# Impedance question

Status
Not open for further replies.

#### throbscottle

##### Well-Known Member
Am I correct in thinking that if I have a signal source and I want to determine it's impedance, I can do so by connecting a variable resistance to it and adjust it so the voltage is half the o/c voltage, and the resistor will now equal the source impedance?

#### OBW0549

##### Active Member
Yes, but you don't have to load the signal source that heavily. Any load resistance that causes a measurable reduction in the signal amplitude will do; just treat the load resistor and the unknown impedance as a voltage divider, and run the math.

#### unclejed613

##### Well-Known Member
yes. that's correct. there's an indirect way of doing it where you connect a resistor across the output, and measure the change in voltage. the formula is Zs=Rl*((v1/v2)-1) where Zs is the source impedance, Rl is the resistor connected to load the source, v1 is the open circuit voltage, and v2 is the loaded voltage.

#### anhnha

##### Member
Yes, but you don't have to load the signal source that heavily. Any load resistance that causes a measurable reduction in the signal amplitude will do; just treat the load resistor and the unknown impedance as a voltage divider, and run the math.
I believe your method is connecting a small load to the signal source and then use the voltage divider formula to find out the source resistance.
This is not the way the OP meant to do. He didn't do any math just adjust the load resistance and observe when the output voltage is haft of the source. At that time the load resistance is equal to the source resistance.

#### crutschow

##### Well-Known Member
That works okay if the source is designed to tolerate a load equal to its source impedance.

But don't try it on a modern audio amp.

#### throbscottle

##### Well-Known Member
I just figured my way was simpler! I actually figured it out after working though a few simple examples in my head (using knife-and-fork values as my maths isn't very good!)
I actually want to know because I need the source impedance for a L/P filter that will go between my OCXO and a gang of 4 op-amp inputs (simple distribution amp) to make sure there's no digital noises sneaking back down the line.
Now it'a good point about what the source is designed to tolerate and it's a complete unknown. It's a used Trimble OCXO with no data available, and it looks like it's going to be in the region of 200R. But maybe these things are designed to operate as voltage sources?
I'm reluctant to include another O/A just to drive the filter, I'd sooner just miss the filter out.

#### unclejed613

##### Well-Known Member
That works okay if the source is designed to tolerate a load equal to its source impedance.

But don't try it on a modern audio amp.
yeah, true. most modern audio amps have an output impedance measured in milliohms.

there was a time when input and output impedances for audio gear was an important factor. the standard impedance for audio gear was 600 ohms. as solid state devices and especially op amps were introduced in audio equipment, it became more important to "voltage match" audio equipment, because you would have equipment with 50k or higher input impedances, and low (from about 20 to 100 ohms) output impedances. so, the term "line level" no longer means 6 milliwatts into a 600 ohm load (0db in a 600 ohm system), but now is interpreted as 1.5Vrms (0db in a modern audio system). that definition isn't really a very strict one either, as i've seen "line level" interpreted by various manufacturers in different ways. one place where that definition is held to is in audio power amplifiers. most modern amplifiers have their gain set so that 1.5Vrms at the input drives the amp to just barely below clipping.

#### throbscottle

##### Well-Known Member
Well, you learn something new every day!
My OCXO output turned out to be 50 ohms BTW, so the filter has nice values.

#### dr pepper

##### Well-Known Member
At 10 mc the load impedance is reasonably important for a clean signal, if the load is 50 ohm the trimble may well o/p a clean enough signal.

#### Grossel

##### Well-Known Member
With the use of a resistor for a given frequens, you'll get the absolute value of the impedance.

Meaning - if a capacitor is part of the current path, then impedance will vary with the frequenz, and also different load types (the reactance part) will have impact on your voltage reading, which in return may trick you to do the math wrong if assuming output impedance only have a resitive part.

#### Diver300

##### Well-Known Member
A lot of devices that are designed to drive transmission lines have their output resistance designed to be approximately equal to the characteristic impedance of the transmission line.

That is done so that if there is any return signal travelling the wrong way, it will be absorbed by the output resistance and not reflected.

#### throbscottle

##### Well-Known Member
Ok. So the next stage after the L/P filter (CLC Chebyshev) is 4 op-amps with parallel inputs (as a simple distribution amp), each input is 200R. Am I better to give each amp it's own coupling cap, or should I connect them directly together and use a shared coupling cap? Separate caps have much less impact on impedance but is there some other effect to watch out for, especially considering it's fed from a filter?

#### unclejed613

##### Well-Known Member
are you feeding inverting or noninverting inputs? it makes a difference because noninverting inputs are high impedance, but the inverting input impedance is equal to the input resistor (the inverting input is a virtual ground).

#### throbscottle

##### Well-Known Member
Inverting. The non-inverting inputs are tied to a supply splitter. It seemed the best way...
Thinking about it, separate caps do have one advantage - if something pulls the output of one of the amps it will stop it affecting the others. The outputs are DC coupled.
Thinking some more about it, maybe it would be better if the inputs were DC coupled and put the caps on the output, since as it is the output of each amp will be sitting at around 2.5vdc. Then just use fairly large caps like 220n or something. So the filter will then have a purely resistive load.

#### dr pepper

##### Well-Known Member
If you dc couple the osc to the op amps you need to be certain there is either no dc on the o/p of the osc, or if there is some dc that it wont mess up the op amps, if the filters have gain then they'll amplify the dc possibly causing clipping if the amp runs out of voltage swing.

#### throbscottle

##### Well-Known Member
It's just a 3rd order Chebyshev filter to keep out any noises from the digital side Good point about DC from the oscillator. So I need caps at both ends. I must have thought of it originally and then forgotten when I started asking questions here...Maybe it should be an input cap to the filter instead.

Last edited:
Status
Not open for further replies.