Difficult, understanding Input and output impedance - Page 1
Forum section: electronics / beginners

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 01:12:24 am
content: Hi,It's a bit difficult for me to understand the concept of input and output impedance. Iknow input and output impedances are thevenin equivalents. I have a picture attached. Is the input impedance of the voltage divider seen from Vcc, R1+R2? And is the output impedance seen from the load, R1 in parallel wit R2 (R1//R2)?

author: Audioguru again
author_rank: Regular Contributor
timestamp: August 18, 2019, 01:20:15 am
content: Quote from: robsims on August 18, 2019, 01:12:24 amNo, the input impedance is not R1 + R2 because R2 has the load in parallel with it.Yes, the output impedance is R1//R2 because the Vcc power supply usually has a very low impedance.

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 01:30:48 am
content: Ok with the load, the input impedance is (Zload//R2) +R1. Is that correct?

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 02:34:53 am
content: What you stated "Audioguru again" is not true. Just watched a video:Input impedance is R1 + R2Output impedance is R1//R2I think that's it

author: vk6zgo
author_rank: Super Contributor
timestamp: August 18, 2019, 04:40:47 am
content: Audioguruhas included the load, as per your diagram--- the video talks about the input impedance (actually resistance in this case) without the load connected.Try it experimentally---- build up a voltage divider as shown on the video .Place your DMM across the input terminals.--- you will now see R1+R2 in series as your reading.Now place another resistor (let's call it R3) in parallel with R2 (a "load").If we haven't shifted in time & space to another dimension, your DMM will now read a value equal to R1+R2//R3.Now move R3 so it is now in parallel with R1.You will now read a resistance value equal to R1//R3+R2.I might lie to you,Audiogurumight lie to you, the Professor's video might lie to you, but a real physical circuit measured with a DMM Ohms range is very, very, unlikely to do so.

author: fourfathom
author_rank: Super Contributor
timestamp: August 18, 2019, 04:44:45 am
content: Audioguru is correct. The load is part of the circuit, and has to be considered when calculating input impedance. It is R1 + (R2 || Rload).

author: magic
author_rank: Super Contributor
timestamp: August 18, 2019, 06:55:13 am
content: Imagine what happens when R1=1Ω, R2=1kΩ and Rload=1Ω.Is the input impedance closer to 2Ω or to 1kΩ?

author: Audioguru again
author_rank: Regular Contributor
timestamp: August 18, 2019, 12:33:58 pm
content: The video shows an actual amplifier circuit (the box) but the OP's circuit has no amplifier.

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 12:59:02 pm
content: Wasn't my intention to be rude to Audioguru. Sorry for that. The video doesn't show an amplifier. It's just an ordinary voltage divider. The box with input and output is the voltage divider. Everything has an input and an output. A simple resistor also has an input and output. Ok but we all agree on the input impedance. That is the impedance that the voltage source sees. You decouple the source and calculate the impedance of the circuit, the source sees. In this case the impedance the voltage source sees is R1 + R2But do you really include the impedance of the load when calculating the output impedance of a circuit. In my opinion, the output impedance is the impedance what the load sees. You have to decouple the load, to calculate the output impedance of the circuit.Theory says that for maximum output, the output impedance of a circuit must match the impedance of the load. From a math point of view, this means that if you include the impedance of the load in the output impedance of a circuit, you can never match the output impedance of that circuit with the impedance of the load. That's also why the professor in the video didn't include the load in the output impedance of the voltage divider.Am i correct???

author: Audioguru again
author_rank: Regular Contributor
timestamp: August 18, 2019, 01:08:55 pm
content: The output impedance of an amplifier or any other REAL circuit has nothing to do with its input impedance. This circuit is unreal.

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 01:15:08 pm
content: Of course the output impedance has nothing to do with input impedance, but that's not my point now. My point is do you include the impedance of the load when calculating the output impedance of a circuit? I want an answer based on theory. Refer to a vid or book i can read?

author: StillTrying
author_rank: Super Contributor
timestamp: August 18, 2019, 01:44:23 pm
content: Quote from: robsims on August 18, 2019, 01:15:08 pmNo you assume the load is not there at all.1st few minutes of this vid looked OK.

