input impedance output impedance

If an input is "high impedance" it means that is easy to "drive" i.e. easy to push around, in the sense that powered steering is easy to handle, you don't need much muscle power to move it.

So generally speaking, inputs will tend to be high impedance because then they will have more tendency to do what they are told as opposed to a lower impedance input e.g. unpowered steering.

A "high impedance" output is a "weak" output, a weak drive. If you used a high impedance output to drive a low impedance input (old man trying to turn non-powered steering) then the movement transferred will be reduced or to use the jargon, the ouput will be loaded by the input, and subsequently the signal will be loaded and reduced in size.

So it is usual to have e.g. the output of a HiFi amplifer (0.0001 Ohms) drive a loudspeaker of 4 Ohms. By this design, the loudspeaker will do exactly as it is told being driven by an impedance very much lower than its own.

Now a big confusion is "matched impedances". When a strong guy turns powered steering, he does not expect to get tired and he won't because his low impedance is not loaded by the high impedance reaction of the wheel. There is no intention to depend on the man's strength. The power is provided by the petrol. But what if there is no petrol? As in e.g. in a bicycle now it is important to get *power* from the man to the machine. When the road speed is low, a gear selection is required to match the maximum output of man to machine i.e. the impedances are "matched" but as he road speed increases so the peddle speed reaches a maximum that the man can manage and no power is transferred (the impedances have become mis-matched) so a new gear is selected to match the impedances again, to reduce the pedal speed to the "power zone" of the man and so on, up through the gears.

Under certain circumstances it is necessary to match impedances to get the maximum power from input to output. Different requirements might need maximum voltage transfer (low driving high) while others require maximum current transfer ( high driving low ).

Cheers Robin

The "ideal" opamp would be one with infinite input impedance and zero output impedance. There is no intrinsic difference between inverting and non inverting inputs on the integrated circuit - if you look at the data sheets and the internal circuit you will see that both inputs are identical - same type of differential amplifier.

Op amps have very high input impedance - or to state it another way - very little power is required to impress a signal on the input.

Op amps have relatively low output impedance. They have the ability to transfer more power to the output load than the input consumes. Some "power" op amps can even drive loads like loudspeakers or motors directly.

In a circuit there will/may be some impedance difference due to the gain setting resistors and capacitors - so you'd have to be more specific with your question if that's the case.

"Impedance matching," in the context of op amps, is harder to grasp. Typically, with a series of op amps used to amplify the same signal, you have a low impedance output driving the high impedance input of the next stage . . . a mismatch but desirable in that case since the input won't overload and change the signal of the output stage driving it.

Matching an output stage to a load is more critical, but even there, with low frequency signals the output stage will be lower than the load as a rule.

You couldn't drive a low impedance load with a high impedance output - most of the available power is dissipated in the op amp as heat, the load will see little power.

A little ohms law will show you how power transfers between a source and load. Arbitrarily assign an impedance to a source (battery, opamp, signal generator, whatever) and see how much energy (power) goes from the source to the load, with the load one quarter of the source, the same and four times the source impedance. How much power is dissipated in the source versus the load for each scenario?

The rules (or perhaps the goals) change in some cases when dealing with radio frequencies - in those cases you usually want a close match. Source, transmission line, and load (antenna) at the same impedance or matched with transformers. With RF, the source that isn't matched to the load - doesn't drive the load, but signal is reflected back to the source creating high voltages or currents along the way, and causing problems in the system or source. With many RF applications you pay a premium in power and parts cost to generate it in the first place, so wasting it makes no sense.

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Which completely ignores how they're used in practical circuits.

Graham

Let's start at the beginning.

Do you know what impedance is ?

Graham

Just pretend its input resistance and output resistance and do all the analysis using DC. Rules of thumb: A low output resistance puts out the same constant voltage no matter how much load you hang on it. (example: audio amp) A high output resistance can be loaded down by a load resistance less than about 10 times the output resistance. (Example: 150 ohm microphone) They call a high load resistance a 'bridging load'. You get maximum power transfer from source to load (watts) when the input and output resistance match.

