OK, apologies, a new user posted a question on bootstrapping, but double posted - so I deleted one of his posts, and BOTH disappeared?.
So, to answer his question:
Bootstrapping to increase input impedance basically removes the impedance added by the biasing resistor.
It works by simple ohms law - 'V = I x R'. The resistor feeding the base will have a DC voltage drop across it, due to the current through it - the resistance of this can easily be calculated by R = V / I.
Now what bootstrapping does is feed a 'copy' of the input signal back to the other end of the resistor - so BOTH ends of the resistor change together. As both ends change together there's no change in current in the resistor, as we talking AC impedance here, no change in current means no AC current flowing - which effectively makes the resistor look like a very high impedance.
So bootstrapping effectively removes the added impedance of the input resistor, but doesn't reduce the actual input imedance of the transistor it's driving - but you design for the transistor to have a high input impedance.
So, to answer his question:
Bootstrapping to increase input impedance basically removes the impedance added by the biasing resistor.
It works by simple ohms law - 'V = I x R'. The resistor feeding the base will have a DC voltage drop across it, due to the current through it - the resistance of this can easily be calculated by R = V / I.
Now what bootstrapping does is feed a 'copy' of the input signal back to the other end of the resistor - so BOTH ends of the resistor change together. As both ends change together there's no change in current in the resistor, as we talking AC impedance here, no change in current means no AC current flowing - which effectively makes the resistor look like a very high impedance.
So bootstrapping effectively removes the added impedance of the input resistor, but doesn't reduce the actual input imedance of the transistor it's driving - but you design for the transistor to have a high input impedance.