Electronic Devices - Diode Applications - Discussion
Discussion Forum : Diode Applications - General Questions (Q.No. 5)
5.
What is the voltage measured from the negative terminal of C4 to the negative terminal of the transformer?
Discussion:
18 comments Page 1 of 2.
Madan said:
1 decade ago
I need explanation for this one.
Rahul said:
1 decade ago
This circuit is known as the voltage doubler.
Srinivasan said:
1 decade ago
This circuit is called voltage quadrupler. 4 times the input voltage which is 20 but the capcitor polarity is given in opposite direction, so -20V
TONMOY BANIK said:
1 decade ago
If it has another capacitor brunch C5 then it would be what 5 times of the transformer voltage.
Banashri said:
1 decade ago
Thanks srinivasan
Gayu said:
1 decade ago
Its a voltage quadrupler circuit which quadrupples the voltage.
Sayali &sarika said:
1 decade ago
This is quadruple circuit but output voltage increases up to two times only i.e. 2Vo.
Ganesh said:
1 decade ago
How does voltage becomes 4 times?
Avnish sharma said:
1 decade ago
What is the working of quadrupples?
Sumanta said:
1 decade ago
Consider this as two voltage doubler. For +5V C1 does not charge. But C2 charges to -5 V. So the negative potential becomes -10V. Now consider the second half with -10V source. Similarly C3 does not charge but C4 charges to -20 V (-10 + (-10V) <- charged capacitor).
Capacitor is charged with current flow - to +. Which is not possible for odd indexed capacitors :-).
Capacitor is charged with current flow - to +. Which is not possible for odd indexed capacitors :-).
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