I started to study more these days about Audio Amplifiers in class AB and I came across about something called bootstrap connection.

As far as I know bootstrap is used to pull the signal higher than the supply voltage and lower than the ground (in this case) since this schematic is based on a single-ended power supply. enter image description here

The book only mentions that R8,R9 and C5 is a bootstrap connection used for optimisation but I don't really get why is this so.

Could someone explain me how does it work?


2 Answers 2


The C5 bootstrap capacitor performs two functions.

a) it raises T3 collector load to increase its gain
b) it provides extra voltage headroom on R9 to drive the output pair

T3 gain is limited by the collector load, here being R8 and R9. Ideally the collector load will be infinite. However, the T3 amplifier needs a fairly high current in the collector to source a decent current into the output pair, which would require very low values for R8 and R9.

One way to provide that would be to replace R8/9 with a current source.

This amplifier designer has chosen an different way. Without C5, R8/9 would simply present an impedance of 2.2k. However, C5 connects to the output of the amplifier, whose voltage follows that of T3 collector. C5 drives the mid point of R8/9, so that the voltage across R9 is now constant, at least at AC frequencies where the impedance of C5 is small. If the voltage across the resistor is constant, it's behaving as a current source.

When large positive outputs are required from the amplifier, C5 can drive the R8/9 junction above the rail, to continue to supply sufficient current into the output pair through R9.

The arrangement where AC-coupled feedback from a voltage follower is taken back into a bias arrangement to raise its effective impedance is called bootstrapping, and is also often found raising the impedance of input bias resistors at emitter follower or op-amp buffer inputs.

The arrangement where an output drives a capacitor to provide bias voltages above the normal rail is also called bootstrapping, and is also found in high side N channel FET drivers, to provide gate voltages above the normal positive rail.

This circuit qualifies for both types of bootstrap.

  • \$\begingroup\$ I enjoyed the entire answer. I wish I might have written as clearly and directly. \$\endgroup\$
    – jonk
    Mar 27, 2018 at 19:27
  • \$\begingroup\$ @jonk Thanks, I appreciate that. I strive for clear. \$\endgroup\$
    – Neil_UK
    Mar 27, 2018 at 20:22

No capacitor charges nor discharges instantneously. C5 and the mentioned resistors are selected to make the voltage of C5 so slowly changing that at audio frequencies even during lowest bass notes C5 doesn't charge nor discharge substantially during one cycle of the audio signal.

But why? Because the base current of T5 comes through R8 and R9. When the instantneous Uout to the speaker is nearly +24V, there would be not much voltage over R8 and R9 without C5. That would severely limit the base current of T5, which causes distortion (T5+T6 are a voltage follower, Uout wouldn't rise if T5's base current pulled the voltage down at the base)

C5 retains its charge, so it lifts the voltage of the connection R8&R9 as Uout grows. Thus there's enough voltage over R9 for sufficient base current of T5.

The same problem is nonexistent for negative signal cycles, because T3 can well pull down quite high currents.


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.