Transistor Capacitances

EEE 131 THX2/Y

Lawrence Quizon

Electrical and Electronics Engineering Institute
University of the Philippines
Diliman

2025-11-19

General Frequency Response

Most amplifiers have a frequency response that looks like the following:

Everything is a capacitor

If something holds charge, it is a capacitor.

\[C = \frac{dQ}{dV} \]

Hence, everywhere in semiconductor devices where there are holes and electrons moving in and out there is capacitance.

BJT Parasitics

BJT Parasitics

There are two main parasitic capacitances in a BJT:

• Base-Emitter Capacitance \(C_{\pi} = C_b + C_{je}\) (Base Charging Capacitance + Base-Emitter Junction Capacitance)
• Base-Collector Capacitance \(C_{\mu}\)

The base charging capacitance \(C_b\)

Before we start using the BJT, there is effectively “no charge” in the base region.

However, as we turn the BJT on, the emitter starts to inject carriers (charge) into the base.

The base charging capacitance \(C_b\)

The amount of carrier charge entering/exiting the base per second is \(I_{C}\)

If the amount of time a carrier spends in the base is \(\tau_F\), then the total charge stored in the base is: \[ Q_B = I_C \tau_F \]

So then a change in \(V_{BE}\) can cause a change in \(Q_B\) through \(I_C\):

\[ C_b = \frac{dQ_B}{dV_{BE}} \]

\[ C_b = \frac{dQ_B}{dV_{BE}} = \frac{d(I_C \tau_F)}{dV_{BE}} \]

\[ C_b = \frac{dQ_B}{dV_{BE}} = \frac{d(I_C \tau_F)}{dV_{BE}} = \tau_F \frac{dI_C}{dV_{BE}} \]

Hence,

\[ C_b = \tau_F g_m \]

The BE Junction Capacitance \(C_{je}\)

A depletion region is a region with charges.

As we discussed before, the width of the depletion region changes with applied voltage. It is known that \(W_{dep} \propto \sqrt{V_{bi} - V_{BE}}\).

The BE Junction Capacitance \(C_{je}\)

Hence, since \(Q_{dep} \propto W_{dep}\), we have that some constant of proportionality \(\alpha\) exists such that:

\[ Q_{dep} = \alpha \sqrt{V_{bi} - V_{BE}} \]

\[ C_{je} = \left|\frac{dQ_{dep}}{dV_{BE}}\right| = \left|\alpha \frac{d(\sqrt{V_{bi} - V_{BE}})}{dV_{BE}}\right| = \left|-\frac{\alpha}{2\sqrt{V_{bi} - V_{BE}}}\right| = \frac{\alpha}{2\sqrt{V_{bi}}\sqrt{1 - V_{BE}/V_{bi}}}\]

We can treat \(\frac{\alpha}{2\sqrt{V_{bi}}}\) as a constant \(C_{je0}\):

\[ C_{je} = \frac{C_{je0}}{\sqrt{1 - V_{BE}/V_{bi}}} = \frac{C_{je0}}{\sqrt{1 - V_{BE}/V_{j,BE}}} \]

and we see that \(C_{je0}\) is base-emitter junction capacitance measured at \(V_{BE} = 0\) V. We also typically refer to \(V_{bi}\) for the BE junction as \(V_{j,BE}\).

The BE Junction Capacitance \(C_{je}\)

\[\frac{C_{je0}}{\sqrt{1 - V_{BE}/V_{j,BE}}}\]

The BC Junction Capacitance \(C_{\mu}\)

Likewise, the BC junction’s depletion region also has a capacitance. We call this \(C_{\mu}\).

Since it’s usually reverse biased, we replace \(V_{BE}\) with \(V_{CB}\) and negate the sign in the denominator:

\[ C_{\mu} = \frac{C_{\mu0}}{\sqrt{1 + V_{CB}/V_{j,CB}}} \]

Note

We call it \(C_\mu\) and not \(C_{jc}\) because of its position in the small-signal equivalent. \(\mu\) is traditionally used to represent voltage feedback from output to input in the hybrid-pi model.

BJT Parasitics Summary

• \(C\mu\) - Junction capacitance between B and C
• \(C_\pi = C_b + C_{je}\)
  • \(C_b\) - Capacitance from carriers crossing the base
  • \(C_{je}\) - Junction capacitance between B and E

Tip

\(C_\mu\) and is typically around 5fF-10fF.

\(C_{je}\) is a little higher, about \(>10fF\).

\(C_b\) is typically around hundreds of fF.

BJT Small Signal Model with Capacitances

\[C\pi=C_b+C_{je}\] \[C_b=\tau_Fg_m\]

\[\frac{C_{je0}}{\sqrt{1 - V_{BE}/V_{j,BE}}}\]

\[ C_{\mu} = \frac{C_{\mu0}}{\sqrt{1 + V_{CB}/V_{j,CB}}} \]

MOS Parasitics

MOS Parasitics

The MOS has three sources of parasitics:

• Drain/Source to Bulk Junction Capacitance \(C_{db}\),\(C_{sb}\)
  - Same as earlier.

