# Homework Help: Input Vs. Output Impedance

1. Oct 29, 2012

### zoso335

I'm having a really hard time understanding input vs. output impedance and also the usefulness of the Thevenin model.

Let's say we have a simple voltage divider consisting of two equal resistors. The output voltage is between the resistors. If the input voltage is 10v the output voltage has to be 5v. What is the point of the Thevenin resistor then? I know that the Thevenin resistor is the output impedance of the circuit but I don't understand why. Also, for a random circuit, how do you calculate the input impedance and output impedance, and how does this relate to "looking into" a circuit?

2. Oct 29, 2012

### Staff: Mentor

Stick a load resistor across the voltage divider's "output". What's the voltage now?

3. Oct 29, 2012

### CWatters

It would only be 5V with nothing connected to the output. As Gneil said, what happens when you connect another resistor or perhaps another potential divider to the output. Try making 5V and 2.5V by chaining two potential dividers together. See what happens.

Perhaps some other examples would help...

Real world batteries are not "ideal" voltage sources. You can't draw infinite current without affecting the voltage because they have a built in output impedance (although it's normally called the internal resistance) due to things like the chemistry of the battery cell.

Real world volt meters are not ideal meters either. They have an input impedance that's not infinite. When you connect them to a circuit they affect the voltage they are trying to measure. Although some modern meters do have a very high input impedance (10MOhms or more) so the effect of the meter on the circuit can sometimes be ignored).

4. Oct 29, 2012

### zoso335

If you connect a load resistor, you would calculate the equivalent resistance and then from that calculate the output voltage. It will be greater than 5v. And Cwatters, I don't get why one of those examples is input impedance and the other is output. Both have some fixed resistance which causes a voltage drop across them, so why is one called output impedance and one input?

5. Oct 29, 2012

### SunnyBoyNY

Output impedance describes output voltage behavior depending on the output current.

Input impedance describes input voltage behavior depending on the input current.

In most cases the impedance could be modeled as a pure resistor.

In other words, in a system where ones only charges a battery, the battery ESR is the input impedance. In a system where ones only discharges a battery, the battery ESR is the output impedance. Thus, a battery has a symmetrical Z-matrix (http://en.wikipedia.org/wiki/Impedance_parameters#The_Z-parameter_matrix).

A typical opamp draws no current and thus has infinite input impedance (no amount of input voltage causes input current) and has zero output impedance (output voltage stays the same regardless of output current).

6. Oct 30, 2012

### aralbrec

As mentioned, input impedance happens where you decide the input is and output impedance happens where you decide the output is. It's just a name but maybe it's easier to understand in terms of an amplifier.

Real amplifiers are not ideal elements. They load the input source and cannot deliver output that is not affected by whatever is attached to it. If we model the amplifier as a box, it has a two-wire input port and a two-wire output port. At the input we call the Thevenin resistance of the amplifier the input impedance. It affects whatever we use to drive the amplifier at the amplifier input. It's the impedance seen looking into the input of the amplifier. At the output, we may attach speakers. The amplifier behaves like a voltage source in series with a thevenin impedance at the output. That impedance is called the output impedance because it is at the output of the amplifier. You see, it's all about what direction signals are travelling.

A diagram is attached below.

Some uses from that diagram:

The input source could be a microphone. Microphones are very weak signal generators -- they act like small voltage sources connected to large output resistances. So Vs is tiny and Rs is large (note I am using a thevenin model for the microphone; Rs is called the output impedance of the microphone because it appears at the microphone output). If the input impedance of the amplifier Rin is small, the amplifier sees an even tinier signal because Rin/(Rin+Rs) is small. This signal would be swamped by noise. So you can see the thevenin resistances at the input summarize how well the microphone can drive signals and how badly the amp loads an attached driver.

At the output, speakers are attached. This is where a lot of power is consumed to turn electrical energy into acoustical energy. You want to do this efficiently. With a little calculus, you can determine that energy is transferred to the speaker at maximum efficiency if the output impedance of the amplifier is equal to the input impedance of the speaker. So you may want to add circuit elements to adjust the output impedance of the amplifier to get that max efficiency.

In the real world, devices don't behave as resistors -- their behaviour is frequency dependent and this means an accurate model must include capacitors and inductors. Some of these will be added deliberately to get desired behaviour but most will be parasitic, making our perfect circuits do bad things. At the output of the amplifer for example, the speaker may have poor frequency response at the low end. We model the speaker as a high pass filter and its thevenin impedance will be complex (as in s-domain). We can perhaps compensate for that in the amplifier by emphasizing bass (the equalizer) but how much? Well all that is summarized in the speaker's input impedance.

Here's another one. Your circuit analysis assumes currents and voltages appear instantly at all points in the circuit when they change. That, of course, doesn't happen in real life because EM fields are limited by the speed of light. It turns out if the frequencies your circuit operates at have a wavelength approaching the dimensions of the circuit, the lumped parameter model (what you assume when voltages appear instantly) no longer applies. You then have to analyze the circuit taking the speed of light into account. This effect can especially be seen in long wires, for example transmission lines, ethernet cables, usb cables, etc. It takes time for the signal to propagate from point A to point B. It turns out that if the impedance seen by the wave at the end of the cable does not match impedance of the cable where the wave is coming from, reflections occur. So this means you have to make the impedance looking into your receiver circuit match the impedance of the source cable.

Another one. In designing analog filters, we are normally doing so from a desired transfer function. If the transfer function is written in a certain form, we can implement it is a ladder network (series L, parallel C, series L, parallel C, etc). We decompose that transfer function by extracting the series L (T(s) = sL + T'(s)). We can then see the remaining impedance looking into the rest of the circuit is a capacitor in parallel with soemthing else, etc.

There are so many applications. You just haven't seen anything complicated yet.

Suppose Thevenin is true, then the circuit can be summarized by:

+------- Rth -------+ A
|
Vth
|
+-------------------+ B

If you measure the voltage across A,B it will equal Vth. If you set Vth to zero, the resistance seen looking into the circuit will be Rth. If you attach a test voltage to AB and measure the current flowing in at A, you can calculate the resistance.

Setting Vth to zero means shorting independent voltage sources and opening independent current sources. This comes from the superposition principle.

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Last edited: Oct 30, 2012
7. Oct 30, 2012

### CWatters

It will actually be less than 5V but that's another matter.

Suppose the curcuit was more complicated than a simple voltage divider and the load varied a lot. Perhaps the circuit is a network of 100 or more resistors. Instead of working out the equivalent resistance for the whole circuit each time the load changed you could model the circuit it as a 5V ideal voltage source plus a series output resistance. That would make it much simpler to work out the actual output voltage when the load changed. That resistor is called the output impedance.

Think of it from a systems perspective..

The role of a battery is normally to deliver power to some other part of the circuit. So it's terminals are generally considered to be outputs rather than inputs. (Althought I agree current flows into one of them!).

If we were talking about a rechargable battery then when the battery is on charge it might well make sense to refer to the battery as having an input impedance (even though current comes out of one terminal!).

A volt meters terminals are generally considered inputs. They take a voltage signal and, after processing, feed it to the meter display where it is output for you to read. A simple volt meter never feeds a signal back into the circuit.