**Bootstrapped Darlington Circuit Operation and its Equivalent Circuit:**

The maximum input resistance of a Darlington circuit is limited to 1/h_{ob} ≡ 2 MΩ, as 1/h_{ob }is the resistance between base and collector. However, the input resistance can be largely increased by bootstrapped darlington circuit through the addition of capacitor C’ between the first collector terminal C_{1} and the second emitter terminal E_{2}, as illustrated in Fig. 19.56.

The noteworthy point is that R_{C} is essential because in its absence R_{E} would be shorted to ground. If the input signal changes by V_{in} then E_{2} changes by A_{v}V_{in} (assuming that reactance of C’ is negligible) and the collector changes by the same amount. Hence 1/h_{ob }is now effectively increased to 1/h_{ob}(1-A_{v}) ≡ 400 MΩ for a voltage gain of 0.995.

An expression for the input resistance R_{in} of the bootstrapped Darlington circuit can be had by using the equivalent circuit shown in Fig 19.57.

The effective resistance R_{eff} between terminal E_{2} and ground is R_{C} || R_{E}. If h_{oe}R_{eff }≤ 0.1, then transistor Q_{2} may be represented by approximate hybrid model. However, the exact hybrid model, as shown in Fig. 19.57, must be used for Q_{2}. Since 1/h_{oe1} >> h_{ie2}, h_{oe1 }may be omitted from this figure. Solving for V_{in}/I_{b1}, we have

The above equation shows that the input resistance of the bootstrapped Darlington circuit is essentially equal to the product of the short-circuit current gains and the effective emitter resistance. If the transistor with the current gains of the order of magnitude of 100 are used and effective resistance as 5 kΩ, then input resistance obtained would be of the order of 50 MΩ.