Timeline for Input Impedance of capacitive amplifier
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 16 at 13:54 | comment | added | Kenlucius | Thank you very much. I got it now :) | |
| Aug 13 at 23:11 | comment | added | Raonoke | Use either I2 or I3 to plot Zin. With differential input Zin = 2 * Xcin for the fully diff amp version with caps and Zin = 2 * Rin for the version with resistors. | |
| Aug 13 at 2:50 | comment | added | Kenlucius | @Raonoke Thank you very much for your reply. So, for the Differential Mode (Fig1), what I should correct for Zin is the denominator part because the two caps are virtually shorted and, therefore, equivalently two caps in series from the input ports' point of view. Therefore: Zin = (V1P - V1N) / I2. Is the explanation correct? To extend the idea. Let's look at the resistive fully differential amplifier in your first comment. The Zin in that circuit would be 2*Rin, where Rin is the input resistor of the amplifier circuit. Could you please help me see if these two understandings are correct? | |
| Aug 12 at 15:30 | comment | added | Raonoke | The problem I see is in the denominator of Zin. As the two 6p caps are virtually shorted (not grounded), I2 and I3 is actually the same current just with opposite direction. What you have is I - (-I) = 2I and as a result the plot shows only 1/2 Zin. On a side note, for normal operation, a 0.5V AC input voltage is rather high considering the large gain of the amplifier and the 1.8V DC supply voltage. | |
| Aug 12 at 14:08 | comment | added | Kenlucius | @Raonoke Thank you very much for your reply. I have a question here. Is my way of calculating Zin for the differential mode correct (Zin = (V1P - V1N) / (I2 - I3) )? V1P and V1N are 0.5V ac magnitude and 180 deg out of phase. I2 and I3 are the current drawn from the sources. | |
| Aug 11 at 18:25 | answer | added | qrk | timeline score: 2 | |
| Aug 11 at 18:18 | comment | added | Raonoke | This should be no different to a fully differential amp with resistors. Differentially you have a virtual short between the +/- inputs of the amplifier which means the input impedance would be two 6pF caps in series or 2.653 MOhm @ 20kHz. In single ended mode, the input impedance converges to 2.653 MOhm @ 20kHz (6pF/2) for gain aproaching infinity and 1.326MOhm @ 20kHz (6pF) with gain aproaching zero. Your gain is far from infinity but still large so the 2.6137 MOhm in single ended mode looks right. | |
| Aug 11 at 3:28 | history | asked | Kenlucius | CC BY-SA 4.0 |