input / output resistance/impedance questions:

My question might be so basic for you. But I believe I dont really get the fundamental understanding on the following:
1) what is input resistance/impedance? How do you measure it? Which is preferable for input? ( high or low )
2) what is output resistance/impedance? How do you measure it? Which is preferable for output? (high or low )

I hope you guys can help me.

In simple terms, lets take an op-amp, input resistance is the measured resistance between the input pin and ground, impedance, the same thing but measured with an AC signal, which will vary with frequency. Output resistance/impedance is the same thing but measured from the output terminal. It can be determined by driving a signal, say a 1K sine wave with no load, then applying a known load and using the difference in amplitude to calculate the output impedance, a simple resistor divider. So, at audio frequencies, to get good signal transfer, you need a low output impedance driving a high input impedance. Because they form a resistor divider network. The impedances are fixed by the circuit configuration. Most op-amps have very high input impedance and a low output impedance. A rule of thumb, for audio, is to have an input impedance at least 10 times the output impedance. But not too high, as high impedances are more prone to noise pickup, but not too low, low impedances are wastefull of power, between 22K and 47K for audio is about right. Things get a lot trickier at higher frequencies!

I have seen that impedance is a very important aspect in electronics. Quite simply it is the change in voltage over the change in current. Good luck with this loop because I have seen circuits that have feedback and the impedance is nothing but a large array of these things. Any time you have a transistor network such as an opamp notice the impedance will normally include at least two impedances that are based on this. So if you sit down and work it out you will find that the voltages have to jive with the base resistor because you can only have a certain change in current. And finally the impedance is what determines the gain.

To measure input/ output impedance

put all active sources to zero (short circuit voltage source , Open Circuit current source)

Put a Voltage source at the Input/output. Measure/calculate the current
then the impedance = V/I

High or low requirement depend upon the applications ..

1) what is input resistance/impedance? How do you measure it? Which is preferable for input? ( high or low )
2) what is output resistance/impedance? How do you measure it? Which is preferable for output? (high or low )

Click to expand...

All electronic circuits have some characteristic impedance. Impedance is the "AC resistance" of a circuit, and mathematically it is the sum of the resistance, the inductive reactance, and the capacitive reactance. It is frequency dependent because the inductive and capacitive reactance are functions of frequency.

Impedance is usually represented by the symbol Z; the formula is [Z=R+j(XL-Xc)]. This is a vector sum because it also takes into account the phase angle of the current through each of the circuit elements - that is the reason for the symbol j above: j represents the square root of -1 which is a vector mathematics operator.

The desired input impedance of a circuit stage is the same as the output impedance of the preceeding stage. For example, if you have a microphone with an ouput impedance of 1500ohms that is connected to the input of an amplifier, then the desired input impedance of the amplifier is also 1500ohms. The reason for matching impedance is to maximize the transfer of energy, and minimize energy reflections that can distort the signal.

The desired output impedance of a circuit stage is the same as the input impedance of the next circuit stage. For example, if you have 8ohm speakers connected to the output of an amplifier, the output impedance of the amplifier should be 8ohms. The reason is the same as above - to maximize energy transfer and minimize reflections that can distort the signal.

Ideally, every stage of a circuit is impedance matched to the output of the stage before and the input of the following stage.

Well House_Cat, you are half right. It is important to match impedances for very high frequencies and things like low impedance low output microphones for maximum power transfer, but at audio frequencies, things are a lot easier. It is just a simple resistor divider, so if you had a power amp with an 8ohm output impedance, swinging 28Volts RMS, into an 8ohm load, only 14 Volts would appear across the speakers, the damping factor would be 1, and half the power would be wasted in heat in the output stage! The output impedance for an amp needs to be as low as possible for maximum volts swing and maximum damping factor, (the ability of the output to control the load.)

