# 50 ohm terminator — voltage divider?

OK, obviously I'm misunderstanding something.

I have a RedPitaya, which is an oscope+generator. Its output impedance is 50$\Omega$.

If I send a signal from its output to its input (which is 1M$\Omega$) and use a BNC T with a 50$\Omega$ terminator (to match impedance), I expect a signal drop of about half, since this is a voltage divider: 50$\Omega$ out, in series with 50$\Omega$ in, with the 1M$\Omega$ in parallel not affecting the signal much.

However, I do not see this on the trace. With the terminator, a 1Vpp (nominal) signal is almost exactly 1Vpp (1.003Vpp). Without it, it is about 1.05Vpp.

This makes no sense to me. The only thing I could think of is perhaps the RP "knows" when the terminator is connected and is doubling the voltage trace? Or, perhaps I don't understand how terminators work.

• Weird... try 10 ohms, 100 ohms on the terminator... maybe 1 kohm. Find out how "smart" it is. – George Herold Jan 16 '15 at 19:17
• What frequency signal are you generating, and how long is the cable between the generator and 'scope? – The Photon Jan 16 '15 at 19:22
• @GeorgeHerold I only have 50ohm terminators. – Paul L Jan 16 '15 at 19:37
• @ThePhoton This applies to any frequency. The cable is short (<1meter) and very good and shielded (LMR) – Paul L Jan 16 '15 at 19:38
• It sounds like the generator has a low-impedance output, but the specs on the website are unclear. Is there a schematic of the generator circuit available? – The Photon Jan 16 '15 at 19:47

The signal generator we use in our university behaves in a similar manner: if you load it with $50\Omega$ the output waveform amplitude is exactly what you dial in the SG. That's because our signal generator has a settable load impedance: you can set it to $50\Omega$ or to $\text{HI}_\text{Z}$, the latter meaning "high impedance".

What does the SF actually do? When you set the load to be $50\Omega$ it doubles the output amplitude so that on your line you have what you set, but if you set the load to $\text{HI}_\text{Z}$ the output amplitude is what you dial in.

You can see if that's your case removing the termination: since the scope input impedance is very high, and negligible as you suggest, you should see a doubled wave. If you can dig up the setting you can try to set it to $\text{HI}_\text{Z}$ and then connect the cable with the termination: you should see a wave that is only half in amplitude.

If you are not sure that this doubling/halving is done SG side but you're afraid it's something scope side, that is very, very unlikely, take your multimeter and measure the output rms voltage.

• You understand, don't you, that measuring the RMS voltage will not produce the same result as measuring the P-P voltage? – Hot Licks Jan 16 '15 at 23:14
• Yes I do, I don't see any problem though. – Vladimir Cravero Jan 16 '15 at 23:25
• downvoters should please explain the reason(s) they are downvoting. – Vladimir Cravero Jan 17 '15 at 13:33
• You say "you should see a doubled wave". This is untrue. When you remove the load all bets are off. – Hot Licks Jan 17 '15 at 13:38
• When you remove the load all bets are off iff the frequency is high enough. With reasonably low ff the wave should double indeed. Reasonably low is under 100kHz for me. If removing the load does not produce any change, as op says, the load is probably open AND the frequency is low. – Vladimir Cravero Jan 17 '15 at 13:42

The manual of this Redpitya says:

"Note: The output channels are designed to drive 50 Ω loads. Terminate outputs when channels are not used. Connect parallel 50 Ω load (SMA tee junction) in high impedance load applications"

Which sort of suggests that the output is indeed 50Ω. Generators with a low output impedance are also very rare and virtually always switchable between 50Ω and low-Z.

Have you ever checked this terminator with a multimeter? It is not that uncommon to find an open terminator in a school or lab environment, most of the time caused by incidental DC from a power supply.

This is actually pretty common. The signal generator output is calibrated assuming a 50 ohm load. So if you set it for 5 volts, the "raw" output internally is set to 10 volts. Since the output impedance is 50 ohms, you are correct in thinking that you will get 5 volts at a 50 ohm load.

If you remove the terminator, you'll see the apparent output double.

It's just a convenience, so you don't have to mentally compensate for the output impedance.

• Removing the terminator doesn't double the output. That is the source of my confusion. – Paul L Jan 16 '15 at 21:15
• You were right up until when you talked about removing the load. If the signal generator were a purely passive 50 ohm source then what you say would be true, but it almost certainly utilizes some sort of op amp on the output, and the load curve is hence nonlinear. When you remove the terminator all bets are off. – Hot Licks Jan 16 '15 at 21:25