Why is a Resistor Needed in an Emitter Follower Bias Current Circuit?

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SUMMARY

The discussion centers on the necessity of a resistor in an emitter follower bias current circuit, as outlined in "The Art of Electronics." The resistor RB is crucial for providing a DC path for base bias current to ground, especially when the signal is capacitively coupled. Without this resistor, the capacitor charges to the peak value of the input signal, leading to a negative voltage at the base, which biases the transistor off and results in no output. The presence of the resistor allows the capacitor to discharge between input cycles, enabling base current flow and preventing distortion.

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nylonman
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Hi, I'm reading The Art Of Electronics and I saw this in the emitter followers section:

"Warning:
You must always provide a dc path for base bias current, even if it goes only to ground... if the signal is capacitively coupled, you must provide a resistor to ground (Fig. 2.18)."

Here is the figure: http://imageshack.us/photo/my-images/38/figurep.png/

Could someone explain me why the resistor RB is needed?

Any help will be appreciated!
 
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If it were missing where would the current path be for the bias current?
 
ok, I see it, it was simpler than I thought. Thanks!
 
That circuit is used rarely because it generates a lot of distortion. To avoid this, a base bias resistor is usually placed between the +ve supply voltage and the base.

However, if you did use this circuit, and omitted the resistor to ground, then the capacitor would charge up on the first few cycles of input (via the base-emitter junction) and the base would have a negative voltage on it of about the peak value of the input signal.

This would bias the transistor off and there would be no output.

Using the resistor allows the capacitor to partly discharge between cycles of input and base current would flow on input voltage peaks.

The output will be very distorted and this is sometimes used in frequency multiplier circuits.
 
ok, so in other words you are saying that we need the resistor in order to discharge the capacitor since the transistor emitter allows the current to flow only in one direction, right?
 
Yes, the base emitter junction acts like any other diode and charges up the capacitor to the peak value of the input signal. After that, the transistor cannot be driven into base current, so there is no output.
 
Most likely this can only be answered by an "old timer". I am making measurements on an uA709 op amp (metal can). I would like to calculate the frequency rolloff curves (I can measure them). I assume the compensation is via the miller effect. To do the calculations I would need to know the gain of the transistors and the effective resistance seen at the compensation terminals, not including the values I put there. Anyone know those values?

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