It is just a simple resistor divider, so if you had a power amp with an 8ohm output impedance, swinging 28Volts RMS, into an 8ohm load, only 14 Volts would appear across the speakers, the damping factor would be 1, and half the power would be wasted in heat in the output stage! The output impedance for an amp needs to be as low as possible for maximum volts swing and maximum damping factor, (the ability of the output to control the load.)

Click to expand...

What is "just a simple resistor divider"? I don't know what kind of audio stages you design, but ever since the days of vacuum tube power amplifiers, impedance matching has been done with minimal loss circuits. In vacuum tube and some transistor amps the impedance match was done with a transformer. Some audio output stages are capacitively coupled. Some stages are direct drive - but ALL are impedance matched. Your confusion probably comes from the relatively new idea of defining impedance as "rated impedance" instead of true output impedance.

Not all audio stages are simple class A - some are class B, some, are class AB, and some are class D. My example was class A to keep it simple for a person new to electronics. Each type of amplification has its own output impedance matching methods and problems.

From your statement, you are mixing terms. OUTPUT IMPEDANCE is the actual impedance, RATED IMPEDANCE is the minimum impedance the amplifier can drive without overload, and DAMPING FACTOR is the ratio of actual impedances that gives an indication of the ability to damp mechanical ringing of the speaker. Mechanical ringing of the speaker cone occurs because of the inertia of the cone and the inductance of the speaker voice coil. There are a multitude of ways to damp ringing - filtering, accelerometers, mechanical resonance of the speaker and/or enclosure, etc. Most frequently, damping is accomplished by controling speaker size and mass, and controlling the air column in the enclosure - NOT by allowing the speaker inductance to kick back into the PA output, or by trying to overpower the ringing by overdriving the speaker.

hi guys,
i am a bit confused. my doubts are considering digital circuits (basic gates).
now do we need to make the output resistance of one stage equal to the input resistance of the next stage for maximum power transfer? or is that only applicable for audio(analog) applications?
as far as i recall we generally keep output resistance very low and input resistance very high(can someone help me understand why they do it?).
now can we say that the output and input are mismatched? will it cause any sort of reflections considering the output imped and input imped are not equal?
regards
Calvo

It isn't power transfer that you are concerned with in digital circuits. You match impedance in digital circuits to prevent reflections that can distort leading and/or trailing edges of digital pulses, and can cause false pulses through additive and subtractive effects.

In high frequency circuits, such as modern digital circuits, you want the load to absorb all of the EM signal traveling down the traces of a PCB from one stage to another. Digital circuits are actually broadband RF circuits - fast digital pulse risetimes can contain frequency components into the gigahertz range. If the energy is not transferred completely, you can get signal reflections. Maximum transfer of energy (load signal absorption) occurs when the source, transmission path, and load all have equal impedance.

You need to stop thinking in terms of only pure resistance. The only time you can ignore everything except resistance is with steady state DC. All electronic circuits have resistance, inductance, and capacitance. The vector sum of those three elements make up the impedance. Impedance is important anytime you work with AC signals - including digital signals.

A wire or PCB trace has inductance, and capacitance, along with a small resistance. Even though a PCB trace has only a few tenths of an ohm resistance, the PCB designer frequently will choose the width of the trace and the distance of the trace from the signal return path in such a way as to give 50, 75, or 100 ohms total signal impedance to the trace at the dominant frequency in use. The exact value of that impedence is matched to the output of one digital stage and the input of the next to keep the digital signal as clean as possible.

Hi ,
Let me mention one more thing. When i say digital circuits i am referring to ic design and mos transistors used to build gates hoping ur answer wasnt realted to board level.
why is it that we need to have low output impedance and high input impedance. how does it help.
and at the same tim u say that we need to match the impedance so as to prevent any reflections which can cause the digital signal to have shoots.

