How to Find Input Resistance?

[k_noe]

Hello, I have a question about how to find the input resistance of an transistor amplifier of the first circuit below. It says to make an equivalent circuit shown in the second circuit so that the input resistance is R1 in parallel with R2. So, the book I have tells me that to an ac signal at the input, the power supply acts like a short to ground. Why is this so? I understand the role of R2 but why would R1 be a path to ground? Wouldn't a signal appearing at the input travel to the ground through R2 and ignore R1 completely so that the input resistance is just R2 only? This is confusing me, appreciate it if anyone can help. Thanks

 

[Nigel Goodwin]

In almost every circuit, there are decoupling capacitors from +ve to ground - this means that your second circuit is pretty well correct, and R1 and R2 are effectively in parallel. The transistor is also in parallel with them as well, which rather ruins the nice simple calculation though

Personally I would simply measure the input impedance, it's reasonably simple to do - just requiring a signal generator, an AC millivoltmeter (or oscilloscope), and a resistor.

 

[Russlk]

If you replace the power supoly by its output impedance, probably .01 ohms, you will see why R1 and R2 are in parallel. The input impedance of Q1 is in parallel also, you can't ignore that. You should compute the bias voltage on Q1, the impedance could be very low.

 

[Josh]

you could put a pot in parrallel with the input, and adjust it until the output signal is half. then disconnect the pot and measure it. this will be equal to the input resistance.

Josh.

 

[k_noe]

I forgot to take the power supply decoupling capacitor into account. That would provide a path to ground for ac signals. Thanks for the replies.

 

[Roff]

you could put a pot in parrallel with the input, and adjust it until the output signal is half. then disconnect the pot and measure it. this will be equal to the input resistance.

Josh.

This only works if the impedance of your signal source is infinite, i.e., a current source.

Put the pot in series (AC coupled, to avoid upsetting the bias conditions). When the output amplitude is half the original, the input resistance is equal to the value of the pot plus the source resistance of the signal generator.

This only works at low (audio) frequencies. Stray reactances (mostly capacitive) will cause erroneous results at higher frequencies.

 

[Roff]

I forgot to take the power supply decoupling capacitor into account. That would provide a path to ground for ac signals. Thanks for the replies.

A true voltage source, such as an ideal battery, has zero impedance. The decoupling capacitors are primarily used to minimize the effects of stray inductances, resistances, and battery or power supply internal resistances (which are pretty small, but can be significant with high load currents).

 

[Nigel Goodwin]

Put the pot in series (AC coupled, to avoid upsetting the bias conditions). When the output amplitude is half the original, the input resistance is equal to the value of the pot plus the source resistance of the signal generator.

That's pretty well the way I do it, sometimes I use a series pot, other times a series resistor - where you can simply calculate the impedance, rather than measuring the pot.

However, where I vary slightly is to measure the signal either end of the series resistor (rather then the output of the amplifier), this takes account of the output impedance of the signal generator. Obviously you 'may' have a problem with the impedance of the measuring device for a very high input impedance amplifier, but a x10 probe is usually plenty high enough, if not use an extra resistor to feed the scope input.

I think it's important to bear in mind that accuracy isn't very important in a case like this anyway, all you really need to know if it's 1K, 10K or 100K, you don't generally need to know if it's 1K or 1.1K.