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I have an input (one-shot) pulse that has to be attenuated by 1/2 (half). The input pulse would be of variable width in between 1us and 100ms and would have a variable amplitude: one voltage level would always be ground (0V) while another level would vary within ±3V (the height would vary within ±3V). So I saw a compensated attenuator as explained here and tried to simulate it. Searching a bit online, I came to the conclusion that R1*C1 = R2*C2 should be satisfied for a square output.

After some trial and error, I could find a value (of R and C, also making sure R1C1=R2C2) that seems to work in LTSpice but I am not sure if this would work for the entire range. Also, the attenuator output is buffered by a voltage follower as well. Below is the subcircuit. 'V_out' is the original pulse and 'op_out_bias' is the output of the opamp. The output of the attenuator is (almost, the difference can't be seen at this zoom level) exactly the same as the output of the opamp.

So first of all, is this the right way to achieve attenuation (by a fixed factor) of a single pulse? How to formally (or informally) make sure that the circuit works for the above-mentioned parameter range (of width and amplitude)?

Note: I have tried different parameters (within the above range) in LTspice and so far it seems to work just fine.

enter image description here

Edit: I am adding the rest of the circuit. I didn't think it was that relevant (I apologize for that). The source is an output of an analog switch (SPDT, ADG619) that is being controlled by a pulse generator. The inputs of the switch are ground and a DAC output (didn't decide on the DAC) through another buffer. Below is the picture:

enter image description here

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    \$\begingroup\$ Why aren't you just using a voltage divider? Why do you need a capacitor at all? \$\endgroup\$ Jan 10 at 22:15
  • \$\begingroup\$ @ElliotAlderson Since it is a pulse with a very small width, so it is (I think) an AC signal. If I remove the capacitors, wouldn't the output get distorted? \$\endgroup\$
    – paulplusx
    Jan 10 at 22:53
  • \$\begingroup\$ I just tried removing the capacitors and it didn't make a visible difference (didn't test it extensively though) and I am getting a similar output. I am a bit surprised here, isn't it (pulse) supposed to be an AC signal? and according to electronics.stackexchange.com/a/272609/193677 shouldn't I use a compensated attenuator? – \$\endgroup\$
    – paulplusx
    Jan 10 at 23:35
  • \$\begingroup\$ What is your source? If it is a 50 ohm output I would consider buying a 6db attenuator: minicircuits.com/WebStore/dashboard.html?model=HAT-6%2B \$\endgroup\$ Jan 11 at 0:31
  • \$\begingroup\$ The problem with this question is the source is unspecified and designing a pad is all about impedance matching source, path and load to the ratio you need for full BW. It looks pretty bad at the input \$\endgroup\$ Jan 11 at 20:32
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The compensated attenuator is used when the goal is to minimize the loading on the source, which is what you need in a scope probe. Notice that the resistor values (in the example you linked) are very large.

A pulse would not typically be called an "ac signal" although it has some high frequency ac components. Your pulses are fairly wide and the edges aren't terribly fast so your situation isn't as tricky as designing a scope probe. A scope probe is also at the end of a relatively long cable so the probe has to compensate for the capacitance of the cable...you don't need to do that.

In your case you aren't worried about loading the source, you just want to reduce the amplitude of the signal. So you can use resistors with much smaller values, and your 10pF capacitors aren't doing much for you.

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  • \$\begingroup\$ Thank you for your answer. I am not worried about loading (in case you are talking about the current requirement) the source (the DAC output) since I am using another buffer which then goes through the switch to the voltage divider. I have edited the question to add more details. If there is something more you would like to add, please feel free to do so. \$\endgroup\$
    – paulplusx
    Jan 13 at 11:43
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How to make sure that my compensated attenuator would work for certain pulse widths?

The thing about connecting your compensated attenuator to an op-amp is that should take into account the input capacitance of the op-amp. For the LTC6228, that capacitance is 3.5 pF and so, ideally the capacitor across R4 should be 3.5 pF: -

enter image description here

enter image description here

But, there will be some variability in this value and that is why o-scope probes usually use a trimmer-capacitor in C4's position.

To avoid using a trimmer you can physically make C3 and C4 swamp the input capacitance value of the op-amp and you'd probably consider starting at about ten times 3.5 pF i.e. 36 pF for C3 and C4. However, this may have a poor loading effect on the signal you are trying to measure so, it can be ineffective.

is this the right way to achieve attenuation (by a fixed factor) of a single pulse? How to formally (or informally) make sure that the circuit works for the above-mentioned parameter range (of width and amplitude)?

It's got nothing to do with a single event or pulses - it's about trying to provide a circuit that "overcomes" parasitic capacitance (whether it's a PCB issue or an op-amp input issue) so that pulse edges are not unduly slewed or distorted. But you need to also ensure that what you do add as extra capacitance doesn't overly distort the actual applied pulse due to excessive loading caused by those extra capacitors.

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  • \$\begingroup\$ Thank you for your answer. I have edited the question to add more details. I am not entirely sure what you mean by loading here. If you mean the current capacity of the output then, as you see in the (edited question), the signal is coming out of a similar buffer (input to the buffer is a DAC output) through an analog switch. So would there be any problems if I go for 36pF capacitors or more specifically do I even need capacitors in the first place? \$\endgroup\$
    – paulplusx
    Jan 13 at 11:49
  • \$\begingroup\$ @paulplusx check the data sheets for the analogue switch and DAC. Evolving questions tend not to be tolerated. \$\endgroup\$
    – Andy aka
    Jan 13 at 11:56
  • \$\begingroup\$ I understand now that I should have mentioned the full circuit from the start. It was my own ignorance(or fault). I did learn something from your answer, so that is a plus. Thank you again. \$\endgroup\$
    – paulplusx
    Jan 13 at 12:13

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