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OP Amp with 50 Ohm Termination and 50 Ohm Load

Hi, I'm designing an OP amp circuit. I would like to terminate the output with 50 ohm because the device will be a measurement device with an oscilloscope or DUT on the iutput.

Left side is basically my PCB right side is outside of it to the DUT/oscilloscope.

Because of the voltage divider from the 50 ohm termination and 50 ohm load, I thought maybe an Idea would be to add a buffer/UnitGain-OPAmp on the output of the PCB. (because input high-impedance, output low-inpedance -> "no" voltage divider)

Now my question, what does the signal from the output see if it looks to the direction of the OPAmps. Does it see 50 ohm or more/less?

Thanks

winny
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R098
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  • Which point in the circuit is the output of your device under test? You should annotate 3 separate regions. They are 1) the DUT, 2) the imagined, but not yet built maybe helpful auxiliary circuits 3) the measuring equipment. I must guess where the region borders are. Others seemingly have already done it. What's the freguency range under your interest and how long signal cable you are going to have in the input of your measuring equipment? What kind of cable it will be? Know that 50 Ohm resistor load will pull the outputs of ordinary opamps to knees if the signal is more than tens millivolts. –  Jan 31 '22 at 22:09
  • It was actually on purpose that I just showed the basic concept of my circuit. I wanted the general explanation how to tackle this problem. Termination was always a thing that I understand in theory, but in practice I have some problems with it. – R098 Jan 31 '22 at 22:29
  • Here some answers to you questions if you want to check that I did everything correct. 1) the right side, next to the dotted line. 2) "I can upload it if you want", but it's basically it. I just measure the Impedance of the DUT over the 50 Ohm. (Hard to explain but here: https://www.omicron-lab.com/products/vector-network-analysis/accessories/b-amp-12-amplifier that's the thing you want to build. 3) Bode100 Network Analyser. Frequency Range: DC-50MHz. Length of Cable: ~50cm. Thanks for noticing that the 50 Ohm would pull some OPAmps on knees, I think my OPAmp should be alright. – R098 Jan 31 '22 at 22:30
  • It's the THS3491. It even says "into 100 Ohm Load" in the Datasheet, and you can see it in the example Circuits. My output is ~11,3V. – R098 Jan 31 '22 at 22:31
  • If you have Bode 100 analyzer and you are going to measure impedance with it you should use the methods, auxiliary devices and connections shown in Bode 100 application notes and here https://youtu.be/T2OqewIUL3M. Inserting opamps or other your own (=not shown in application notes) buffers between the analyzer and DUT is pure nonsense. Very few opamps make anything except exist and consume current if your signals are in MHz range. –  Feb 01 '22 at 00:17
  • Figure 25 of datasheet ... see output impedance versus frequency. – Antonio51 Feb 01 '22 at 08:19
  • The reason why it is seen "into 100 Ohm load" is obvious. One need a 50 ohm for the output (Output impedance < 1 Ohm ... +50, then "outputs Z" is ~50 Ohm) until 100 MHz ... and then the 50 Ohm load ... – Antonio51 Feb 01 '22 at 08:22

3 Answers3

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You can do all of those things, but you should not.

schematic

simulate this circuit – Schematic created using CircuitLab

If you are intending your measurement device to ultimately drive a 50 ohm load, then it should have a 50 ohm output impedance.

This means both ends of any 50 ohm coax you use to connect them will be properly matched.

It means that your device output amplifier will be driving 100 ohms, not 50. It will also be short-circuit proof.

Most generators are built like this. They tend to be calibrated assuming the 50% signal reduction due to the 50 ohm load. A standard quick test of the output impedance of such a generator is to remove the 50 ohm load and see if the output voltage doubles into a high impedance load.

