# Design an emitter follower to operate in the audio range (20 Hz - 20 kHz)

I am now self-learning with the "Art of Electronics" book by Holowitz and Hill, and I encountered a problem understanding one example from the book. Here it is: And here is the explanation for the calculation of the circuit:

A. Emitter follower design example

As an actual design example, let’s make an emitter follower for audio signals (20Hz to 20 kHz). VCC is +15V, and quiescent current is to be 1 mA.

Step 1. Choose VE. For the largest possible symmetrical swing without clipping, VE = 0.5VCC, or +7.5 volts.

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

Step 3. Choose R1 and R2. VB is VE +0.6V, or 8.1V. This determines the ratio of R1 to R2 as 1:1.17. The preceding loading criterion requires that the parallel resistance of R1 and R2 be about 75k or less (one-tenth of 7.5k×β ). Suitable standard values are R1 = 130k, R2 =150k.

Step 4. Choose C1. The capacitor C1 forms a highpass filter with the impedance it sees as a load, namely the impedance looking into the base in parallel with the impedance looking into the base voltage divider. If we assume that the load this circuit will drive is large compared with the emitter resistor, then the impedance looking into the base is β RE, about 750k. The divider looks like 70k. So the capacitor sees a load of about 63k, and it should have a value of at least 0.15μF so that the 3 dB point will be below the lowest frequency of interest, 20 Hz.

Step 5. Choose C2. The capacitor C2 forms a highpass filter in combination with the load impedance, which is unknown. However, it is safe to assume that the load impedance won’t be smaller than RE, which gives a value for C2 of at least 1.0μF to put the 3 dB point below 20 Hz. Because there are now two cascaded highpass filter sections, the capacitor values should be increased somewhat to prevent excessive attenuation (reduction of signal amplitude, in this case 6 dB) at the lowest frequency of interest. C1 = 0.47μF and C2 = 3.3μF might be good choices.

My questions are: 1) they only mention high-pass filtering (below 20 Hz). But what ensures filtering above 20 kHz, namely where is the low-pass filter here?

2) Is this modification will do the job (adding C3 to form a low-pass filter with the Re)? (sry for terrible drawing) 3) If no, than how to add high-frequency filtering properly here and calculate the values of additional components? (if needed).

• @KevinWhite Agreed. I think Spehro's last two sentences are the right thing. It doesn't take much series resistance and capacitance across $R_2$ to set a useful upper bound. Of course, the OP wants a calculation too. But it's just the obvious RC low pass calculation, so the OP should be able to readily find it on the web. Just make the added bypass C small compared to $C_1$ and figure out the series R from there. – jonk Jan 16 '20 at 1:14