# input impedance output impedance

Discussion in 'Electronic Basics' started by [email protected], Jul 12, 2007.

1. ### Guest

I'm a newcomer to analog electronics.
Currently, what I'm studying is Op-amp circuits.
The term I'm confused about is Input & Output impedance concept.
For example, input impedance is differenf for inverting amplifier and
non-inverting amplifier.
How do you define the input impedance or output impedance?
and what does the term "Impedance matching mean?"

2. ### RobinGuest

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

3. ### defaultGuest

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

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

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.

4. ### EeyoreGuest

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

Graham

5. ### EeyoreGuest

Let's start at the beginning.

Do you know what impedance is ?

Graham

6. ### BobGGuest

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
(Example: 150 ohm microphone) They call a high load resistance a
(watts) when the input and output resistance match.

7. ### Guest

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?

Impedance between output terminal and ground ?

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

8. ### EeyoreGuest

Input voltage / input current

That's normally a wire !

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

Graham

9. ### John PopelishGuest

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
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.

10. ### John FieldsGuest

---
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--> /
+----->>------+
| |
| |
+----->>------+
|
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

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.

11. ### John PopelishGuest

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.

12. ### defaultGuest

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.

13. ### neon

1,325
0
Oct 21, 2006
on a dif amps there are usualy two bases of transistors whereby their inpedance charatheristics are he same. the differential voltage is nill or virtualy the same. the output the amplifier must be relatively low otherwise you could not drive any load pluss any F/B will bereduced by output inpedance drop to the input source.

14. ### PeteSGuest

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

15. ### PeteSGuest

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  