Resistor values for basic bootstrapping circuit

cake4all

Joined Oct 19, 2018
6
Hello, I am currently reading the section about bootstrapping (page 112) of the book "The Art Of Electronics (3rd edition)". The core concept of bootstrapping is explained nicely, however I'm struggling to understand how to choose the resistors $$R_1$$, $$R_2$$ and $$R_3$$ in the attached circuit.

On page 84 there is a design receipt printed for the standard emitter follower:

Step 1. Choose $$V_E$$.
For the largest possible symmetrical swing without clipping, $$V_E=0.5 \cdot V_{CC}$$, or +7.5
volts.

Step 2. Choose $$R_E$$
For a quiescent current of 1mA, $$R_E=7.5k$$

Step 3. Choose $$R_1$$and $$R_2$$.
$$V_B$$ is $$V_E+0.6V$$, or $$8.1V$$. This determines the ratio of $$R_1$$ to $$R_2$$ as 1:1.17. The preceding loading criterion requires that the parallel resistance of $$R_1$$ and $$R_2$$ be about 75k or less (one-tenth of 7.5k×β ). Suitable standard values are $$R_2$$=130k, $$R_2$$=150k.

Everything is easy until step 3. But how to choose now (with the bootstrapping concept) $$R_1$$, $$R_2$$? And $$R_3$$? As far as I know, I need to know the base current of the transistor.
$R_3 = \frac{U_{R_3}}{I_{R_3}} = \frac{U_{R_3}}{I_{B}} = \frac{U_{R_3}}{\beta \cdot I_C}$

But in the general case $$\beta$$ is not known. How to proceed?

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WBahn

Joined Mar 31, 2012
26,156
The data sheets will give you an idea of the range of beta values. If you need to be less than about one-tenth of something times beta, then design for a beta that is at the bottom end of its range. If that choice meets your needs, you are done. If it doesn't, then you have to start delving deeper.