# Bootstrap in power amplifier

So I know the theory behind a bootstrap current source in power audio amplifier and understand its operating principle but some things remain uncertain to me.

First - how to divide the collector resistance in voltage amplification stage? I did run some simulations and it's clear, that $$\frac{R_1}{R_2}$$ ratio actually matters.

When $$\R_1>R_2\$$:

• Gain drops
• Crossover distortion becomes apparent (despite the biasing)
• voltage across $$\R_2\$$ is not stable so current from current source does not maintain its value very well (unstable current source)

When $$\R_2>R_1\$$:

• Gain raises
• Less distortion
• Curernt source very stable

I don't understand why it happens. From the theoretical point of view, it should not matter which resistance is bigger.

Second - how to calculate the open loop gain after applying a bootstrap? Or perhaps I should ask - what's the actual value of collector resistance then? I need this to choose the correct value of $$\R_{\text{in}}\$$.

• Quiescent current - $$\3\text{mA}\$$
• Desired max power - $$\5\text{W}\$$
• Input - $$\1\text{V}_{\text{pp}}\$$
• Total collector resistance - $$\3.6\text{k}\Omega\$$

• Do you keep R1+R2 constant while varying R1/R2? If you don't, DC current will change... – bobflux Dec 28 '20 at 11:17
• As mentioned previously in an answer, you need a resistor from the base of T1 to ground; you cannot rely on the value of 380 kohm to set just the right amount of base-bias-current for T1 to obtain perfect mid-rail quiescent operating conditions. – Andy aka Dec 28 '20 at 11:25
• Of course I keep the sum constant. Total collector resistance is 3.6k and $R1+R2$ never exceeds nor recede this value, regardless their ratio. – Dawid W Dec 28 '20 at 11:43

Take a look at this example circuit:

As you can see without any input signal (DC condition) the bootstrap capacitor ($$\C_1\$$) is charged to $$\7.85V\$$. Also notice that the circuit time constant is very large ($$\t = R_X||R_Y * C_1 = 0.544s\$$) compared to the audio signal period ($$\1/20Hz = 0.05s\$$). Thus, we can tell that the capacitor will act just like a $$\7.85V\$$ DC voltage source (the input signal is changing way too fast to be able to charge/discharge the $$\C_1\$$ capacitor).

Now let us see what will happen if the output voltage is at a positive peak and equal to $$\+10V\$$.

The situation is shown here:

As you can see the voltage at $$\V_X\$$ node is now higher than the supply voltage. And this is why we can achieve a larger voltage swing at the output for a positive cycle.

Because now the maximum positive voltage at the output can reach:

$$\ V_{max} = (V_{CC} - V_{CE(sat)}) ≈ 14.8V \$$.

But we have another benefit from the bootstrap capacitor.

Notice, that now the voltage across $$\R_Y\$$ is almost constant and equal to:

$$\ V_{R_Y} = V_{C1} - V_{BE} ≈ 7.15V \$$.

And this means that $$\R_Y\$$ acts just like a constant current source.

For the AC signal, the $$\R_Y\$$ resistor is seen as a bigger resistor due to Miller's effect. But this time we have positive feedback (non-inverting stage) and the amplifier gain is less than one (voltage follower). So we called it a bootstrapping.

so, the new $$\R_Y\$$ value is:

$$R_Y = \frac{R_Y}{1 - A}$$

Where $$\A\$$ is an output stage (voltage follower) gain.

The typical value of output stage voltage gain will be around $$\0.9\$$. Thus the $$\R_Y\$$ a resistor will be seen by the VAS stage as a ten times larger resistor.

As for the selection, we typically choose $$\R_X = R_Y\$$ or sometimes we pick $$\R_Y = 2...5 \times R_X\$$ to further increase the VAS stage gain.

But the $$\R_X\$$ value cannot be too low because this increases the size of a bootstrap capacitor.

$$\C_1 > \frac{0.16}{F_{LOW} \times R_X||R_Y}\$$

Also for the AC signals the$$\R_X\$$ resistor will appear as a resistor in parallel with $$\R_L\$$. So another restriction related to the minimum value of a $$\R_X\$$ resistor, this is why $$\R_X >> (20...100)R_L\$$

• That's another fantastic answer from You! So, contrary to what @Andy aka stated above, collector resistance and thus gain rises significantly, way above simple $Rc/re$ value. Thank You again. – Dawid W Dec 28 '20 at 14:10
• The voltage gain will increase but not for DC. Because the bootstrap is a "dynamical current source". Thus, only for the AC signals, the open-loop gain will increase his value. – G36 Dec 28 '20 at 15:21
• @G36 yes this is a great (and convincing) answer so I shall delete mine. – Andy aka Dec 28 '20 at 18:18