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Question regarding Vout of a charge pump

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Willt

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Hi experts,

I have a question regarding charge pump.

Vout of charge pump = (n+1) Vdd where n is the # of stages

Q1. In practice, Vout never reach (n+1)Vdd. What are the causes?

Q2. When the # of stages increases, Vout does not increase PROPORTIONALLY. Why??

Your comment is highly appreciated.
Thanks.

Will
 

Charge Pump Question

Could you show a schematic.

Is it the one with diodes and caps? If that's the case, the voltage stored in every stage loses the voltage drop over the diode from the previous stage..

so you get (n+1)vdd - nvD where vD is the diode ON voltage drop.
 

Re: Charge Pump Question

Hi jonashat,

Thanks for your comment. ^.^
Sorry for being unclear.

That's the one with MOS switch & capacitor.

(1) I want to know the causes leading to Vout != (n+1)Vdd in practice where n is the # of stages.

(2) Also, Vout does not increase PROPORTIONALLY when increasing the # of stages. What are the causes?

Thanks.

Will
 

Charge Pump Question

because of the threshould voltage of the switch, vout!=(n+1)vdd. Since the vth increase with the stage increasing, vout does not increase proportionally
 

Re: Charge Pump Question

I still need to see the circuit...
:roll:
 

Re: Charge Pump Question

Here is the circuit ~

pfd001: I think you got one of the points ~

The switches are MOS transistors with bulk connected to gnd.

Thanks.
 

Charge Pump Question

I'm presuming that all the transistors are NMOS transistors (since you said bulk is connected to ground).

The transistors that are connected to a high voltage (that's about 2/3 of your switches) should be PMOS, with bulk to VDD. An NMOS has a limit on conducting a high voltage.
 

Re: Charge Pump Question

the loss is due to:

1) threshold voltage of the switch

2) charging and discharge of any parasitic capacitance

3) load (resistive)
 

Re: Charge Pump Question

A few reason for voltage losses

1. The switch resistances caused voltage losses.
2. Parasitics.
 

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