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I have an analog circuit which has an input voltage range Vin of 5V - 10V and output voltage Vout.

One can assume that the circuit has been simulated and a DC sweep of Vin with a range from -10V to +10V has been performed and the corresponding Vout=f(Vin) has been plotted.

Is it correct in that case to compute the DC offset of the circuit just by obtaining Vout at Vin=0V even though this is outside of the allowed input voltage range?

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  • \$\begingroup\$ This has a strong smell of homework. If so then you need to show your work. Hit the edit link below your question ... \$\endgroup\$ – Transistor Apr 29 '19 at 21:25
  • \$\begingroup\$ Nope. No homework. Real world problem of someone without electronics background. \$\endgroup\$ – Maxwell1919 Apr 29 '19 at 21:31
  • \$\begingroup\$ Typically input offset voltage is defined for a differential amplifier as whatever differential voltage is required between the two input terminals to make Vout = 0. That means the amplified Vin voltage is the difference between two terminals rather than an absolute voltage. Since you are describing a single-ended amplifier (it seems), I think in your case you would want to see what Vin makes Vout be at the center of its operating range (0V), and how far off this is from the center of the Vin input range (7.5V I suppose?) \$\endgroup\$ – jbord39 Apr 29 '19 at 21:39
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    \$\begingroup\$ I was going to ask if you could supply a graph of the transfer function. What happens when the input goes outside 5 to 10 V. If it never does then why do the DC sweep from -10 V to +10 V? \$\endgroup\$ – Transistor Apr 29 '19 at 21:56
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    \$\begingroup\$ Let's step back from this. Why do you care what the "offset" voltage is? You know the transfer function, what difference does it make that you choose some voltage and call it the offset? \$\endgroup\$ – Elliot Alderson Apr 29 '19 at 22:19
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When a circuit has a (non-zero) DC offset, it behaves as follows:

You apply a certain input voltage and expect a certain output voltage.

However, due to offset, the circuit does not output the expected voltage, it outputs a (slightly) different voltage.

For example, let's say I have a voltage amplifier with 10x gain. It has an input voltage range of:

-1 V < Vin < + 1 V.

When I apply exactly 0 V at the input (short it) I would expect 0 V at the output.

However my amplifier has a DC offset so I measure +10 mV at the output.

Offset is usually referred back to the input we have to divide that + 10 mV by the gain.

So my amplifier has a +1mV offset at the input.

In your case the input range does not include 0 V, that does not matter we can simply "shift" the voltages. Suppose we have a -1x amplifier then a +5 V input should give an output of -5V. Now suppose it is -4.95V, that means +0.05V (50 mV) offset at the output (the output voltage is higher, more positive, than expected) so divide by gain: 50 mV / -1 = -50 mV offset at the input.

To fully understand offset you have to know that offset has a static component and a statistical component.

Suppose you measure (on a bench, not simulate!) many amplifiers to collect offset data. Then you would expect that the average offset centers around 0 V meaning "no offset". Of course there will be no amplifier having zero offset, some are around +1 mV, others around -1 mV etc. But that all offsets average out at 0 V.

That would mean the expectation value of the offset is 0 V. This means the design is good and well balanced (it has no imbalance).

In a design that's not so good the expectation value could be shifted to for example + 10 mV. Then most offset voltages could be between + 9 mV and + 11 mV. Note how that reduces the chance of finding an amplifier with an offset close to 0 V or even a negative offset voltage!

Then there is deviation meaning how far from the average offset do the values deviate. Obviously offsets between -1 mV and + 1 mV is generally "better" than values between -10 mV and + 10 mV. That +/- 10 mV means the offset has more "spread" or a higher deviation.

Now offset simulations:

In a simulator you usually only simulate an ideal version of a circuit. If the design is good it will show no offset at all. If the design isn't perfectly balanced it will have a static offset, you will see that in a simulation.

The offset deviation is more challenging to simulate. First you will need statistical models of all relevant circuit components. This can include statistical models of feedback resistors for example.

When you have all the statistical models in place a "Monte Carlo" Simulation needs to be done. This then runs your normal (DC or whatever you use) simulation many times each time with slightly (statistically) changed component values. Afterwards all results are combined and you can see the average offset \$\mu\$ and the simulated deviation \$\sigma\$.

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  • \$\begingroup\$ This is great. Thank you very much. One more thing. How could I compute the gain in my case if it is unknown at the beginning? \$\endgroup\$ – Maxwell1919 Apr 30 '19 at 12:35
  • \$\begingroup\$ Gain is simply: output change / input change and that's simply the derivative of your Vout = f (Vin) curve. So in the simulator plot the derivative and that's your gain. \$\endgroup\$ – Bimpelrekkie Apr 30 '19 at 12:53

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