Reactive impedance is nothing more than ac resistance, that varies with frequency. No matter what fancy name you give it.
If I had an amplifier with an output impedance of 100ohms, and I tried to drive a load of 1ohm, clearly, a lot of current would flow, but very little voltage would be generated, conversly, if I tried to drive a load of 1000ohms, I would generate a lot of voltage, but very little current would flow. So, if my aim is to transfer maximum power, my ideal load would be 100ohms, a matched load, then I would get maximum current and voltage. When would I want to do this? if my amp was an arial and I wanted maximum power transfered to the air, which has a characteristic resistance of 377ohms.
Now we come to my small signal amplifier chain, My opamp is driving a stage with an input impedance of 10K, I wish to swing 10 volts into this stage, that is a current of 1mA. My opamp can drive 20 times this current with no distortion, my aim is for maximum voltage transfer, so I need a low output impedance driving a high input impedance because they form a simple resistor divider. If I made my output impedance 10K, I would half my voltage amplitude and increase the noise. Now, why would I want to do that?

Reactive impedance is nothing more than ac resistance, that varies with frequency. No matter what fancy name you give it.

Click to expand...

No, there is much more than just the "fancy name". Inductance and capacitance are energy storage devices. As they store energy in electromagnetic and electrostatic fields, they modify the relationship between the current and the voltage in the circuit. The signal reflection that has been mentioned several times is really the return of energy from those energy storage elements in the circuit. If you don't match the characteristics of the driver, the transmission path, and the load, you get energy returned back toward the source as the input signal level falls. That "reflected" energy wavefront is signal distortion because it happens after your origninal signal has begun to change and is added or subtracted to whatever part of the next cycle is coming from the source.

In the simple world of DC and audio, circuits are more forgiving. Your ear can't hear all of what happens to the signal, and you can't see that changes have been made to the real signals. If you care about the real world of electronics and physics, you can put a scope on the device and see that science works. It isn't all about getting the maximum voltage or maximum power all the time - you want the best results for the application. For example: In a digital circuit you want maximum integrity of the digital signal because all the information is encoded in that signal - you don't care as much about maximizing power. In an audio output stage you want maximum power delivered to the voice coil of the speaker because it takes power to move air - you don't care as much about perfect signal integrity (you can't hear it anyway).

You can ignore the real world and treat everything as "fancy names", or you can study a bit more and learn how to make everything you design work the way you want, and expect, it to work. The "fancy names" are there for a reason.

mohdfaaf said:

1) what is input resistance/impedance? How do you measure it? Which is preferable for input? ( high or low )

I hope you guys can help me.

Click to expand...

You will not be mad if I drop the term resistance?

Ok now depends on application and frequency.

a) In application depends if you would consider impedance of this wire.

What a) is also depend is frequency. In well matched circuit if you are talking for audio frequencies forget it find out what is Z0 characteristic impedance of wire you are using from manifacturer datasheet and write it in a piece of paper. That is all.

If you would consider frequency to be relatively high do this:

1) if load is matched to Z0 characteristic impedance of wire then Zin input impedance is Z0. Normally this is not the cases usually but if you have a matched load it is safe to assume it in circuit.

2) Take for example you run at 6GHz frequency. CHaracteristic impedance is 100ohms. And you have a wire 10meters. And let suppose you have a load of 20+j*10ohms.

Here is what you do: input impedance Zin = 20+j*10ohms when you measure at 2.5cm in the line. It is at 2.5cm it is at 5cm it is at 7.5cm, 10cm down the line and so on until you go up to 100m.

Wave will induce the same voltage accross the load same voltage and current throught it as the line is not there. This happens for every distance (n*llambda)/2 n=1,2,3,..infinit and is always safe to assume it. Then remove propable loss which adds to that value.

This is theoretically. So it can be done when you have a short/open circuit let say in stubs for example. 2 Fourmulas find them in RF books. Not worth to write them here.

"Technologically" you sweep in frequency using a network analyzer in the line and you do interpolation to values you measure from input to output. In waveguides you use slotline measuremeants.

After I read all the feedback from you guys, I get a better understanding but still a little bit confused. Anyway thanks.

When ppl talk about frequency... how high is considered as high freq and vice versa ?

If any of you guys can give me some links or documents which will help me to understand more I will appreciate it.

Thanks.

