What is a looking into resistance?

[muhandis]

What is a "looking into" resistance?

I'm studying transistor circuits at the moment and my textbook has mentioned the "resistance" looking into one of the terminals several times. Also mentioned is the resistance "seen" by a component. I've seen this in previous textbooks (especially with Thevenin) as well but just ignored it since it seemed I could do without it, but now it's just gotten to be too annoying.

They never explain what it means and just pop it in out of nowhere and it's difficult to understand (at least for me) from the context. I also searched the Internet but surprisingly there isn't a thing.

 

[Phrak]

This means that you pretend to cut the circuit in two. When I've seen this, it's always two nodes like ground and some other wire.

The two wires are connected to the terminals of the divice loaded by whatever you've cut-away. Now look at the effective resistance or impedence of the cut-way part. This is what the device 'sees' as it's load resistance, assuming it's driving into it.

I say 'assuming it's driving into it', because this is the usual problem you get in textbooks. You can also look into the driver--the other half, and calculate it's impedence.

 

[waht]

Basically, introduce a voltage source at the input of a circuit , and measure current flowing into it. The resistance is given by the ohm's law R = V/I

So that's the resistance as it appears at the input. If you connect some other circuit to the input, it would be the same as if it were connected to a resistor R.

Output impedance is similar. Short the output of a circuit, and measure voltage and current. R_out is V/I

Thevenin and Norton theorems go one step further, and can reduce a complicated circuit with gazillion of components (as long as it's linear) into a voltage or a current source with an appropriate output resistance. Which is better than having to worry about gazillions of components in a circuit.

 

[Phrak]

Basically, introduce a voltage source at the input of a circuit , and measure current flowing into it. The resistance is given by the ohm's law R = V/I

I think you're in danger of confusing the student. There is no 'input' until you define one.

 

[vk6kro]

See attached diagram.

If you were looking into the input of this amplifier, you would just see a resistor and a FET.
Humans can't see impedance.

But if you were a multimeter measuring ohms, you would see 100 K ohms in parallel with the almost infinite input resistance of the FET.

If you were an impedance bridge you would see this 100 K but you would also see a lot of Miller capacitance which at a high enough frequency would be enough to make this amplifer appear to be a low impedance input.

So, it is just an expression used as a way of explaining something.

 

[muhandis]

I understand it much more clearly now. Thank you very, very much for your help everyone, I really appreciate it.