# Charge pump circuit with one MOSFET

This is an example of charge pump from MOSFET Models for SPICE Simulation: Including BSIM3v3 and BSIM4 by William Liu.

Could anyone explain why $v_D = \frac{(v_{IN} - V_T)}{2}$ here?

Why $V_T$ is involved in the drain voltage?

Thank you for help. ## 2 Answers

In this example only intrinsic capacitances are used. The capacitance from the gate to the channel is 1pF or 0.5pF from gate to drain and 0.5pF from gate to source.

The capacitors that are attached to the drain and source are 0.5pF as well. So basically we have a capacitive voltage divider with a 1pF gate-channel and a 1pF channel-ground capacitor.

Before all this can happen a channel is required, so we need a threshold voltage to invert the area under the gate. Then the excess voltage (Vin-VT) is split equally between the two 1pF caps, resulting in (Vin-VT)/2 at the drain (and the source) side.

Maybe I should add, that the shown circuit is NOT a charge pump. It is merely used to demonstrate that shortcomings of non-charge-conservative transistors models that exhibit a charge pumping effect. A charge conservative model (and physically correct model) will show no charge pumping effect whatsoever.

To better understand why VT has to be subracted a Q-V curve for the MOS capacitor is shown below. Below the threshold voltage the charge stays almost constant, the transistor has only a very small capacitance (almost zero). Above VT the relationship between the voltage and the charge is linear. The transistor acts as a capacitor.

• Thank you. Is it possible to know where the VT is dropped? Apply KVL, Vin = VT + Vcap but I can't imagine from the picture where VT is dropped. – anhnha Jun 12 '16 at 2:16
• @anhnha -- The voltage VT is required to get a channel underneath the gate. The channel is the bottom-plate of the gate-source/drain capacitance. – Mario Jun 12 '16 at 5:01
• Yes, I understand that but is it possible to apply KVL for this? – anhnha Jun 12 '16 at 6:08
• Since VT has to be subtracted from the input signal you can think of it as a voltage source and apply KVL. – Mario Jun 12 '16 at 6:27
Why $v_D = \frac{(v_{IN} - V_T)}{2}$ here?:


It has to do with the gate channel capacitance of the MOSFET. Until the channel is established, the capacitance is minimal and as a result the gate can swing without too much movement of the S/D. So in reality the first little excursion of the gate up to $V_{th}$ is for free and then you get the ratiometric (voltage dividing) happening above that. That is why you need to subtract the $V_{th}$ from $V_{in}$.