Differential Amplifier Output Voltage Issue

[azadfalah]

hi everyone ,

I’m working on a differential amplifier circuit and would like to know how to calculate its gain. Here is the configuration:


How is the gain of this circuit calculated?


If nothing is connected to the inputs of the HCPL-7840 IC, its output voltage will be approximately 2.56V, meaning both outputs of the IC will be at this voltage. Under these conditions, I am observing an 8V output at the op-amp. How can this be calculated?

Thank you.

 

[FvM]

Vout = 7.5 V + 5 * V(7,6)
V(7,6) is U2 differential output voltage

 

[Easy peasy]

The voltage gain is (R3 / R1) applied to ( Vpin7 - Vpin 6 ) - because the whole is configured as a diff amp, R4a & b in // form 10k to gnd ( effectively ) .

caveat: R1 = R2, R3 = R4a//R4b for the above to apply.

--- Updated Dec 2, 2024 ---

FvM meant to type V(7,6 )

 

[azadfalah]

Thank you friends, for your guidance.


Why does the equation change if I replace the 20k resistor with a 10k resistor for R4A and R4B?
R4A = R4B =10k
Vpin7 = Vpin6
Vout = 11.6V

 

[barry]

Thank you friends, for your guidance.


Why does the equation change if I replace the 20k resistor with a 10k resistor for R4A and R4B?
R4A = R4B =10k
Vpin7 = Vpin6
Vout = 11.6V

the equation doesn’t change, the result changes. A + B changes if the value of A changes, doesn’t it? You’re changing the effective resistance as explained in post #3.

 

[FvM]

Resistor symmetry, R1 = R2, R3 = R4a || R4b is necessary for differential amplifier operation.

 

[azadfalah]

the equation doesn’t change, the result changes. A + B changes if the value of A changes, doesn’t it? You’re changing the effective resistance as explained in post #3.

So, even though R4A is connected to 15V, we consider it in parallel with R4B in the calculation, and the rest of the analysis is similar to a standard differential amplifier ?

--- Updated Dec 3, 2024 ---

Thank you, friends. The issue with the first question has been resolved.

In this method, I encounter an issue: when I connect the signal, after using a resistor divider, to the ADC, the offset causes me to utilize only half of the ADC's voltage range, which reduces accuracy. What do you think is the best way to bring the entire waveform above 0V without any offset, allowing me to use the full range of the ADC?

 

[KlausST]

How is the gain of this circuit calculated?

Generally: The gain is always: V_out / V_in

In the meaning of output signal (amplitude) divided by input signal (amplitude).

Now it depends on your application how the signal is defined.
In your case I guess (that´s all I can do) you are NOT interested in DC signals. (maybe you are, since the circuit can amplify the change in DC voltage)

If this is true you need to calculate AC gain ... at a frequency of your choice.
Yes, frequency is your choice. (when talking about frequency in amplifier circuits we usually refer to sine wavefrom)
The problem is, that with all amplifier circuits the gain depends on frequency.

When talking about DC gain here .. the formula is: Gain = delta_VOut / delta_Vin.

If nothing is connected to the inputs of the HCPL-7840 IC, its output voltage will be approximately 2.56V, meaning both outputs of the IC will be at this voltage.

This 2.56V is NOT your signal. It is just an offset of both signals, also known as "common mode voltage".
If both signals are at the same level ... the difference between them is zero. (OK, ... one can argue that zero is no signal at all)
--> but still: [the difference between the signals] IS your input signal.

The same is true for the output.

I am observing an 8V output at the op-amp.

With zero (differential) input ... this 8V also is NOT your signal. It again is the offset. .. The value you have to reference your "output signal" to.

The DC gain of your circuit is considered with "zero input" to act as reference (8V output)
For gain calculation:
You have to consider a second step: When the input is not zero.
--> how much moves the output (delta) when the (differential) input changes a known value (delta).

Let´s say you change the differential input voltage from 0V to 0.1V ... then the output will change from 8V to 8.5V (or 7.5V)
Delta_Out = 0.5V .. delta_in = 0.1V --> gain = 5

***
Mind:
* You may change the output offset voltage by changing R4A and R4B. When you do so you need to satisfy R4A || R4B = R3
* you may change the gain by changing R1 and R2. But always keep R1 = R2.

Klaus

 

[FvM]

In this method, I encounter an issue: when I connect the signal, after using a resistor divider, to the ADC, the offset causes me to utilize only half of the ADC's voltage range, which reduces accuracy. What do you think is the best way to bring the entire waveform above 0V without any offset, allowing me to use the full range of the ADC?

HCPL-7840 has bipolar signal range, the present circuit is obviously designed to process both signal polarities. If you don't need the feature, you can change the offset by changing R4a:R4b ratio, keeping 10 kohm parallel resistance..
I suggest to review HPCL-7840 datasheet, there are also application circuits with offset 0. An amplifier with unipolar supply voltage can't output zero signal accurately due to saturation voltage, so you'll either implement a small offset (can be as low as e.g. 0.1V) or bipolar amplifier supply voltage to process signal near zero without error.