0
\$\begingroup\$

This lecture (page 11) shows how to calculate input impedance of a differential amplifier.

As you can see from the image below the input capacitance is calculated through the steps shown at the right.

Could anyone explain the principle or how the method shown at the right works?

enter image description here

\$\endgroup\$
  • \$\begingroup\$ The big assumption here is that the transistors are matched (and therefore Cgs are matched also). The gyrations on the right are algebraic simplifications based on that assumption. \$\endgroup\$ – pgvoorhees Jul 12 '16 at 14:19
2
\$\begingroup\$

The method on the right describes what happens if you have a balanced circuit with a balanced (differential) input signal.

The top right capacitors with the ground in the middle directly represent the input capacitances of the PMOSFETs. Since the middle node is grounded (actually Vcc but for small signal that is ground) the voltage on the node is zero, as it is grounded DUH !

Yes but as the input signal is differential, the voltage at the middle node would still be zero even if you did not ground the node. Why is that ? Because the input signal is differential, if the voltage on the left input increases x Volt then the voltage on the right input decreases x Volt.

And because the two capacitors have the same value, the voltage in the middle will not change. For small signal it (behaves as if it) is grounded !

If the capacitors confuse you, just imagine if the capacitors were resistors, the same principle will still hold.

The 3rd picture on the right shows that you can replace the 2 series capacitors as one of half their value. This gives you the differential input capacitance.

The 4th picture with the ground on one side wants to explain what the input capacitance is when applying a single ended input signal.

\$\endgroup\$
  • \$\begingroup\$ Thank you very much for the answer! I was confused about removing ground symbol. I've got it now. Btw, are you from POSTECH? \$\endgroup\$ – anhnha Jul 12 '16 at 15:09
  • 1
    \$\begingroup\$ No I'm not from POSTECH, I'm just an IC designer working in one of the EU countries. \$\endgroup\$ – Bimpelrekkie Jul 12 '16 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.