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Say the above OPAMP is normally operated with supply of +15V and -15V and output is said to swing from +15V to -15V. But I changed -15V power supply {of OPAMP} to -14V and Vi to:

(a)1V

(b)-15V

(c)1/2V

What should be the output and regions at which the OPAMP operate in each case?

Actually in Sergio Franco's book it was written that if we do so we will have to change the reference voltage of the circuit to average of supply and I am a bit confused about it.It is not meant as practical implementation I am just trying to learn. Also consider below examples (they are not homework just trying to be exhaustive)

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Welcome to EE.SE. \$\endgroup\$ – Sparky256 Mar 1 at 5:34
  • \$\begingroup\$ You can add a better looking schematic using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. \$\endgroup\$ – Transistor Mar 1 at 7:26
  • \$\begingroup\$ What I think : for instance I applied 1V and 0V with respect to actual ground as OPAMP don't knows where ground is it finds mean of power supply and consider it as ground so for OPAMP ground is 1/2V vn is 1/2V while vp is -1/2V also output should be -3/2V higher then 1/2V which is its ground so answer should be -1V .Correct me \$\endgroup\$ – Buzz bee Mar 1 at 13:46
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For an ideal op amp, which has zero input offsets and infinite gain, nothing happens.

For real op amps, the output will change slightly with changes in the power supply voltages. The ratio between the supply changes and the output changes is called the Power Supply Rejection Ratio (PSRR)or Supply Voltage Rejection Ratio (ksvr). The ratio will vary with op amp, and to make things more complicated the number for any op amp will vary with frequency. Higher frequency supply variations will produce a greater output variation than lower frequencies.

In general, for low frequencies, the rejection ratio will be on the order of 80 to 100 dB.

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Not a thing will happen as long as both power rails are beyond the minimum stated voltage. If the op-amp is not the rail-to-rail output type, then the output voltage range is typically between Vcc-1.2 volts and Vee + 1.2 volts.

Due to built-in constant current sources and sinks, shifts in power rail balance can mostly be ignored, but rails with less than full voltage lower the common mode range. In some cases a -6.2 volts maybe applied to a good quality op-amp so its input and outputs can have zero volts as a common mode range, though the application may only require positive values to work, such as servo-loops or simple peak detectors.

On a more serious note you need to check the minimum load an op-amp will drive. 1 ohm is absurdly low. Even UHF op-amps have a 25 ohm to 50 ohm minimum load. Typical DC and audio op-amps have a 2.2K ohm minimum load. The feedback resistors counts as part of a load, as the (-) input strives to stay at zero volts using this topology.

An all around safe load for full voltage swing is about 10K ohms, with a feedback resistor no less than 10K ohms.

Op-amps vary greatly in output current, so read the almighty datasheet in detail before drawing up a design based on a particular op-amp.

Due to their internal topology, many RF and video op-amps do require supply rails with a fixed study voltage, such as +/- 5 volts.

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  • \$\begingroup\$ Actually in Sergio Franco's book it was written that if we do so we will have to change the reference voltage of the circuit to average of supply and I am a bit confused about it.It is not meant as practical implementation I am just trying to learn. \$\endgroup\$ – Buzz bee Mar 1 at 5:51
  • \$\begingroup\$ I have run a OP220 op-amp at +24 and -5 volt, so a perfect balance is not always mandatory. \$\endgroup\$ – Sparky256 Mar 1 at 6:08
  • \$\begingroup\$ I have also run TL074 op-amps at +12 volts and -6.2 volts, for fan speed control to cool large heatsinks. \$\endgroup\$ – Sparky256 Mar 2 at 0:19
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Modern OPAMPs are designed to have a \$0\mathrm{V}\$ output and input reference level, when powered from a symmetric power supply, i.e. \$V_0=0\mathrm{V}\$ if \$V_+=V_-\$ in the ideal case, when no other errors (offsets etc.) occurs: this is due to the fact that, in doing so, the input and output voltage ranges are maximized in the sense that you can go over and below respect to the given reference level by the same amount of voltage.
Thus, referring to the following schematics, where \$V_\mathrm{ref}\$ is an arbitrary reference voltage for the inputs, every asymmetry in the supply voltage would appear, again in the ideal case and when no feedback is applied, as an output offset voltage \$V_{o_\mathrm{off}}\$

schematic

simulate this circuit – Schematic created using CircuitLab

$$ V_{o_\mathrm{off}}=\frac{V_++V_-}{2}\quad(\text{in our case}=0.5\mathrm{V}\text{ since }V_+=15\mathrm{V},\:V_-=-14\mathrm{V}), $$ which assumes also the role of a new reference level for the inputs.

