# ADA4098-1 what is the meaning,Overdrive voltage

#### pinkyponky

Joined Nov 28, 2019
270
Hi,

Please can you open the datasheet of this component and go to the page 4 under the section of 'Output Characteristics'. In which, I'm not understanding of what is the meaning of the "**Overdrive voltage (Vod) = 30mV**", at both for Output Voltage Swing Low and High.

Is overdrive voltage is the Supply Voltage (OR) Output Voltage?.
What will be the output voltage swing for both High and Low when +5V and -5V voltage is applied?.

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#### Ian0

Joined Aug 7, 2020
5,161
I believe that it is the difference between the input+ and input- voltages.

#### pinkyponky

Joined Nov 28, 2019
270
I believe that it is the difference between the input+ and input- voltages.
Hi Ian0,

Thank you for quick replies for my all posts.

You would say that the Overdrive voltage is the voltage between the input voltages of the non-inverting and inverting terminals. If this is the case, then If used this component in High Side Current Sensing, then we have a low input voltage difference between the input terminals, then we can consider what you said above.

On the other hand, if used this component in Low Side Current Sensing, then we have a high voltage difference between the input terminals (for example: +5V applied to +IN terminal and GND to -IN terminal, therefore we have a +5V difference between the two terminals). In this case, how we can consider the Output voltage swing on Low and High when supply voltage applied as a +5V and -5V.

#### Ian0

Joined Aug 7, 2020
5,161
As it's an op-amp, and you are using it as such (and not as a comparator), then there will be very little difference between the input voltages because feedback would ensure that they are at almost the same voltage, provided that the output voltage is in the linear region.
The linear region is everywhere apart from voltages within 50mV of each power supply.
e.g. for a 5V power supply, anywhere between 50mV and 4.95V
What it's trying to tell you is that if you want to get the output voltage within 50mV of the power supply, you need rather more voltage between the inputs than you normally would have for an op-amp.
It's all a bit irrelevant, as you wouldn't normally want the outputs closer than 50mV to the supply voltage.

• dcbingaman

#### pinkyponky

Joined Nov 28, 2019
270
Hi all,

In general, In the datasheets, talk about the Common mode input voltage or Input voltage range. And also says that the supply voltage Vs or V+ = 5V, the common mode voltage is max ((V+)-1.5V) . Does it mean that if 5V applied to the one of the inputs of the op amps, then the op amp will be damage, because the data sheet is stated Vcm = 5V - 1.5V = 3.5V. See below datasheet is attached.

https://www.analog.com/media/en/technical-documentation/data-sheets/2057f.pdf

For example:
The supply voltage is +5V and -5V. If IN+ is applied approx. 5V and IN- is applied approx. 5V. Then, what is the common mode voltage here?.

The supply voltage is +5V and -5V. If IN+ is applied approx. 5V and IN- is applied to GND. Then, what is the common mode voltage here?.

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#### Ian0

Joined Aug 7, 2020
5,161
All datasheets are written to a certain code.
If you exceed any specification in the ABSOLUTE MAXIMUM section, you are likely to damage the device.
If you exceed any specification elsewhere in the datasheet, it will do no damage, but it may not work.
So, "Maximum input voltage = V+ +0.3V", means that on a 5V supply, allowing the inputs to exceed 5.3V may damage the device.
(Only "may" because "V+ + 0.3V" is likely to mean that there is a protection diode, and if the current is limited then the protection diode will do its job and protect the device, but don't rely on it!)
So Vcm(max) =V+ - 1.5V, means the maximum common mode voltage on the inputs should not exceed 3.5V on a 5V supply. If you do, it may not work. i.e. the output voltage would not be what you would predict from normal op-amp theory.

As an example, the good old TL072 had a common mode voltage limitation on the negative supply. If you exceeded it, there was a phase reversal and the output jumped to the opposite supply.

#### pinkyponky

Joined Nov 28, 2019
270
The maximum limitations I have understand.
Please, can you explain the example which I have asked in the above post?. What value is the common mode voltage according to the above example.

#### Ian0

Joined Aug 7, 2020
5,161
Assuming that your op-amp is working as an op-amp, then both inputs will be at very nearly the same voltage, and the output will be determined by the minute voltage between them multiplied by the huge open-loop gain.

The common mode input voltage is the voltage between either input and the power supply voltage, or some other reference point such as ground. (It doesn't matter which input, as they will both be at almost the same voltage)
The supply voltage is +5V and -5V. If IN+ is applied approx. 5V and IN- is applied approx. 5V. Then, what is the common mode voltage here?
Common mode voltage is "approx" +5V.

The supply voltage is +5V and -5V. If IN+ is applied approx. 5V and IN- is applied to GND. Then, what is the common mode voltage here?.
There is no meaningful common mode voltage here, as the op-amp cannot operate correctly. It cannot amplify (5V - 0V) * Open_Loop_Gain.

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#### pinkyponky

Joined Nov 28, 2019
270
The supply voltage is +5V and -5V. If IN+ is applied approx. 5V and IN- is applied to GND. Then, what is the common mode voltage here?.
Sorry, Above I was described wrongly. Please follow below description and please let me know what is common mode input voltage.

The supply voltage is +5V and -5V. If IN+ is applied approx. 5V and IN- is connected to output of the amplifier as a feed back, and also this configuration called it as buffer circuit. Then, what is the common mode input voltage here?.

#### Ian0

Joined Aug 7, 2020
5,161
Sorry, Above I was described wrongly. Please follow below description and please let me know what is common mode voltage.

The supply voltage is +5V and -5V. If IN+ is applied approx. 5V and IN- is connected to output of the amplifier as a feed back, and also this configuration called it as buffer circuit. Then, what is the common mode voltage here?.
It's 5V again. Because if the buffer is operating correctly, the output will be the same as the + input, and so will the - input, because it is connected to the output.

Think about the differential amplifier circuit with a gain of 1: all four resistors are the same value, working on a ±5V supply: V out will be the difference between V1 and V2. If V1=4V and V2 = 5V, the ouutput will be 1V. If V1 = -5V and V2 = -4V, then the output will also be 1V.
The op-amp is working correctly, the inputs will both be at the same voltage. That voltage is the common mode voltage. It will be 2.5V for the first case and -2V for the second case.

If V1=20V and V2=21V, then the output should still be 1V, but it might not be, because the common mode voltage will be 10.5V, and that will be outside the common mode input voltage for most op-amps running on a 5V supply.

#### pinkyponky

Joined Nov 28, 2019
270
It's 5V again. Because if the buffer is operating correctly, the output will be the same as the + input, and so will the - input, because it is connected to the output.
According to this formula your saying that the Input common mode voltage will be 5V. Right?- If so, then I need to choose the amplifier which has the input common mode voltage higher than the 5V. Am I right?.

#### Ian0

Joined Aug 7, 2020
5,161
If so, then I need to choose the amplifier which has the input common mode voltage higher than the 5V. Am I right?.
Yes, exactly. Most rail-to-rail op-amps will have common mode input ranges which extend to the supply rails. Some of them will have common-mode ranges that will extend 0.5V above and below the supply rails.

• pinkyponky