Thanks for everyone who answered to my question. Yeah..I know what impedance is.And I know the voltage source and current source concept a little bit. But input impedance and output impedance are new concepts to me.

It seemed that input and output impedance concepts are related to those "source" concept. For example.. Let's suppose that there is a circuit that has input and output( ex) amplifier)

How do you define input impedance? Impedance between input terminal and ground? or Impedance between signal source and input terminal?

How about output impedance? Impedance between output terminal and ground ?

I was confused about that. I also can't know why value of impedance changes amplifier's function. Is it really related to some "source" concept "?

Input voltage / input current

Yes.

That's normally a wire !

Yes.

It's mostly related to the resistor values used around the op-amp.

Graham

First lets get impedance, in general, out of the way. It is a concept closely related to resistance. It has the same units, volts per ampere, as resistance, but includes the frequency dependence. But, in simplest terms, high impedance is a lot like high resistance, and low impedance is a lot like low resistance.

When applied to an amplifier circuit made with an opamp, input impedance refers to how many volts it takes to drive an ampere of current through the input of the amplifier (or microvolts to drive a micro ampere, etc.). Output impedance is how many volts the output signal is loaded down by, if you pull an ampere out of it. A high input impedance reacts to a signal without requiring much current from the signal source, while a low output impedance drives a wide range of load resistances without the output voltage drooping much.

The impedance of the amplifier, configured with programing resistors is a combination of the opamp impedances and the resistor impedances. The reason the inverting configuration has a lower input impedance is that the output (which swings in the opposite direction from the inverting input) is connected back to that input through a feedback resistor. This causes the inverting input to be held at an almost fixed voltage (a voltage that matches whatever you have connected to the non inverting input. So, whatever resistor you connect between the signal source being amplified and the inverting input is what determines how many volts per ampere (ohms) the signal source must drive, since the entire input signal voltage will be dropped across that resistor.

If the opamp is configured as a non inverting amplifier (lets say, a follower with a gain of 1 and the output connected directly to the inverting input) the inverting input is forced to match whatever voltage you apply to the non inverting input, by the output, so there is no other path for input current than just what the opamp inputs leak. The opamp is designed to keep that input leakage current as low as possible, so the amplifier input impedance (volts required to force an ampere of current through the input) is very high.

I have defined input resistance and output resistance, but I'll leave it there, for the moment as an approximation of input and output impedance. The full definition involves including how the volts per ampere changes with frequency.

It is a signal concept that generally does not apply to opamps, where lots of gain is available and gain precision is important. It is an energy transfer concept. If a signal source is loaded with a load that has a matching impedance, the largest possible energy transfer is achieved. This can be a useful thing to strive for when the signal energy is th limiting factor. For instance, if you are trying to force the maximum energy out an antenna you might try to impedance match the antenna impedance to the amplifier output impedance. In that case, you are not trying to get some specific voltage across the antenna terminals, but are just trying to radiate as much power as possible from a given amplifier.

There are other reasons to be concerned with source and load impedances that do not involve matching. Some components, like transformers, have a frequency response that depends on the impedances of the sources and loads connected to them. So you have to design the rest of the circuit to produce the impedances that will give the transformer the best frequency response.

--- Impedance is the opposition to the flow of current.

In a DC circuit or in an AC circuit which is resistive, resistance and impedance are the same. In a reactive AC circuit (a circuit containing capacitive and/or inductive elements) the impedance is more complex than that.

For the moment, let's say we have a circuit with a resistive load and a resistive generator that looks like this: (View in Courier)

10VRMS 10mA--> / +----->>------+ | | [GENERATOR] [LOAD] | | +----->>------+ | GND

In this case, since there is 10 volts across the load and 10mA passing through it, and it's resistive, we can use Ohm's law to find the impedance of the load, like this:

E 10V R = Z = --- = ------- = 1000 ohms I 0.01A

Now suppose that instead of a resistor the load was an amplifier with a resistive input, and that with a 10VRMS signal into it the input took 10mA of current from the generator.

Using Ohm's law again:

E 10V R = Z = --- = ------- = 1000 ohms I 0.01A

We see that the results are identical to the case with a resistor, and since that was the load the amplifier's _input_ presented to the generator, we say the amplifier has an input impedance/resistance of

1000 ohms.

