# common base amplifier

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#### teachlit

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I am interested in a circuit which will amplify RF in the first stage of a Kitchin regenerative receiver which will receive frequencies between 4 and 7.5 mHz. The circuit I have uses this circuit for stage one. The top of the L1 inductor attaches to a nine volt power source. The design requires a low impedance input for this stage. I have limited knowledge of impedance and would appreciate it very much if someone can show me how to calculate the input impedance of this stage. I am sure someone out there can help. thanks in advance.

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This configuration is most often used when only a very low source voltage is available (1.5 volts or less). You might want to put a voltage divider on the base to allow the transistor to operate a little further outside of saturation. That said...

With a supply of 9.0 volts, you will have 9.0 volts on the base and about 8.4 volts on the emitter. Ie = 8.4/2.2K = 3.8 mA. The transistors effective AC impedance of the base emitter junction is 26mV/Ie = 26 mV/3.8 mA = 6.84 ohms. Since the base is AC grounded through C2, that makes the input impedance about 6.84 ohms in parallel with 2200 ohms = 6.8 ohms.

Adding that resistor voltage divider to the base and setting the base voltage to about 1.7 volts would give you an emitter current of 0.5 mA and an input impedance of near 50 ohms.

With a supply of 9.0 volts, you will have 9.0 volts on the base and about 8.4 volts on the emitter. Ie = 8.4/2.2K = 3.8 mA.
Sure, this is because the threshold on base to emitter is overcome at 0.6 volts, which leaves 8.4 volts at the emitter. I follow this okay.
The transistors effective AC impedance of the base emitter junction is 26mV/Ie = 26 mV/3.8 mA = 6.84 ohms.
This part is not clear to me. What is the 26 mV?

The effective AC resistance of the B-E junction varies with the amount of emitter current. Not sure where it is derived from, but the formula normally used is
r'e = 26mV/emitter current in mA. So if the transistor is biased to 1 mA of emitter current, the AC B-E resistance is 26 mV/1 mA = 26 ohms. If the emitter current is 10 mA, the AC resistance is 26 mV/10 mA = 2.6 ohms.

Found it, if you are good with math....

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Found it, if you are good with math....
Thanks for the link. I wasn't able to do the maths, but thanks for the effort you put in to find it. Maybe one day...
I see you computed the parallel resistance of the emitter resistor and base-emitter reactance with the straight-forward parallel resistance formula. But wait a moment, are these two components parallel to each other?

Yes. For AC purposes, the 2.2K resistor goes directly to ground. R'e is the equivalent B-E reistance and the base is bypassed to ground through C2, so for AC purposes, r'e and the 2.2K are directly in parallel.

Yes. For AC purposes, the 2.2K resistor goes directly to ground. R'e is the equivalent B-E reistance and the base is bypassed to ground through C2, so for AC purposes, r'e and the 2.2K are directly in parallel.
Thank you. This has made it very clear to me. My circuit has the low input impedance I wanted and I have made a bit of progress in my learning about radio - and particularly common base amplifiers. One question: would this same formula - 26mV/Ie - work on a common emitter amp?

Yes. It would appear as an equivalent AC resistance in series with the emitter.

Thanks. Learning radio is taking me quite a bit of time but each step like this one takes me further than I was when I started.

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