I've used bootstrapping several times, either to boost the input impedance of an amplifier or to increase its gain. However most of the times I can see the beneficial effects of bootstrapping without actually knowing what's the actual input impedance of an amp due to bootstrapping.

Consider the following emitter follower stage which I got from a textbook.


simulate this circuit – Schematic created using CircuitLab

According to the book from which I got that circuit, the input impedance without the 10kΩ load (\$R_2\$) is 500kΩ and the input impedance with the load is 350kΩ. However the book doesn't provide any formulas or explain how any of those values were derived. So I'm trying to figure it out. This is what I've got:

Just to compare, suppose there is no bootstrap capacitor C2, transistor MPSA42 has a beta of 70, so the input impedance would be roughly \$\beta (R_2\parallel R_1) \parallel R_3\$ (I didn't consider R4 for the calculation since \$R_4 << R_3\$), substituting values results in an input impedance of 77.7kΩ.

Now consider what happens when the bootstrap capacitor is in:

The emitter current of the transistor is approx. \$\frac{24-0.7}{\frac{R_3+R_4}{\beta}+R_1}=2.01mA\$

Which results in an incremental emitter resistor of \$r_e=\frac{V_T}{I_C}=12.93\Omega\$

The voltage gain of the CC stage is roughly:

$$A=\frac{R_1 \parallel R_2 \parallel R_4 }{R_1 \parallel R_2 \parallel R_4 +r_e}=0.996$$

Now suppose that 1V is applied to the input. That would mean that the current through \$R_3\$ is \$ \frac{1-0.996}{100k}=40 nA \$ which, compared to the current through the 100kΩ resistor without bootstrap is \$\frac{40nA}{1/100k}=4x10^{-3}\$ times the current without bootstrap. This means the resistance of 100kΩ actually looks like a resistor of \$\frac{1/100k}{40nA}=250\$ times greater value, or \$100k\times 250=25M\Omega\$. This will be in parallel with the resistance reflected from the emitter to the base, which is roughly \$\beta (R_1 \parallel R_2 \parallel R_4)=233k\Omega\$, that in parallel with the 25MΩ resistor gives 230kΩ, as opposed to the 350kΩ mentioned in the book. I checked with a simulator and the input impedance is indeed close to 350kΩ.

What am I doing wrong, is my line of reasoning correct?

  • \$\begingroup\$ I'm with you on your reasoning, bootstrapping removes the loading effect of R3, but places an addtional 10k load on the emiiter. What does your simulator give as input impedance without C2? What about without C2, but an additional 10k (so three total) loads to ground from the emitter? \$\endgroup\$ – Neil_UK Dec 27 '17 at 8:36
  • \$\begingroup\$ Without C2, the simulator gives a 89k, vs the 84 kohm theoretical value, and without C2 but with an additional 10K resistor the simulator gives 82K VS 75k theoretical value. \$\endgroup\$ – S.s. Dec 27 '17 at 18:10
  • \$\begingroup\$ my bad, you did what I asked, but that wasn't what I meant. I meant without C2, with r3/r4 driven by a different but perfect bootstrap source, say another copy of the input voltage source, with 2 or 3 10ks on the emitter. This was to see what the tranny beta effect alone was. However, I see from the other answer, you were using the wrong value for beta. Duh! \$\endgroup\$ – Neil_UK Dec 27 '17 at 21:25
  • \$\begingroup\$ Yes and no, the beta stated in the data sheet is 40, I was using a beta of 70 which is the one stated in the SPICE model of the transistor, and apparently the author used 100 for no particular reason, but this is all speculation because no mathematical derivations are included in the book. \$\endgroup\$ – S.s. Dec 28 '17 at 0:30

Well, the equation for the Zin and voltage gain looks for the emitter follower with the bootstrap (positive feedback) at the input


simulate this circuit – Schematic created using CircuitLab

looks like this:

$$Z_{IN} = \frac{(r_eR_E + R_3(r_e + R_E))\cdot (1 + \beta)}{R_3 + (\beta+1)r_e }\approx \frac{R_3}{1 - A_V}||(r_e+R_E)(\beta+1)$$

$$A_V = \frac{(R_3+r_e) R_E}{r_e R_E + R_3r_e + R_3R_E} = \frac{R_E}{R_3||r_e + R_E} \approx \frac{R_E}{r_e+R_E}$$

Of course \$R_E\$ is for your circuit equal to

$$R_E = R1||R2||R4$$

So, in conclusion, your reasoning is correct

| improve this answer | |
  • \$\begingroup\$ Thanks for the input, the other thing im thinking of, is that perhaps the beta selected by the author of the book is not the same as the beta I selected (which is 70), and beta has a high impact on the reflected impedance \$\endgroup\$ – S.s. Dec 27 '17 at 20:36
  • \$\begingroup\$ And the name of the book is? \$\endgroup\$ – G36 Dec 27 '17 at 21:07
  • \$\begingroup\$ And yes if we ignore R3 Zin = beta*Re . So if the beta is 100. The Zin=100*5k = 500k and if we include the load resistor we will get 100*10/3 = 334k author rounds this value to 350k \$\endgroup\$ – G36 Dec 27 '17 at 21:18
  • \$\begingroup\$ Yes, I think that is correct. The name of the book is Small signal audio design by Douglas Self, great book, but unfortunately no math, which makes things harder if you ask me, like these figures that he brings out of the blue. \$\endgroup\$ – S.s. Dec 28 '17 at 0:22

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