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Voltage Divider Bias BJT clarifications

electronicsLearner77

Jul 2, 2015
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I am studying the voltage divider biasing of a transistor. I came across few terms which i do not understand."When the emitter resistor is viewed from the base circuit, the resistor appears to be larger than its actual value because of the dc current gain in the transistor. That is RIN(Base) = VB/IB"
What is this RIN(Base)? why is this new term defined? I thought that the resistance is only RE which is visible.
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Audioguru

Sep 24, 2016
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A transistor has a base to emitter impedance as shown on the datasheet of a 2N3904 transistor. When an emitter resistor is added then the hfe gain of the transistor increases the effective value of the emitter resistance which increases the transistor input impedance since it is in series with the base-emitter impedance.
It affects the input impedance, not the voltage divider biasing.
 

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electronicsLearner77

Jul 2, 2015
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Thank you for the reply. Ok now i understand that, when the book refers to the initial transistor characteristics, it never refers to the internal resistance of the transistor, only when voltage amplification is defined, the internal resistance is defined. So, i assume it is when an AC signal is applied to the transistor then i have to consider re'? In case of DC voltage the emitter resistance becomes 0? The other strange thing is there is no mention of 0.7V in voltage amplification.Why?
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Audioguru

Sep 24, 2016
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In the case of DC voltage, the emitter resistor has the collector current plus the tiny base current in it. Ohm's Law says that the voltage across the emitter resistance is the emitter resistor value times the current in it.

The emitter voltage will be about 0.65V lower than the base voltage.
The voltage divider sets the base voltage so be so that the emitter resistor has your desired DC voltage and the collector has about half the remaining supply voltage across the collector resistor.

If the transistor is used for audio and has no emitter resistor or has a bypass capacitor parallel with the emitter resistor then the gain will be high but the distortion will be extreme. The top of the waveform is squashed.
I show a transistor with high voltage gain, then with its emitter resistor unbypassed for some negative feedback to reduce the gain and reduce the distortion:
 

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