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Circuit Working

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ericgibbs

Well-Known Member
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Hi all,

I need to understand how this circuit works for AC and DC signal upto 10V.
hi,
Its fairly easy to explain, but I would like to see how you believe the circuit works.:)
 
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Vikky

New Member
Yes,

I know it very simple to understand. What i have understood is as follows

Their would be a potential difference across R3 nd R4. When the +V is given to R1 and ground to R2, the diode D2 and D3 will conduct providing again a potential difference at the end of the ciruict which could be used for the difference amplifier.

When the +V is given to R2 and ground to R1, D4 and D1 will conduct giving potential difference at the end.

The point i stuck is when the +V is given to R1, D2 also conduct. So what would be the voltage at that point? what would be the equation for that??
 

ericgibbs

Well-Known Member
Most Helpful Member
I know it very simple to understand. What i have understood is as follows

Their would be a potential difference across R3 nd R4.
* thats correct, the Vs1 = 1V according to the drawing

When the +V is given to R1 and ground to R2,
*the R2 is not grounded, it goes to ground via R6 and R8

the diode D2 and D3 will conduct providing again a potential difference at the end of the ciruict which could be used for the difference amplifier.
*The drawing shows Vs1 is +1V.

* With the centre points of the diodes connected to 0V ground via 1k resistors R7 and R8 and to the 1V source via R1/R5 and R2/R6 which are 100k resistors, the voltage at the junction of the diodes will be less than the turn on voltage [0.7] of the diodes, so the diodes will not be conducting.

When the +V is given to R2 and ground to R1, D4 and D1 will conduct giving potential difference at the end.

The point i stuck is when the +V is given to R1, D2 also conduct. So what would be the voltage at that point? what would be the equation for that??

* with voltage drop across R1/R5 and R2/R6 the diodes are off and can be ignored, so just calculate the circuit using the resistor voltage drops.


hi,
Hope you follow ok.
IMO its a poorly designed question.
 
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