Neil_UK
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  • Aaah, now I understand why it says "at 100-Ω loads" in the Datasheet of my Amplifier. (And than the examples with 50 Ohm Termination Resistor, and 50 Ohm Load.). Also the Short-circuit-Protection makes sense, didn't think about this. Ok than you just have to take this into account and it's common practice to just design it with dobble the gain. Interesting – R098 Jan 31 '22 at 21:42
  • Is this also the reason why we use parallel Termination on the Input (after the cable) and Series Termination on the Output (bevor the cable). That it's seeing everytime the 50 Ohm load before or after the Cable? Or is there a better explanation when to use Parallel Termination and when Series Termination? Because I thought to myself, "why not just use parallel Termination before and AFTER the cable, but it's most likely not correct". – R098 Jan 31 '22 at 21:47
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    @R098 An amplifier output is low impedance, it needs a series resistor to match to the cable. If we have an impdeance controlled bus, for instance a PCI backplane, then that's parallel terminated at both ends, and current driven, that is from a high impedance. – Neil_UK Feb 01 '22 at 05:18
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Answering the direct question first - The feedback means that the op amp can output a fairly large current in order to maintain a desired voltage at the output node. This means that the effective impedance at that node is near zero, and likely even less than the open-loop output impedance of the op amp itself.

However - the point of 50-ohm termination is to impedance-match to transmission lines, which are commonly made to a standardized 50 ohms in various forms such as microstrip, coaxial, etc. To that end, you actually want a 50-ohm output impedance which you can achieve by not including the op amp buffer at all.

You do end up with a voltage division, but you also avoid dealing with issues such as signal reflections.

In fact, it's common that some function generators can be configured for 50 ohm loads in this manner - they end up with 50-ohm output impedances, and their displayed values take this voltage division into account - when it says "1V" on the screen it's driving its output in a way which would deliver 1 V into a 50-ohm load. A high-impedance measuring device would display 2 V (while creating signal reflections) in that scenario.

You may want to consider simply using 50 ohm matched impedances throughout your system and then correcting for any voltage division numerically.

nanofarad
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  • Usually you can toggle a function generator between 50-ohm and high-impedance output. I'm not sure what exactly that changes other than the displayed numbers, though. – Hearth Jan 31 '22 at 21:17
  • @Hearth Edited. I don't recall ever toggling the ones I used in undergrad *out* of 50 ohms, if it was even possible on the models in our lab. – nanofarad Jan 31 '22 at 21:19
  • It may be something that's only become common recently. – Hearth Jan 31 '22 at 21:20
  • Ah okay, so you just live with it and just have to design it for 50 Ohm with double the gain. Okay :/. As you said, now I can remember that the setting in the Oscilloscope really change the displayed number. Do you also have a explanation why when use Series Termination on the Output and not Parallel Termination? – R098 Jan 31 '22 at 21:54
  • @R098 Parallel termination is not useful - the op amp is closer to a voltage source than a current source. A parallel termination resistor is useful when the output has a very high impedance, much greater than 50 ohms. – nanofarad Jan 31 '22 at 21:55
  • Can you explain it more detailed please why the current source and voltage source play a role in this? Or do you have a link to a explanation? It matches up with the Theorie (Input - High Impedance-Side - Parallel Termination) (Output - Low Impedance-Side - Series Termination). But I don't get the reason what does the current and voltage do that you deside that it is like it should, sorry . – R098 Jan 31 '22 at 22:16
  • @R098 A *series* resistor in series with a *voltage* source drops some voltage. A series resistor in series with a current source does nothing except waste energy - the current is still whatever the source outputs. A resistor in parallel with a voltage source does nothing except waste energy since the voltage is the same regardless, while a resistor in parallel with a current source actually changes the effective output impedance by shunting some current. Reading about Thevenin and Norton theorems/equivalents can help clear this up. – nanofarad Feb 01 '22 at 01:56
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Now my question, what does the signal from the output see if it looks to the direction of the OPAmps. Does it see 50 ohm or more/less?

The 50Ω load looking back on 3, the 50Ω load sees R1 and R2 and the opamp output. The output could be modeled as a voltage source with series resistance. In many datasheets the output impedance will be listed or this can be calculated from short circuit current. In many cases opamps can be found with output impedance will be tuned to 50Ω.

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Source: https://www.electronics-tutorials.ws/opamp/opamp_1.html

Voltage Spike
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