# How is a 50Ω "back-termination" accomplished on a scope probe line?

When probing high frequency signals, the standard way to allow an arbitrary length of cable between the device under test (DUT) and the scope, is to make the scope 50\$\Omega\$ input impedance, and use 50\$\Omega\$ cable.

In the ideal world, that will be good enough, Because the cable is terminated by the scope correctly, no reflections will occur at the scope, so no reflections will make it back to the driven end of the cable. The input to the cable will present a 50\$\Omega\$ load to the device being measured. We can choose to drive that load how we like.

However, in the real world, both scope and cable have a tolerance, and there will be some reflection. At very high frequencies, that could be quite large. Making the drive to the cable approximately 50\$\Omega\$ absorbs whatever does come back, improving the frequency response dramatically.

The 'tidiest' way to make this happen is to arrange for your DUT to have a 50\$\Omega\$ output impedance, to a connector. If the source of signals is low impedance, like the output of a power supply for instance, then a series 50\$\Omega\$ resistor will do nicely. If it's not convenient to use a connectered jig, then solder a 50\$\Omega\$ in line at the end of the cable.

Knowing what I did about matching, I was then surprised on my first day in a microwave lab to be shown how they probed circuits. A 50\$\Omega\$ cable, with a 470\$\Omega\$ carbon resistor soldered to the end. This was the -20dB probe.

Remember I said the input to a cable properly terminated by the scope looks like 50\$\Omega\$. The 470\$\Omega\$ resistor in series with this gives a roughly 10:1, or -20dB pot-down. It doesn't need to be matched at the sending end. It would have a flatter frequency response if it were, but another 50\$\Omega\$ resistor at the probe end would complicate the probe (obviously the cable ground is grounded to the circuit at the 'same' point, size matters!), and decrease signal or increase circuit loading for the same pickoff. For most measurements it was flat enough, and was the right price!

Let me take a slightly different approach. As Neil_UK has stated, an unterminated transmission line will produce reflections at the load end when driven by an AC signal. To get rid of these it's necessary to match the source and load to the characteristic impedance of the transmission line (cable). There are two simple ways to do this. The first (and most common)is parallel or load termination. This is done by putting a terminator at the load end of the cable, like

simulate this circuit – Schematic created using CircuitLab Note that the entire signal voltage appears across the termination/load resistor. When looking at the output of a power supply, this may not be too great an idea. A 12 volt supply, for instance, will dissipate nearly 3 watts in a 50 ohm termination.

There is, as I've indicated, another way. This is called series, or back termination, and it looks like

simulate this circuit

For best results, this requires an infinite load resistor. Since this is engineering, and perfection does not apply, any load greater than about 10 times the cable impedance will work. Larger is better, of course. If a 10x load resistor is used, you obviously get a 10% signal reduction, but this is not ordinarily significant.

This has the considerable benefit that, since the load resistance is very high, not much DC power is drawn, and in this case it's the high-frequency AC signals that matter.

The absolute best results are produced using both methods, series AND parallel termination at the same time, and most high-speed function generators will use this. To see this, take a function generator and connect it to a scope. Now switch the scope to 50 ohm input, or put a 50 ohm load on the output, and the output will drop by half. It does produce a 50% signal reduction, but as long as you know about it in advance you can compensate.