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I have the following problem: -

schematic

simulate this circuit – Schematic created using CircuitLab

For the amplifier shown in circuit 1, the open circuit voltage gain at low frequencies was calculated to be \$A_{voc}= 29.2 \text{dB}\$.

In circuit 2, the input of the amplifier is connected to a signal source with \$R_s=100\text{k}\Omega \$. Furthermore, a capacitor \$C_p = 5\text{pF} \$ is connected between the input and output of the amplifier. It can be assumed that the frequency response of the amplifier has a dominant pole caused by \$R_s \$ and \$C_p\$.

What is the 3-dB frequency caused by the dominant pole?

Transistor values: \$\mu_nC_{ox}(\frac{W}{L})_n= \mu_pC_{ox}(\frac{W}{L})_p=300\mu\text{A}/\text{V}^2, \: V_{tn}=-V_{tp}=0.7 \text{V}, \: \lambda=0.1 \text{V}^{-1}, \: V_{DD} = V_{SS} = 1.5\text{V}\$.

My attempt

My first instinct was to draw the small-signal model of the entire circuit, calculate the small-signal values and find \$\frac{V_o}{V_{in}} \$ of the amplifier stage, with \$R_s \$ and \$C_p \$ included. This is doable, but somewhat laborious. I therefore simulated the circuit with LTSpice, and found a 3-dB frequency of 10.8kHz: -

enter image description here

However, in the actual solution they use something called "Miller transformation". They write:

Since the dominant pole is caused by \$R_s \$ and \$C_p \$ the 3-dB frequency can be calculated $$f_p = \frac{1}{2\pi} \cdot \frac{1}{R_sC_p(1+|A_{voc}|)}=10.7\text{kHz} $$

The equation they are using I have not been able to find in my book or on the internet.

My question is: where does this equation come from? Can it be derived? Why does the 3-dB frequency depend on the gain \$A_{voc} \$ at low frequencies?

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    \$\begingroup\$ where does this equation come from? Can it be derived? Yes, from the small signal equivalent model. It is really not that complex. The gm of both MOSFETs will end up in parallel, same for their output impedances. See sheet 15 of this presentation: web.mit.edu/6.012/www/SP07-L12.pdf \$\endgroup\$ Nov 7, 2021 at 12:26

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"My question is: where does this equation come from? Can it be derived? Why does the 3-dB frequency depend on the gain 𝐴𝑣𝑜𝑐 at low frequencies?"

The equation you are asking about is due to the very well known miller effect. In short, when a capacitance is connected in feedback between output and input nodes of a MOSFET (or any amplifier for that matter) transistor, it is seen by the gate (input) as stepped up by one plus the gain of the device.

enter image description here

The dominant pole in your schematic is the input node. So time constant * is $$\tau = {R_{in}C_{M}} = R_{in} C_{p}(1+|A_{vo}|)$$ and $$f_{p} = \frac{1}{2\pi \tau} = \frac{1}{2\pi R_{s} C_{p}(1+|A_{vo}|)} $$

  • Total device input capacitance is assumed << \$C_{m}\$, here.
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This is an approximate approach to the answer.

The gain at low frequencies is 29.2 dB or, in real numbers it is 28.84. At low frequencies we can forget about the capacitor and, we can model the low frequency operation as an infinite gain amplifier with negative feedback. Because we know the gain is 28.84, the "effective" feedback resistor is 28.84 times Rs. This is a value of 2.884 MΩ and, it's that effective value of resistance that forms the 3 dB point when in parallel with the 5 pF capacitor hence, the 3 dB point is 11.037 kHz. Circuit: -

schematic

simulate this circuit – Schematic created using CircuitLab

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