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Is it possible to create a capacitive voltage divider where one of the capacitors is a varactor whose capacitance can be manipulated/varied according to a signal? A varactor is a two port device; what would the circuit look like so the voltage control signal could vary the capacitance?

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Typical varactor circuit has the control signal isolated by a high-value resistor or inductor, and a series capacitor (or another varactor, the opposite way) to block DC. There are dual varactors made for the purpose.

For example:

enter image description here

Where C2 and the BB804 dual varactor form a voltage divider, and V1 controls the capacitance.

(The reason for using a dual varactor rather than a single varactor is that the capacitance will vary with the signal voltage, causing unwanted side effects such as distortion if the signal voltage is relatively large. By using two diodes, that variation tends to cancel out, at least to the extent that the diodes are both matched and linear).

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  • \$\begingroup\$ I expect the current that could be output by such a circuit would be tied to the absolute size of each "capacitor"? \$\endgroup\$ – Jordan McBain Jun 12 '18 at 18:57
  • \$\begingroup\$ The output impedance of the divider is the usual X1*X2/(X1+X2) where X = 1/(2*pifC). So, yes. \$\endgroup\$ – Spehro Pefhany Jun 12 '18 at 19:11
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Yes as long as the net impedance ratio is not frequency dependant on other factors of inductance and load impedance sensing.

Actual impedance of every wire, source , load and stray impedance must be known with applied voltage across Varicap as modulation will average the attenuation in a log manner. So it is not a linear attenuating capacitance voltage transformer.

Switched cap bank with CMOS may be more linear, so it depends on use.

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