author: rstofer
author_rank: Super Contributor
timestamp: August 18, 2019, 01:53:43 pm
content: You don't usually include the load (which is often application specific) in the output impedance calculation. The circuit would be considered a 3 terminal 'black box' (see bottom link).It is important to know that maximum power transfer occurs when the output impedance and the load impedance are equal. For the idea to make any sense (and it's an important concept), the output impedance must not include the load impedance.I wouldn't include the load in the input impedance calculation either simply because the load impedance isn't always known or is likely to vary with application.https://www.electronics-tutorials.ws/dccircuits/dcp_9.htmlProblem 1 here:http://pleclair.ua.edu/ph102/Homework/Sum09/HW5_21Jul09/HW5_magn_21Jul09_SOLN.pdf

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 02:16:22 pm
content: Very nice rstofer. That's what i thought. The professor is right too. You gave an answer based on theory and logic. "OUTPUT IMPEDANCE MUST NOT INCLUDE THE LOAD IMPEDANCE". See attached picture. This is from the book: Grob's Basic Electronics, Mitchel E. Schultz, 11th Edition, page 920. From a math point of view it's impossible to include he load impedance when matching output and load impedances. I think we all learned again. Many thanks

author: magic
author_rank: Super Contributor
timestamp: August 18, 2019, 03:07:10 pm
content: Output impedance includes impedance of the source.*Input impedance includes impedance of the load.Not the other way around.Simple*Yes, that Vcc thing you have drawn will have some output resistance in practice, which adds to R1.

author: fourfathom
author_rank: Super Contributor
timestamp: August 18, 2019, 03:20:15 pm
content: Quote from: rstofer on August 18, 2019, 01:53:43 pmThe OP's circuit is a passive attenuator. In a passive circuit you absolutely do include the load impedance when calculating the input impedance. If you don't know the load impedance then neither the input impedance nor the attenuation can be determined.With an amplifier in the circuit then things may be different. In some RF designs (for example) the load impedance is reflected in the input impedance, even with an amplifier between input and output. In these cases you do need to include the load when calculating input impedance. In other cases the amplifier provides essentially perfect isolation so the load has no effect in input impedance

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 03:21:19 pm
content: Magic, of course Vcc has output resistance. That's the internal resistance of Vcc which is very low for a good Voltage source.

author: rstofer
author_rank: Super Contributor
timestamp: August 18, 2019, 04:14:22 pm
content: As long as you state whether the impedances calculated include a load or some source impedance, neither being a part of the 'black box', you're still good. In a practical world, of course the load is important. In a classroom, maybe not. It depends on what is being taught.If you have the 3 terminal characteristics, it is easy enough to add in source impedance and load impedance. The important thing in a classroom is to get the answer the instructor wants you to get. Their game, their rules.

author: chrisl
author_rank: Regular Contributor
timestamp: August 18, 2019, 06:18:12 pm
content: The input impedance is what the source sees (in this case the Vcc) and includes everything after the source.The output impedance is what the load sees and it includes everything right before the load.

author: GerryR
author_rank: Frequent Contributor
timestamp: August 18, 2019, 07:38:23 pm
content: I think if the OP would have drawn the circuit as a "Two-Terminal-Pair Network" it would have been a little clearer. See the attachment from R. E. Scott's. Linear Circuits.

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 08:34:00 pm
content: Seems like everyone has a different understanding about calculating input and output impedance's. It doesn't matter how the circuit is drawn. There are always two input terminals and two output terminals. Even for a single resistor (see attached picture). Do you decouple the source and the load when calculating input and output impedance's. What is the right answer?

author: fourfathom
author_rank: Super Contributor
timestamp: August 18, 2019, 08:42:43 pm
content: Quote from: robsims on August 18, 2019, 08:34:00 pmRinput = R + RLRoutput = R + RiJust that simple.

author: GerryR
author_rank: Frequent Contributor
timestamp: August 18, 2019, 09:14:13 pm
content: If you draw a box around R1 and R2 in your initial drawing, you have your answer. When trying to load-match the output impedance to the load, you can't include the load in the calculation of the output impedance. Look at it this way, many instruments have a 50 ohm output impedance and tell you to match that with your load. The load isn't connected to the instrument until you put it there, so it must have a 50 ohm impedance before you connect the load. (That is how you get maximum power transfer for an output device (frequency generator, etc.) and the way they spec it to give outputs to a particular value.) If your load is different than the output impedance of the instrument, say in this case a frequency gen, the output will vary from spec. (Of course, there are cases where you want the "load" impedance to be much higher than the output of the device, e.g. a voltmeter.)

author: robsims
author_rank: Frequent Contributor
timestamp: August 18, 2019, 09:20:58 pm
content: Many thanks GerryR. That's the same rstofer mentioned earlier and that's what i also thought. Again, the professor in the vid i posted earlier is right. So i think this will be the end of discussion (EoD). Everything is crystal clear now.