I have a 50-ohm source which is the output from a photodetector, and I'm using a digitizer card which has 50-ohm inputs, reducing the signal voltage by a factor of 2.

The information I care about is encoded in the voltage amplitude. So, does this voltage reduction affect the final SNR negatively?

And if it does indeed affect the SNR negatively, what can I do to obtain maximal voltage transfer? Should I use a voltage follower?


2 Answers 2


Johnson noise density of a 50 ohm resistor is in the region of 0.5 nV/rtHz. So, given the photodetector voltage you're seeing, the bandwidth of interest, and the SNR you need (none of which you've told us) it's easy to check.

For example, if you are seeing 1mV rms signal, your bandwidth of interest is 1 MHz, and you need 40dB SNR ...

Noise voltage is 0.5 * sqrt(1e6) nV or 500nV or 0.5 uV in 1 MHz bandwidth. That's 66dB below your 1mV signal, well below your 40db SNR requirement, so ... no problem.

But if you needed 60dB SNR, then the resistive noise contribution is significant and you'd have to consider the noise budget carefully.


I have noticed your linked question.

50 Ohms symmetrical? That's a curious combination. Only some relatively low-end twisted-pair cabling with high conductor crossection and piss-poor RF properties has that. Any sort of signal transmission twisted pair would have 100 to 125 Ohms wave impedance. If it was 50 Ohms single ended coax, that would make sense :-)

If you can attach your "digitizer" straight to your photo-detector, by a section of cable whose length is insignificant compared to your required bandwidth, you don't need to load the receiving end of the transmission line with the line's nominal impedance. Just set your 'scope for a high input impedance and be done with it.

If OTOH you need to transfer the signal across some distance and prevent distortion of the pulse shapes, you need to impedance-match the transmission line both at the signal source and at the receiving end. The waveform is then razor-sharp, but your signal level is half the open-ended output voltage of the source.

As for "significant length" in the transmission line theory... 1 m of cable corresponds to a reflection coming back in 10 nanoseconds. That implies 100 MHz or thereabouts for some "resonating stub" effects (your stub would be open-ended). Looking at your scope screenshots, the signal's main period is about 400 kHz (2.5 microseconds). If you wanted a useful bandwidth of about 5 MHz to see some of the finer contours of the signal, i.e. 200 ns, that would imply on the order of 20 meters of cable as a "significant length" - but maybe you should start scratching your head at about half that length. Check out the oscillograms at the end of this web page, created using an impedance-matched pulse generator (a "reflectometric probe").

As for "noise contribution" versus voltage: I seem to recall that differential inputs based on BJT's benefit from a lower signal source impedance at the input terminals. Benefit in terms of lower own noise. Also, high-impedance (1 MOhm) signal conductors/inputs are more susceptible to electrostatic EMI pickup, compared to a 100-some Ohm transmission line.

I mean to say that you should not be too shy of properly impedance-matched low-ohm transmission lines, if your actual goal is precise information transfer, rather than cosmetically maximal voltage transfer.

When I need to bring a high-speed signal to my scope with reasonable precision, I put the high-Z adjustable probes aside, and connect the coax transmission line to the scope through an insertable 50-Ohm coax load that "terminates" the line (with pass-through for the coax transmission line).

ADC circuits natively tend to have relatively low-impedance inputs (I even seem to recall ADC's with a differential current input). But if yours is some more complex part, maybe with a programmable gain amplifier on chip to condition the input, and nominally "voltage" input flavour, then the inherent character of the actual ADC stage is shielded away from the outside world.


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