Question about inverting op Amps

hgmjr

Joined Jan 28, 2005
9,027
It is unclear what you are asking. What opamp parameter are you wanting to calculate when the opamp is in saturation?

If your question is "How do you calculate the gain of an inverting opamp when the opamp is in saturation?", then the answer is that once the opamp enters saturation the feedback loop can no longer exercise control over the opamps behavior and the normal gain equation no longer accurately predicts the output signal as a function of the input signal.

hgmjr
 

beenthere

Joined Apr 20, 2004
15,819
A slightly different way to say it is to look at the magnitude of the input signal and figuring from that when the op amp can no longer swing the output far enough to meet the gain requirement. Like an op amp with a G = 1000 will have a hard time with a 1 volt input.
 

recca02

Joined Apr 2, 2007
1,212
sorry for interrupting,
so, does the supply voltage has nothing to do with saturation voltage? (a simple yes or no wud suffice.:))
 

studiot

Joined Nov 9, 2007
4,998
Surely Voltage limitations due to power supply headroom are called clipping not saturation?

As I understand it saturation occurs when the load resistance is so low that the amplifier cannot satisfy the Current demand and so distorts the waveform. The amplifier may burn out or limit in some way.

If the amp has short circuit limiting then you can 'calculate' the max gain, without distortion for a given input and load with reference to the manufacturer's max current figure and Ohms law.
 

Distort10n

Joined Dec 25, 2006
429
Output current limitation should not distort the waveform. You would only see a waveform across the load that would correspond to the current across it:

Vcc = 15, RL= 100Ω, Vo = 10V, Io(max) = 50 mA would show 5V across the load. If it was a sine wave, it should not show any signs of distortion just not the amplitude expected.

Clipping, headroom issues, saturation are all the same. It is not a current limitation due to the load rather the output stage limits imposed by the supply voltage.
 

hgmjr

Joined Jan 28, 2005
9,027
Hello,

How do you calculate when a inveting op amp saturates? Thanks
Upon re-examination of the question asked, I now believe the question was "How do you calculate or predict the voltage at which an inverting opamp stage will saturate?".

This still leaves room for interpretation as to what is to be calculated.

Is it:

a. How do you calculate the input voltage that will result in saturating/clipping the output signal?
OR
b. How do you calculate the voltage at which the output will clip?
If the question is "a." then

The answer to this depends on whether the opamp is one of the older generation opamps or one of the newer rail-to-rail opamps.

For opamps that are not rail-to-rail capable, a careful study of the manufacturer's datasheet will provide information on the maximum output signal swing that can be expect. This output level is dependent on the power supply voltages used to power the opamp. This information is usually provided in the form of a graph.

For rail-to-rail opamps, the output voltage swing is very nearly equal to the positive power rail on the positive signal swing and very nearly equal to the negative power rail on the negative signal swing.

In either opamp type, the input signal peak-to-peak voltage needed to drive the output voltage to its maximum output peak-to-peak voltage is equal to the maximum output peak-to-peak voltage range divided by the magnitude of the gain of the inverting opamp stage.

If the question is "b." then

Once again the answer depends on whether the opamp is one of the older generation opamps or one of the newer rail-to-rail opamps.

In the case of the rail-to-rail opamp, the output peak-to-peak signal swing is very nearly equal to the difference between the positive power rail and negative power rail.

In the case of opamps that are not rail-to-rail capable, the manufacturer's specification will need to be consulted to determine the maximum output peak-to-peak voltage swing.

hgmjr
 

Ron H

Joined Apr 14, 2005
7,063
Output current limitation should not distort the waveform. You would only see a waveform across the load that would correspond to the current across it:

Vcc = 15, RL= 100Ω, Vo = 10V, Io(max) = 50 mA would show 5V across the load. If it was a sine wave, it should not show any signs of distortion just not the amplitude expected.

Clipping, headroom issues, saturation are all the same. It is not a current limitation due to the load rather the output stage limits imposed by the supply voltage.
You might want to think about that a little more. Remember that output current limiting is generally an instantaneous phenomenon. Suppose you had an op amp connected to +/- supplies, +/-50mA current limiting, and a 100 ohm load to GND. With the output signal at the zero crossing, no current flows. At 100mV output, 1mA flows, so no current limiting. At 1V output, the current is 10mA. Still no limiting. At 5V output, the output has reached its maximum current capability. If it tries to go to 6V, or 7V, etc., the output will be limited to 5V. This will cause severe distortion.
 

gootee

Joined Apr 24, 2007
447
Hello,

How do you calculate when a inveting op amp saturates? Thanks
Look at the datasheet, or test the opamp, and find the maximum positive and negative voltage swings (which are often different magnitudes) for the given supply voltages. Use the gain and the input amplitude to calculate whether or not the opamp will try to go past its available voltage swing range.
 

Distort10n

Joined Dec 25, 2006
429
You might want to think about that a little more. Remember that output current limiting is generally an instantaneous phenomenon. Suppose you had an op amp connected to +/- supplies, +/-50mA current limiting, and a 100 ohm load to GND. With the output signal at the zero crossing, no current flows. At 100mV output, 1mA flows, so no current limiting. At 1V output, the current is 10mA. Still no limiting. At 5V output, the output has reached its maximum current capability. If it tries to go to 6V, or 7V, etc., the output will be limited to 5V. This will cause severe distortion.
Very true. Thanks for the correction.
 
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