When a feedback is present, the effect of the asimmetry in power supply voltage is taken into account by considering the reference potential of the input and the output to be \$V_{o_\mathrm{off}}\$ instead of zero: this is the common situation faced when designing single supply (\$V_+=V_{DD}\$, \$V_-=0\$) circuits. In the example proposed we have the following situation:

schematic

simulate this circuit

where \$V_u\$ is the inner output voltage which adds to the offset voltage in order to produce the real output.
Now consider your inverting amplifier circuit, where feedback is applied, the gain is $$ A_v=-\frac{R_{fb}}{R_{i}}=\frac{1\Omega}{1\Omega}=1. $$ By applying the virtual reference pin principle we get $$ \frac{V_o}{R_{fb}}=\frac{V_u+V_{o_\mathrm{off}}}{R_{fb}}=-\frac{V_i}{R_i}\implies V_o=-\frac{R_{fb}}{R_{i}}V_i\tag{1}\label{1} $$

As you can see from formula \eqref{1}, the behavior of the asymmetrically powered circuit is absolutely identical to the balanced one: the only difference (in the ideal case, I stress) is that the input and output range will be asymmetrical, as clipping occurs when \$V_i\ge +14\mathrm{V}\$. This implies that in your first circuit, when

(a) \$V_{i}=1\mathrm{V}\implies V_o=-1\mathrm{V}\$.

(b) \$V_{i}=-15\mathrm{V}\implies V_o=15\mathrm{V}\$.

(c) \$V_{i}=0.5\mathrm{V}\implies V_o=-0.5\mathrm{V}\$

In your second circuit, the non inverting unity gain buffer powered \$V_+=+40\mathrm{V}\$ and \$V_-=+10\mathrm{V}\$, the output with \$V_i= +12\mathrm{V}\$ will be \$V_o= +12\mathrm{V}\$ while it will be stuck at \$V_o= +10\mathrm{V}\$ for every input \$V_i\le +10\mathrm{V}\$.

Notes
The above analysis describes what happens in the ideal case, as opposed tho the real case: in the real world, obviously, there are other issues that cause the output voltage to be different from \eqref{1}, and a few of them are described below.

  • a very important issue pointed out by WhatRoughBeast in his answer is the influence of the PSSR, especially important when the asymmetry of power supply is in the form of a noise/power supply ripple: even in this cas,e the asymmetry voltage, multiplied by the PSSR, is modeled as an additional noise contribution at the inputs.
  • As Sparky256 pointed out, apart from the ideal reduction of maximum output swing, there are other problem related to the output driving capability of the OPAMP related to its circuit structure and characterized, more or less fully, in its datasheet. Also, I would like to point out the changes in the common mode voltage very important in those applications where you need to amplify very small signals.
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    \$\begingroup\$ As said the offset voltage for OPAMP changes for asymetrical inputs so all together as we are applying inputs with respect to ground {both vn and vp} so shouldn't both inputs should be increased by half for instance I applied 1V and 0V with respect to actual ground as OPAMP don't knows where ground is it finds mean of power supply and consider it as ground so for OPAMP ground is 1/2V vn is 1/2V while vp is -1/2V also output should be -3/2V higher then 1/2V which is its ground so answer should be -1V .Correct me.. \$\endgroup\$ – Buzz bee Mar 1 at 13:39
  • \$\begingroup\$ It is not a change in the offset voltage of the OPAMP but, as I said above, a change in the reference voltage: in asymmetric power supply conditions, the reference voltage is no more the \$0\mathrm{V}\$ potential, but \$V_{o_\mathrm{off}}\$ so when one of the input pins is connected to the ground poterntial, the OPAMP sees it as it was connected to \$V_{o_\mathrm{off}}\$. The situation for the non inverting amplifier sees this voltage as applied in series to the feedback resistor \$R_i\$, which goes from the reference potential to the non inverting input. \$\endgroup\$ – Daniele Tampieri Mar 1 at 14:46
  • \$\begingroup\$ But if it is not possible to set/change the reference for instance consider above or any non inverting negative feedback case then what should be output. \$\endgroup\$ – Buzz bee Mar 1 at 15:57
  • \$\begingroup\$ Also for values of output you gave for Vi=1V Vp=Voff=.5==>by virtual short Vn=.5==>Vo=0 but you said 1 ?? \$\endgroup\$ – Buzz bee Mar 1 at 16:06
  • \$\begingroup\$ @Buzzbee. Mmm, let me check. Perhaps I messed something: basically, however, the reasoning is correct i.e. asymmetry in power supply values means change of reference for the two inputs. I have to check the correctness of the equations. \$\endgroup\$ – Daniele Tampieri Mar 1 at 16:47

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