Now, since the generator isn't perfect, it has resistance through which the current into the amplifier's input must pass, and since it's on the generator's _output_, it's called the output impedance.

Modifying the circuit above to show the generator's resistance, we have: 10VRMS 10mA---> / +------>>------+ | | [Rg] | | [1000R] [GENERATOR] | | | +------>>------+ | GND

In order to find out what the generator's output impedance is, we can disconnect the load, measure the generator's output voltage, reconnect the load, measure the voltage again and (knowing the current into the load) calculate the output impedance of the generator.

In this case, let's say that with the load disconnected the generator's output voltage went to 20V, and when the load was reconnected the voltage fell to 10V.

Since, when the load was disconnected, the generator was supplying no current, (except to the voltmeter) its output voltage rose to essentially its no-load value. Then, since its voltage dropped to

10V when the load was connected, 10 volts was being dropped across the generator's internal resistance and, using Ohm's law, the resistance had to be:

E 10V R = --- = ------- = 1000 ohms I 0.01A

---

--- Yes.

For a non-inverting amplifier:

Vin>--------|+\\ | >--+-->Vout +--|-/ | | | +--[R1]--+ | [R2] | GND

The input impedance is going to be, essentially, the very high input resistance of the opamp, while for an inverting amplifier:

+--[R2]--+ | | Vin>--[R1]--+--|+\\ | | >--+-->Vout vref--> +--|-/ | 0V It's going to be the resistance of R1 because Vout will swing to whatever voltage it has to to make the voltage on the + input equal the voltage on the - input. In this case that's 0V, so the input resistance, as far as Vin is concerned, will be R1.

--- See above

---

--- Impedance matching refers to making the output impedance of the source (the generator) the same as that of the load.

-- JF

You measure the impedance by applying a voltage and measuring the current that voltage drives through the input. the ratio of voltage to current is the input impedance.

The definition of input impedance is the ratio of input voltage to input current.

Yes, unless the input is applied between two nodes, neither of which is ground. Balanced differential signal lines do not use ground as one of the signal nodes.

No. If the source and input are directly connected, there is zero impedance in that connection, regardless of the source impedance and the input impedance.

Since the output is a source, you measure the voltage change caused by a load current and divide that voltage change (the drop across the output impedance) by the load current. Ideally, this would be the same ratio you would find if you divided the open circuit output voltage (no load voltage) by the short circuited current, since a short circuit (zero impedance load) drops all the output voltage across the output impedance.

Yes, if the output is ground referenced, and not a floating transformer winding or something else not related to ground.

You will have to think that question through a bit before you can ask it in a way that makes sense.

The output is a source concept. The input is a load concept.

I'm not following you. Your point is what?

We almost never use op amps as they come - typical op amp has an open loop gain of 200K how many times have you encountered an op amp that actually uses a tenth of that? (it is done . . . but not often) - great way to sense current for instance.

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The input impedance (for low frequency signals) for the ordinary inverting configuration is simply the value of the input resistor to the inverting input. That's because the effective potential at this point (provided the device is in it's linear range) is the same as at the non-inverting input (It's all because of the feedback loop - you really need to understand the basics of feedback theory before you'll understand an op amp or virtually any other amp for that matter). Normally, that's ground (at signal frequencies). This is a basic part of op amp theory and operation.

Basically, the inverting input (which is the feedback point) is changed to follow the potential at the non-inverting input. If it's fixed, the inverting input will remain (roughly) fixed while the device is in it's linear range.

A non-inverting configuration varies the potential at the non-inverting terminal, forcing the output to drive the inverting input until the difference between the inputs is 0 - the input impedance (at low frequencies) is Vin/Ib (where Ib is input bias current). That's a fairly high impedance.

There's a whole lot more to op amps than this, of course.

Cheers

PeteS

The whole point of the opamp is to deliberately 'throw away' the excess gain to get other desirable characteristics. What characteristics we are after depends on application, of course. A typical cheap opamp with inputs shorted might easily have it's output pinned at one of the supply rails (Vo = Vos x G) which would hardly be a desirable thing ;)

Cheers

PeteS