0
\$\begingroup\$

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?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\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
1
\$\begingroup\$

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

\$\endgroup\$
  • \$\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
0
\$\begingroup\$

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?

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.