# Impedance Measurement Point

I've been doing some work with interstage impedance and I keep getting confused and I can't seem to get the correct answer myself. I know Input Impedance should be 10x the amount of the previous output stage, but I'm not sure where measurements should be taken from.

In the attached image, I’ve got 10 Ω coming out of the previous stage, 100 MΩ into the next input stage and an equivalent resistance (pot at 50%, linear taper) of 2.5k || 1M = 2.93k. Now the query is, does the 2.93k get factored into both input and output impedance or only for the output calculation?

Zout = 10 + 2.93k

Zin = 2.93k || 100M

or

Zout = 10

Zin = 2.93k || 100M

I've seen the source impedance also calculated from the output of the potentiometer. I can't seem to work out where one calculation ends and another begins.

• NB : (2.5k || 1 Meg) < 2.5k ... (=2.494 k) Feb 9 at 7:49
• Yes but isn't it calculated as 5k in parallel with 5k? Feb 9 at 9:06

We'll call the top and bottom resistances of the pot $$\R_{top}\$$ and $$\R_{bottom}\$$, since we can't tell what they're going to be at any given time.

The impedance that the source sees looking into the rest of the circuit will be

$$R_{top} + (R_{bottom}~||~1M\Omega)~||~R_L$$

The impedance the load sees looking into the rest of the circuit will be

$$1M\Omega~||~ (R_{bottom}~||~(R_{top}+R_S))$$

To plug in some numbers, assuming the pot is set at 50%, the source will see an impedance of:

$$5k + (5k~||~1M)~||~100M = 9974.88\Omega$$

Very close to the impedance of the pot.

The load will see: $$1M~||~ (5k~||~(5k+10)) = 2496.25\Omega$$ Around 1/4 the impedance of the pot.

With a 1 V signal the voltage across the load will be: $$1V*\frac{(R_{bottom}~||~1M\Omega)~||~R_L}{(R_{top}+R_S)+((R_{bottom}~||~1M\Omega)~||~R_L)}=1V*\frac{4974.88\Omega}{5010\Omega+4974.88\Omega}=0.4982V$$ So just a little under half the voltage with the pot set halfway, which would suggest that the impedances aren't causing any undue loading effects.

• Thankyou! That's exactly what I needed to confirm. Feb 9 at 20:58

I've seen the source impedance also calculated from the output of the potentiometer. I can't seem to work out where one calculation ends and another begins.

Nor can we. If this is an assignment, then that should give which value shall be calculated.

In general, the impedance is a property of a specific node in a circuit, and you can determine it like this: simulate this circuit – Schematic created using CircuitLab

You inject current into the node of interest and then observe its voltage change: $$\Z_\text A=\frac{\text dV_\text A}{\text dI1}\$$. It is a small-signal quantity and in general varies with the frequency of the injected current.

In your circuit, I assume you want to prevent any signal voltage attenuation, so one minimizes source impedance and maximizes receiver impedance. One should observe that the impedance at each node remains almost unaffected if you connect anything to the right of it (speaking in terms of your schematic). So the impedance of node A shouldn't differ if you connect the center box, and the impedance of node B shouldn't change if you further connect the right box. • That's the application! I assume since anything to the right is generally not affected then it's safe to say that Zout would stay roughly the same while Zin would take the centre circuit into account? Feb 9 at 9:34
• @DylanBenton I don't fully understand your comment. Zin and Zout are both "given" if I understand well, as 100M and 10R, respectively. But the impedance of nodes (A) and (B) will change depending if you connect only their respective left or right or both sides' circuits to them. Feb 9 at 12:37
• Sorry, maybe it was just how I had it drawn. Node A and B are connected to the Zout and Zin, I just trying to show that I had an output z an input z and a circuit in-between. If all these nodes were connected, I was wondering how you'd work out the ratio between output z and input. Feb 9 at 14:07