• Gate-bulk capacitance \(C_{gb}\)
  - The MOS Capacitor. Remember this?

• Gate-overlap capacitance \(C_{gs}\)
  - MOS Capacitor but we accidentally do it to S/D.

MOS Parasitics

Overlap Capacitances \(C_{GS}\) and \(C_{GD}\)

\(C_{GS}\) and \(C_{GD}\) are overlap capacitances caused by an effective MOS capacitor on the overlap of the G region to the S and D regions.

\[C_{GS} = C_{GD} = \frac{\epsilon_{ox} x_d W}{t_{ox}}\]

But if you may remember…

\[C_{ox}=\frac{\epsilon_{ox}}{t_{ox}}\]

\[C_{GS} = C_{GD} = C_{ox}x_dW\]

The Gate-bulk Capacitance \(C_{gb}\)

As we discussed before, the main physical device behind the MOSFET is a MOS capacitor.

\[ C_{gb} = \frac{\epsilon_{ox}}{t_{ox}}WL \]

The Drain-bulk and Source-bulk Capacitances \(C_{db}\),\(C_{sb}\)

\(C_{db}\) and \(C_{sb}\) come from junction capacitances of the PN junction formed by the opposite-type semiconductors of the drain/source and the bulk.

MOSFET Capacitances

MOS Parallel Plate Capacitances

• \(C_{gs}\)
• \(C_{gd}\)
• \(C_{gb}\)

S/D-Bulk Junction Capacitances

• \(C_{sb}\)
• \(C_{db}\)

Effective MOS Small-signal with Capacitances

The bulk is typically connected to the source meaning:

\(C_{gs}\) typically also accounts for \(C_{gb}\).

\(C_{sb}\) is shorted.

So, only \(C_{gs}\), \(C_{gd}\) and \(C_{db}\) survive.

Special Properties

The fundamental gains

Let’s see what happens to the fundamental gains due to the capacitances:

Voltage gain:

\[ \frac{v_o}{r_o || 1/sC\mu} + v_i (g_m + sC\mu) = 0 \]

\[ \frac{v_o}{v_i} = - (g_m + sC\mu)(\frac{r_o}{1+sC\mu r_o}) \]

\[ \frac{v_o}{v_i} = - g_mr_o \frac{(1 + sC\mu/g_m)}{(1+sC\mu r_o)} \]

Current gain:

\[ i_o = v_i (g_m - sC\mu) \]

\[ i_i = v_i (sC\pi + sC\mu + 1/r_{\pi}) \]

\[ \frac{i_o}{i_i} = \frac{g_m-sC\mu}{1/r_{\pi} + sC_\pi+sC_\mu}\]

\[ \frac{i_o}{i_i} = g_mr_{\pi} \frac{1-sC\mu/g_m}{1+sr_{\pi}(C_\pi+C_\mu)}\]

The Transition Frequency \(f_T\)

You’ll notice that the current gain reaches 1 (0 dB), but the voltage gain doesn’t.

Hence, people measure the “speed of the transistor” or the transition frequency as \[ \left| \frac{i_o}{i_i} \right|_{f_T} = 1\]

The Transition Frequency \(f_T\)

\[ \frac{i_o}{i_i} = g_mr_\pi \frac{1-sC\mu/g_m}{s(C_\pi+C_\mu)}\]

\[ \frac{i_o}{i_i} = \frac{g_m-sC\mu}{1/r_\pi + s(C_\pi+C_\mu)}\]

\[ \left| \frac{i_o}{i_i} \right| = \frac{\sqrt{g_m^2 + \omega_T^2C_\mu^2}}{\sqrt{(1/r_\pi)^2 + \omega_T^2(C_\pi+C_\mu)^2}} = 1\]

\[ \frac{g_m^2 + \omega_T^2C_\mu^2}{(1/r_\pi)^2 + \omega_T^2(C_\pi+C_\mu)^2} = 1\]

The Transition Frequency \(f_T\)

\[ \frac{g_m^2 + \omega_T^2C_\mu^2}{(1/r_\pi)^2 + \omega_T^2(C_\pi+C_\mu)^2} = 1\]

\[ \omega_T^2 = \frac{g_m^2 - (1/r_\pi)^2}{(C_\pi+C_\mu)^2 - C_\mu^2}\]

Since \(g_m^2 \gg 1/r_\pi^2\) and \((C_\pi+C_\mu)^2 \gg C_\mu^2\)

Definition

\[\omega_T \approx \frac{g_m}{C_\pi + C_\mu}, f_T \approx \frac{g_m}{2\pi(C_\pi + C_\mu)}\]

The Transition Frequency \(f_T\)

For a MOSFET, everything is the same.

Set \(r_\pi \rightarrow \infty\)

Definition

\[\omega_T \approx \frac{g_m}{C_{gd} + C_{gs}}, f_T \approx \frac{g_m}{2\pi(C_{gd} + C_{gs})}\]

The Transition Frequency \(f_T\)