How can I calculate the slew rate and bandwidth of a typical transistor amplifier such as the following? What are the parameters I need to consider?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Use LTSpice, a free sim tool. \$\endgroup\$ – Andy aka Nov 3 '17 at 17:58
  • 2
    \$\begingroup\$ Well how would LTSpice be calculating it? There must be some set of equations somewhere that it is referencing if it can compute them from the transistor parameters. \$\endgroup\$ – Billy Kalfus Nov 3 '17 at 20:51

To compute the dynamic response of this circuit, you can use the hybrid-pi model to which you add a parasitic capacitance between collector and base. This is a 1st-order circuit and be solved in a variety of ways. I used the fast analytical techniques described here. First, I determine the dc transfer function, \$H_0\$, for which I open the capacitor. Then, I will determine the resistance "seen" by the capacitor when the excitation \$V_{in}\$ is reduced to 0 V (replace the source by a short circuit in the schematic). Finally, I will determine the resistance "seen" by the capacitor when the output is nulled with the excitation back in place. The below diagram shows all these steps:

enter image description here

Each resistance associated with a capacitor forms a time constant. For a 1st-order system, the pole is the inverse of the natural time constant. When you determine the zero, you realize that this is a right-half-plane zero and distorts the phase response. When you assemble all results, you have

\$H(s)=H_0\frac{1-\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$ with:

\$H_0=-\beta\frac{R_2}{R_1+r_\pi}\$ \$\omega_p=\frac{R_1+r_\pi}{C_1(R_1R_2(\beta+1)+r_\pi(R_1+R_2))}\$ \$\omega_z=\frac{\beta}{C_1r_\pi}\$

Plotting the whole thing with Mathcad gives you the dynamic response you need:

enter image description here

Now, regarding slew-rate, I am not sure the term fits this circuit. We usually identify slew-rate when the rate-of-change of the input signal of an op amp unbalances its differential bipolar input and the output cannot follow the input slope (see here). With a simple circuit like here, it is more about the rise time you can obtain I believe when you step the input. Considering the high-frequency zero and the low-frequency pole incurred by the capacitor, the rise time can be approximated as \$t_r\approx 2.197\tau=75\;\mu s\$ as given here. \$\tau\$ is the inverse of \$\omega_p\$. This is an approximation but as the zero is really high, it works ok as shown in the below sim where the input is stepped from -10 to -20 mV:

enter image description here

  • \$\begingroup\$ Very interesting - in the hybrid-pi model, how did you find the value for Rpi? \$\endgroup\$ – Billy Kalfus Nov 6 '17 at 21:50
  • \$\begingroup\$ Hello, this is difficult to find small-signal values for bipolar transistors in data-books. Otherwise, bias your transistor to have the right voltage on the collector and slightly move the \$V_{be}\$ around the operating point. The dynamic resistance you want is \$r_\pi=\frac{\Delta V_{be}}{\Delta I_b}\$. \$\endgroup\$ – Verbal Kint Nov 6 '17 at 22:10

We'll compute the slewrate when Vce is 2.5 volts.

At that operating voltage, Ic is 2.5/10k or 0.25mA.

GM (transconductance) at 0.25mA is 1/(0.026 / 0.00025) = 1/39*4 = 1/156 ohms

Gain is 10kohm / 156 ohm or 10 * 6 = 60x

The Cmiller is 10pF (Cob) * (1 + 60) ~ 10pF * 60 = 600pF.

The input timeconstant is Rbase * 600pF = 600,000 picosecond (ignoring the input resistance of the transistor in linear region).

If input timeconstant is 0.6uS, what will be the slewrate?

  • \$\begingroup\$ The loop including Vin, Rb, and the base acts as an RC circuit. Supposing Vin is 1V, then the equation for charging the base capacitance is 1(1-e^-t/RC) = 1-e^-t/RC. If I assume I want my base voltage to be 99% Vbb to consider it fully risen, then it would take ~5 time constants. 1V/(5*0.6us) = 0.33V/us. However, there is one thing I am a bit confused about - in a regular RC circuit, the capacitor charges up to the input voltage - but we know that a base junction is basically a diode, and will have a constant ~0.7V across it - does this mean the base capacitance can only charge to 0.7V? \$\endgroup\$ – Billy Kalfus Nov 4 '17 at 5:35
  • \$\begingroup\$ Also, where does the 0.026 come from? And where does the 1+ come from in (1+60)? \$\endgroup\$ – Billy Kalfus Nov 4 '17 at 5:36

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