# Using a differential op amp to convert a differential pair into a single ended signal

I need to convert a differential pair into a single ended signal and would like to use a differential amplifier to do this.

Using the above circuit and the ADA4850 I was hoping to achieve this but of course, the DC from each differential pair branch cancel one another out so only the AC components from each remain (which are centred around 0V). We cannot have a -ve supply and have connected VEE to GND. Because of this, whilst this is a rail to rail op amp, the DC bias cancel one another out and the -ve portion of the AC signal at the diff pair hits the GND floor/ limit.

If adding a DC bias to one input so the DC is passed through the op amp as show below (Diff pair input 200mVpp (peak to peak)):

Then the output DC bias is 50% of the input DC Bias with the peak to peak voltage being a quarter (100mVpp) of the expected (400mVpp) output voltage as shown below:

Why in the second circuit is the AC peak to peak voltage a quarter of what was expected?

What is the best way to convert this diff pair into a single ended signal (a transformer is not suitable in this case, nor is dumping the -ve and only using the +ve signal)?

• You need to ask a question. Apr 27 at 11:53
• Note that you could simply connect your DC bias to the upper end of R12 (no need for R15) and get the same effect. Apr 27 at 11:54
• You have a simulator; you should be able to answer these questions on your own by looking at some of the other voltages within the circuit. See if you can work out why each voltage is what it is. Apr 27 at 12:01
• What differential pair that is? Analog audio, digital data? Apr 28 at 4:32

With the resistor values shown, gain is 1. Those 100mV signals in antiphase produce an input difference going between +200mV and -200mV (amplitude is 200mV), so I expect the output (without the coupling capacitors) to have 200mV amplitude also, not 400mV.

With those capacitors in place, forming high-pass filters, and input signals very close to their cut-off frequency, I would expect significant attenuation. This is why you are seeing an output attenuated by 50%, to 100mV amplitude.

If you have to use input capacitors, they will need to be significantly larger to mitigate this attenuation, or the network of resistors should be larger, to set the cut-off frequency much lower than expected input frequency.

As Dave Tweed said, you can offset the output by applying bias using R12:

simulate this circuit – Schematic created using CircuitLab

The blue boxed area above is the Thevenin equivalent of everything in the blue box below, allowing you to create a 1V source and 1kΩ impedance very easily:

simulate this circuit