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I have some doubts trying to understand the "shielding" property of a cascode device.

I understand that a cascode provides a higher output resistance which can help to increase gain. However, in several books it is mentioned that it has a shielding property whereby the input device is protected from voltage changes at the output.

I couldn't find a good explanation for this. How exactly does M2 shield M1 in the circuit below?

enter image description here

Surely, the change at the output voltage is due to a change at Vin which is translated to a small-signal current change. The voltage at the drain of M1 is surely going to change based on the resistance to AC ground at that point (ro1 parallel with 1/gm2). So, how is the cascode device M2 protecting M1?

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If the M1 drain was connected directly to Vout, then it has the Vout voltage swing on it.

As Vb is constant, the voltage of M2 source is very nearly constant, with orders of magnitude less voltage swing on it than Vout. M1's drain is therefore barely moving at all.

This is important as the major bandwidth limiting effect in a single stage inverting amplifier is the capacitance from drain to gate, and its amplification by the gain of the device.

With M1 driving Vout, the Miller effect times Cgd loads the gate, reducing the bandwidth.

In the cascode, the Miller effect for M1 is reduced to nearly 1. The Miller effect for M2 is not a problem due to the low impedance gate drive.

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  • \$\begingroup\$ I see. But what about the resistance seen at the drain of M1? The small-signal current in the branch is the same. So will there not be a voltage change at the M1 drain due to the resistance seen and the small-signal current through the branch \$\endgroup\$ Commented Jul 16, 2020 at 17:12
  • \$\begingroup\$ @AlfroJang80 Yes, M1 drain will move. If M1 and M2 are the same device type, then the gain across M1 is about -1, so the Miller multiplication of Cgd is 2x, as opposed to maybe 100x without M2 present, or 1x if the drain was shorted to AC. M2 reduces the gain swing, it doesn't eliminate it. \$\endgroup\$
    – Neil_UK
    Commented Jul 16, 2020 at 19:56
  • \$\begingroup\$ Okay, am I correct in saying that the M1 drain move is small because the resistance at that node is (ro1 || 1/gm2) which is approximately 1/gm2. Since V=IR, if resistance is small, the voltage shift is small. Is that correct? \$\endgroup\$ Commented Jul 16, 2020 at 22:58
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    \$\begingroup\$ @AlfroJang80 Yes, that's exactly right. I prefer to think of the same thing via gain. That is, M1 has a gain of -gm1, and a drain load resistance of (approximately) 1/gm2, so its gain is -gm1/gm2, which is -1 if the devices are the same part number. \$\endgroup\$
    – Neil_UK
    Commented Jul 17, 2020 at 8:17
  • \$\begingroup\$ Amazing. Thank you very much. \$\endgroup\$ Commented Jul 17, 2020 at 16:37
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If you use two identical FETs, and with operation at the same current, the gm of each will be "matched", and the small signal voltage at the shared node will be a 1:1 copy of the input AC signal, and will be INVERTED.

The input capacitance of the amplifier will be

Cin = C_gate_drain * (1 + Gain_Bottom_FET) = Cgs * (1 + 1) = 2 * Cgs

[ note: gm == transconductance]

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