I found this video which shows how to plot the transit frequency in Cadence Virtuoso. But on another site, instead of using c_gg, the capacitance used was c_gs+c_gd for the ft equation. Which is correct?

Also, the plot I'm getting in Cadence is around the hundred MHz to GHz range, which increases with increasing Vgs (if length is fixed) and decreases with increasing length (if Vgs is fixed). If I want to have an amplifier with GBWP of 10MHz, which transit frequency should the transistor have? Should it be near 10MHz or as large as possible?


1 Answer 1


The transit frequency \$f_t\$ of a transistor is mainly a method to compare the performance of transistors (or more often, the performance of a integrated circuit technology) with eachother. It should not be used as a design parameter, for two reasons:

1) The \$f_t\$ of a transistor is calculated/measured in an idealized circuit: Infinite impedance source and load (current sources)\$^1\$. This is never achievable, and thus without the circuit around it taken into consideration, it's pretty much impossible to go from \$f_t\$ to a useful spec. The practical GBW of a circuit is always going to be lower\$^2\$.

2) The \$f_t\$ calculated at DC (through the gain and capacitance at nodes) will not actually mean that if you build up the AC circuit to test it, you will find the GBW of your transistor to be identical. There are additional capacitances, inductances and resistances not taken into consideration in the simplistic \$f_t\$ equations that will actually result in the measured \$f_t\$ to be lower than the calculated one.

Knowing all this, to answer your question: You want your transistor to have an \$f_t\$ far above your opamp's GBW. After all, the \$f_t\$ is a measure of the maximum possible performance of your transistor in ideal circumstances (no load capacitance, perfect current sources), and the rest of your opamp's circuit will not be ideal. Hence, you should just ignore \$f_t\$ in your opamp design and focus on the actual GBW achieved with the needed load capacitance.

\$^1\$ in the video shown the lecturer does not use current sources because he is calculating them from DC specifications (capacitance and gain). However, the definition of \$f_t\$ is a small-signal parameter that is calculated with current sources as AC source and load, and the transistor operating at the intended bias point.\$f_t\$ is then defined as the point at which the transistor provides no gain, IE, it performs worse than a wire. You could think of the \$f_t\$ as the frequency at which the transistors own internal capacitance presents too much load for the transistor to still provide any gain.

\$^2\$ Note that while the GBW of a classic opamp-style structure will be lower, it is possible to have circuits provide power gain past \$f_T\$. This can be done with matching networks. The maximum power gain frequency is called \$f_{max}\$, and in the case of many modern CMOS processes, \$f_{max}>f_t\$. (In this case, even though the input current might be larger than the output current, the voltage swing will be much larger at the output, resulting in a net larger power at the output).

  • \$\begingroup\$ This clears things up. One clarification: Was the lecturer's method in getting the transit frequency in the video correct? \$\endgroup\$
    – user203020
    Nov 21, 2018 at 6:26
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    \$\begingroup\$ I didn't watch the entire thing, so I can't be sure, but I think what he did is correct in the sense that he got Virtuoso to give him the DC parameters (capacitance and gm) to calculate the theoretical \$f_t\$. \$\endgroup\$
    – Joren Vaes
    Nov 21, 2018 at 6:29
  • \$\begingroup\$ Time_constant of any single-device amplifier is GM / C. Where the C is the total capacitance on the gain (output) node. Once you have the time_constant, invert that and divide by 6.28 for the predicted F3dB. \$\endgroup\$ Nov 21, 2018 at 13:34

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