House_Cat said:

In high frequency circuits, such as modern digital circuits, you want the load to absorb all of the EM signal traveling down the traces of a PCB from one stage to another. Digital circuits are actually broadband RF circuits - fast digital pulse risetimes can contain frequency components into the gigahertz range. If the energy is not transferred completely, you can get signal reflections. Maximum transfer of energy (load signal absorption) occurs when the source, transmission path, and load all have equal impedance.

Click to expand...

One addition to complement this statement from House_Cat:

All elements have impedances, no matter input impedance is high at amplifier and output impedance must be low for purpose of amplifier itself only!

In distributed circuit elements this impedance discontinuity counts towards power requirements.
Bandwidth is Power.

mohdfaaf said:

When ppl talk about frequency... how high is considered as high freq and vice versa ?

Click to expand...

When talking how high consider this: how long? distance? RF considers a circuit operation as "an evolution" in space and time.

There is no hard rule. But a rule of thumb! Again also depends on application exactness of it.

llambda = (velocity of light)/frequency
(length of wire)/llambda > 0.01 is normally rule of thumb.

If this relation is true for circuit you are using then you better do matching preparations to implement or else bandwidth for example will suffer losses and degradation. You get fired from a Telecomm company if you did not do calculations right.

For Audio Applications do not worry for the above rule. Audio frequency circuits will get affected "approximately" if wires are more than 200meters.

Was this helpful?

I am referring in terms of circuit level in IC. I have interest in analog+digital cmos circuit. Unfortunately my background in design is not that good.

Well that’s a nice lecture on impedance. However my problem is a little different, when it comes to most circuits, calculating impedance is pretty easy, most of the time I just use the Pythagoras method. My problem is ( Zin on transistors). The input impedance on a small signal amplifier ( lets say a common emitter)What is it? (Ib/Ie) or (25*β/Ie) may be (Vb/Ib which equals Rb+Re) most likely none. If you do answer this Question some kind of logical formula will be nice and also is the impedance on all “two port Networks” calculated the same?

The input impedance of a bare transistor is the relatively low impedance of the base-emitter diode - a non-linear relationship that varies with the current. There is also lead inductance and junction capacitance; however, unless the frequency approaches UHF, they are negligible.

The input impedance of a small signal transistor amplifier is the Thevenin equivalent impedance of the base circuit which consists of the base resistor, anycoupling and bypass capacitors, and base biasing network (which may include circuitry in series with the emitter). We ignore the non-linearity of the base-emitter diode because it is biased in a more-or-less linear region for small signal amplifiers - it becomes a fixed resistance for the calculation. Again, at UHF or above, there will be parasitic inductance and capacitance that would also be considered.

Yes, the method of determining the impedance of all linear two-port networks is basically the same. See the class notes at the URL below:

**broken link removed**

Well I haven’t read your document but I saw the direction you were pointing me in (Circuit Analysis, Equivalent circuits and so forth) so I studied “Introductory Short Course on Electronic Circuits” it’s a document that I came across on the internet. Now I see why you didn’t just give me a simple formula. It’s like you said there are two many variables to take into account (Like bypass capacitors, emitter resistance, base biasing network …..). But it seems once you understand all of this the approach is only logical, bypass capacitors, biasing networks… they all boil down to the same thing, Base resistance and Emitter resistance as far as the input is concerned. This resistance comes in two forms Dynamic (small signal AC.) and static (large signal DC.). The static resistance takes into account the biasing networks, Emitter and base resistance. Dynamic Resistance accounts for the resistance resulting from conditions (and they are many conditions to give an example “configuration”). The Dynamic Resistance of the Circuit cannot be simply calculated because it’s not necessarily physical but is there because of consequence. However without getting into the mathematics let me say that they are formulas for calculating the Dynamic Resistance but it’s best to take into account the ways the formulas are derive (because most circuit aren’t the same). My conclusion, when calculating the input impedance of a transistor amplifier the biggest problem is finding the dynamic resistance.