistor .
Assume that the diode drop = 0V, then ......... (2)
From equations (1) and (2) we can write
.
Output voltage in terms of diode current (because op-amp is ideal one then there is no current flow in inverting and non-inverting terminals )
........... (3)
............ (4)
Brightness of the LED is proportional to the .
So from equation (4)
Brightness of the LED is proportional to the source voltage .
From equation (3) trans-conuctance =
 
I hope this helps you........................



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