I have a differential signal ranging from 0 to 2 MHz and need to transform it into a single-ended signal.

I found out that there are 3 methods:

  1. Transformer
  2. Amplifier
  3. simply ground the negative part with a matching load

How can I decide which method to use? I mean I would just take the load-to-ground method but I guess it's not that simple to decide.

Edit: The transformer method would fall out, because DC cannot pass, although this would be no problem.

  • 1
    \$\begingroup\$ Look at this answer on common mode noise and why people use differential pairs to avoid it. electronics.stackexchange.com/questions/79752/… That should show you one reason you might chose option 2 (or a differntial receiver). \$\endgroup\$ Nov 24, 2017 at 14:04
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    \$\begingroup\$ I don't get it. You write that a transformer doesn't work because it blocks DC, but then you write that it is not a problem? \$\endgroup\$
    – pipe
    Nov 24, 2017 at 14:27
  • \$\begingroup\$ Yes this was confusing, sorry. The signal ranges from 0 to 2 MHz but it would be no problem if, let's say, the frequency up to 300 kHz would be not there anymore \$\endgroup\$
    – OcK
    Nov 24, 2017 at 14:29
  • \$\begingroup\$ Given that last comment, you should limit your signal to its useful bandwidth, unused bandwidth = more noise down the line. \$\endgroup\$
    – Sclrx
    Nov 24, 2017 at 15:20
  • \$\begingroup\$ Grounding the negative part either works or not, depending what's driving the signal. A transformer (or a floating balanced driver) you get the full differential V on the other output. A well protected buffer : you get half the differential voltage on the other leg. An unprotected buffer : you get a blown output buffer. \$\endgroup\$
    – user16324
    Nov 26, 2017 at 14:25

3 Answers 3



If your signal is a single, or narrow band, frequency this method is fine. However, anything else and you introduce the transformers frequency response into your signal.

3.simply ground the negative part with matching a load

This method is ok, IF the signal source is close to the receiver. Since you are are throwing away the common mode noise reduction that is the reason the signal was sent differential in the first, this method should be limited to part to part connection locally on a PCB.


A differential amplifier is ultimately your best choice. Not only will it be easier to control the frequency response of the receiver, but you also take full advantage of the common mode noise removal. Further, this method decouples the input signal from wherever the signal is headed to giving you the appropriate output impedance to feed the next stage.

  • 1
    \$\begingroup\$ Thank you for your detailed answer. Because I am going to bring the signal on another PCB with a cable, the amplifier solution will be the best choice! \$\endgroup\$
    – OcK
    Nov 24, 2017 at 14:25

DC to 2Mhz is differential amplifier territory, they can have excellent phase and gain flatness over that sort of range combined with reasonable CMRR (This tends to fall with frequency). Have a look at the high CMRR parts from THAT Corp, they are aimed at audio but IIRC have plenty of GBP and usefully bootstrapped common mode impedance.

300KHz to 2MHz is easy (and compact) for a transformer to handle, excellent CMRR and common mode range, and over a mere decade of bandwidth phase and gain should be flat. These are however best in fairly low impedance circuits, so not what you want if trying to measure a high Z source. Minicircuits have what you need among others.

The resistor to ground is usually something you see at the line driving end rather then the receiver as it buys you no CMRR at all in the receiver while helping with the impedance balance in the driver.

What to do depends on your design parameters (Such things as source impedance, common mode range, required CMRR...).


To add to existing answers, a typical differential signal has the same signal with opposite phase on both signal lines. Your solution 3 has the disadvantage of giving you half the signal amplitude which a differential signal would see, because you only have one signal line.

Professional audio has a workaround for this, because audio applications frequently need to drive single-ended sources such as guitar amplifiers. One half of the differential signal has the actual signal, and the other half is simply driven at 0V - that's driven by the same output impedance (typically the same op-amp) as the "signal" half, not simply connected to 0V. The screen of course is connected directly to 0V. A differential input will subtract the 0V reference signal line from the signal line, which removes common-mode noise from the signal but does not otherwise affect the signal amplitude. A single-ended input will simply see the signal line and the 0V, which does not allow noise to be removed of course, but does not change the signal level.

This is fine for your signal. It does have a downside though (as does everything). Equal-and-opposite signals on a twisted pair radiate very little, which reduces their effect on equipment around them. A single "hot" signal line does not have this advantage, and this could be an issue as you head up into MHz territory.


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