Hey Guys,

Attached I have a schematic of a Voltage Doubler. I was wondering if I could get a REALLY detailed explanation as to what exactly is happening in the positive and negative cycles at both nodes v2 and the output. I don't quite understand what happens to the nodes v1 and v2 when the capictor charges, and how ground comes into play and all that. So please, if you could give me a hand, I would really appreciate it.

Thanks Guys

 

i am not very shure for destination of curents but may be they are thrue. this circuit i for low curents and hight voltages /in MW tehnologies/

 

Assuming the source voltage (VI) is squarewave, when VI is positive V, then V1 is positive V as well. C1 would be charged through the D1 (which is forward-biased), where V2 would be at ~0.7V above ground. If the positive period is long enough, C1 would be charged up to ~ V - 0.7V. There'd be no current flowing through D2 and C2 because D2 is reversed-biased.

When the source voltage switches to zero, VI = 0, V1 = 0, however we can not change the voltage across C1 instantaneously because C1 is still charged (Q = CV). Pulling V1 to ground, thus resulted in V2 to be pulled down by the same differential value to maintain the voltage across C1, thus V2 = -V + 0.7V. When this happens, D1 is reversed-biased and no current would flow through D1. D2, however, is forward-biased and C2 would be charged up by the current flowing through the outer loop (C2, C1, VI) up to negative V2 value + 0.7V = -V + 0.7V + 0.7V after a few periods (depending on the ratio of C1:C2). Thus eventually would be V3 = -V + 1.4V.

In effect, this circuit 'pump' the charge in C1 to C2, which is why it is also known as charge pump circuit.

If VI switches from +V to -V instead of +V to 0V, the process is the same, but instead of getting V3 = -V + 1.4V, it would be V3 = -2V + 1.4V. V2 would switch between 0.7V and -2V + 0.7V. In general, V3 would equal negative value of VI peak to peak voltage difference plus twice the diode voltage drops.