how do you calculate the input/output resistances of an amplifier config?

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swagadon

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What are the input and output resistances of an inverting amplifier configuration
if both resistors in the circuit have a value of 1kΩ?

uberteknik

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(Original post by swagadon)
What are the input and output resistances of an inverting amplifier configuration
if both resistors in the circuit have a value of 1kΩ?

I assume you mean a circuit like this?

The output resistance is assumed to be zero provided the output load does not exceed the current capability of the op amp at full output voltage (typically the supply voltage rail less a volt or so).

You need to consult the op-amp manufacturers data for the given device to find out the maximum output voltage swing and output current capability and then make sure the load resistance/impedance does not exceed that maximum current at full voltage output.

swagadon

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(Original post by uberteknik)
I assume you mean a circuit like this?

why is the output resistance assumed to be 0? and what is an output load?

uberteknik

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(Original post by swagadon)
why is the output resistance assumed to be 0? and what is an output load?

Output load is the resistance of the circuit connected to the output terminal of the op-amp.

The output current capability of low-power op-amps is typically 10mA which places a maximum limit on the resistance that can be connected to the output for a given output voltage swing.

For instance, if the maximum output voltage needed is 12V and the maximum output current capability is 10mA

V/I = 12/10x10^-3 = 1200 ohms, is the minimum resistance that can be connected to the output to maintain correct operation.

If the load resistance is less than this, then the output voltage swing will reduce.

i.e. 10mA x 600 ohms = 6V maximum output.

I should have qualified the 'assumed' zero output resistance is used for general purpose calculations (engineers love rule-of-thumbs).

It's assumed zero because signal op amps are most commonly used as voltage gain devices and not intended to supply large currents for most applications. Because of that, for practical purposes, the output resistance is not an issue provided of course the current limitations are not exceeded.

If a large current is needed, then further discrete power components are added to the external circuit to execute this requirement.

In reality, the actual output resistance is a function of the load exactly because of that current limitation and dependent on the required voltage swing.

So for the circuit in question with a voltage gain of -1, the output resistance will be a function of the input voltage, gain and load resistance:

$\frac{V_{in} A_v}{(R_{load} \lvert \rvert R2)}$

swagadon

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(Original post by uberteknik)
Output load is the resistance of the circuit connected to the output terminal of the op-amp.

The output current capability of low-power op-amps is typically 10mA which places a maximum limit on the resistance that can be connected to the output for a given output voltage swing.

For instance, if the maximum output voltage needed is 12V and the maximum output current capability is 10mA

V/I = 12/10x10^-3 = 1200 ohms, is the minimum resistance that can be connected to the output to maintain correct operation.

If the load resistance is less than this, then the output voltage swing will reduce.

i.e. !0mA x 600 ohms = 6V maximum output.

I should have qualified the 'assumed' zero output resistance is used for general purpose calculations (engineers love rule-of-thumbs).

It's assumed zero because signal op amps are most commonly used as voltage gain devices and not intended to supply large currents for most applications. Because of that, for practical purposes, the output resistance is not an issue provided of course the current limitations are not exceeded.

If a large current is needed, then further discrete power components are added to the external circuit to execute this requirement.

In reality, the actual output resistance is a function of the load exactly because of that current limitation and dependent on the required voltage swing.

So for the circuit in question with a voltage gain of -1, the output resistance will be a function of the input voltage, gain and load resistance:

V_in / (R_load parallel with R2)

so I cant actualy calculate an answer?

and btw is the output load the output resistance?

uberteknik

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(Original post by swagadon)
so I cant actualy calculate an answer?

Before I answer that, what is the context of your question? Is this from an exam paper or are you trying to do something practical with the circuit? What course are you studying so that I know the level to pitch the answer.

(Original post by swagadon)
and btw is the output load the output resistance?

No. The load is the resistance of the load attached to the output.

For instance, the load connected to an audio power amplifier would be the impedance of the loudspeaker.

swagadon

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(Original post by uberteknik)
Before I answer that, what is the context of your question? Is this from an exam paper or are you trying to do something practical with the circuit? What course are you studying so that I know the level to pitch the answer.

No. The load is the resistance of the load attached to the output.

For instance, the load connected to an audio power amplifier would be the impedance of the loudspeaker.

this was from an exam paper, im doing physics

so what is the output resistance? o.O

uberteknik

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(Original post by swagadon)
this was from an exam paper, im doing physics

so what is the output resistance? o.O

Is that A-level physics or undergraduate?

swagadon

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(Original post by uberteknik)
Is that A-level physics or undergraduate?

1st year undergraduate

uberteknik

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(Original post by swagadon)
1st year undergraduate

OK

The op-amp on its own has an output resistance (R_o) measured between the output terminal and the OV supply reference of typically 50 to 100 ohms.

The feedback circuit modifies the effective output impedance the load resistance connected to the output is presented with. (Think Thévenin source equivalent).

In which case:

$Z_{out} = \frac{V_{o}}{I_{o}} = \frac{R_o}{(1 + A\beta)}$

where

$V_{o} = \mathrm {open \ circuit \ output \ voltage}$

$I_{o} = \mathrm {short \ circuit \ output \ current}$

$A = \mathrm {open \ loop \ gain}$

$\beta = \mathrm {closed \ loop \ gain} = \frac{R2}{R1}$

the effect of this is to reduce the output resistance by a factor of $1 + A\beta$

and for typical device parameters of general purpose op-amps, the circuit in question becomes:

$A = 10\mathrm x10^3 \mathrm {\ typically}$

$\beta = \frac{R2}{R1} = 1$

$R_o = 100 \Omega$

$Z_{out} = \frac{100}{[1 + (10 \mathrm x10^3)\mathrm x(1)]} = 0.01 \Omega$

Hence my original statement that the output resistance is assumed to be zero.

uberteknik

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(Original post by swagadon)
1st year undergraduate

Does the question ask you to calculate the input/output resistance or 'derive an expression' for the input/output resistance/impedance?

The latter would be more typical of an electronics engineering course I would have thought.

swagadon

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(Original post by uberteknik)
Does the question ask you to calculate the input/output resistance or 'derive an expression' for the input/output resistance/impedance?

The latter would be more typical of an electronics engineering course I would have thought.

The terminal labelled with a minus sign on an operational amplifier is often referred
to as a ’virtual ground’ when connected in an inverting amplifier configuration.
(i) Explain why the minus terminal behaves like the ground of the input. [2]
(ii) What are the input and output resistances of the inverting amplifier configuration
if both resistors in the circuit have a value of 1kΩ?[2]

this was the question originally, i dont think weve covered output resistances in lectures yet though, weve been treating the op amp as a 'blackbox' and using the rules that no current goes through the 2 inputs and Vout does whatever to keep the potential difference between the 2 inputs 0

uberteknik

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(Original post by swagadon)
The terminal labelled with a minus sign on an operational amplifier is often referred
to as a ’virtual ground’ when connected in an inverting amplifier configuration.
(i) Explain why the minus terminal behaves like the ground of the input. [2]
(ii) What are the input and output resistances of the inverting amplifier configuration
if both resistors in the circuit have a value of 1kΩ?[2]

this was the question originally, i dont think weve covered output resistances in lectures yet though, weve been treating the op amp as a 'blackbox' and using the rules that no current goes through the 2 inputs and Vout does whatever to keep the potential difference between the 2 inputs 0

Ahhh. OK.

The lecturer should take you through the derivation then otherwise the equations won't make much sense.

Not hard, but you should remember the key characteristics as I've mentioned here.

Virtual ground or virtual earth behaviour and why they behave like this is critical to understanding the operation of op-amps. Make sure you understand it.