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I need to create a circuit which converts a pulse to very low voltage pulse (micro volts). This circuit output goes to an amplifier circuit. Right now, I created a circuit using a voltage divider and op-amp (AD8512) HERE IS THE LINK. So I can change the input amplitude and output amplitude varies. This circuit is working fine in actual hardware as well. This Circuit behaviour is fine:

input --> output
5V --> 151mV
2V --> 61mV
500mV --> 15mV
100mV --> 3.46mV

My pulse generator can not go below 100mV pulse but I need for example a few micro-volts for testing. I need to test circuit with a pulse having amplitude from 1uV to 1mV. Is there any better circuit for such kind of voltage change.

Update: Circuit input and output pictures added. R1 = 100KR and R2 = 100R

Voltage Divider Input: -3V pulse enter image description here Voltage Divider Output: enter image description here

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3 Answers 3

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If you make a resistor potential divider using a 1 kohm and a 0.1 ohm resistor you can convert a pulse of peak amplitude 1 volts to an output pulse of peak amplitude of 100 uV. Instead, if you used a 10 kohm resistor your output amplitude peak would be 10 uV. If you want a 1 uV output pulse use a 100 kohm resistor and 0.1 ohms.

If you need variations of these make several - they can all operate parallel driven together because the input impedance is in the region of tens or hundreds of kohm. The output impedance is very low (circa 0.1 ohm).

Don't use an op-amp for this - it will only disappoint unless you are prepared to pay tens of dollars or GB pounds.

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  • \$\begingroup\$ Hi, I tried the circuit with R1= 100KOhm and R2= 10 Ohm. Now the issue is for input pulse with 3us PW, 10ns Rise and Fall time, the output on rise and fall time is only glitch but no pulse shape. I can increase the pulse rise and fall time for testing. \$\endgroup\$
    – user777304
    Jun 1, 2020 at 13:27
  • \$\begingroup\$ How did you measure the output? What was your testing measurement schematic? What oscilloscope did you use? What probes did you use? \$\endgroup\$
    – Andy aka
    Jun 1, 2020 at 13:30
  • \$\begingroup\$ First I tried to measure with amplifier circuit and the output shows high power of same glitches. After that I replaced the R2 with 100 Phm and test the output of voltage divider circuit with 5V input. Same glitches has been observed. Scope = wavepro 7300A. Probe = Lecroy pp005A 500MHz \$\endgroup\$
    – user777304
    Jun 2, 2020 at 5:43
  • \$\begingroup\$ Can you also please tell me the effect of resistors in voltage divider format on pulse shape. \$\endgroup\$
    – user777304
    Jun 2, 2020 at 5:45
  • \$\begingroup\$ Well, there's nothing more perfect in the EE world at dividing voltages than resistors so maybe show an oscilloscope picture of input and output pules to make comparisons. \$\endgroup\$
    – Andy aka
    Jun 2, 2020 at 9:13
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I presume that last one is meant to be 3.46mV

5 / 0.151 = 33.113

2 / 0.061 = 32.787

0.5 / 0.015 = 33.333

0.1 / 0.00346 = 28.902

So you're after approximately a divide by 33, For this your circuit suits, while you do not tell us how short the pulses need to be, you may look into a variable attenuator if the frequencies are rather high.

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  • \$\begingroup\$ Pulse can be any from 1 us to 100 us for testing only. \$\endgroup\$
    – user777304
    May 16, 2020 at 11:08
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Given you use an opamp buffer, I assume your frequency is not fast.

If you want to preserve the pulse risetime, then use the method used in scope probes, and have the two RC time constants be the same.

A number for parasitic capacitance across a resistor is 0.5pf. So for some accuracy, use addition C at least 10x that.

If you have 100,000 ohm and 10 ohm, then place 10 pf in parallel with the 100,000 ohm resistor. And place 0.1uf in parallel with the 10 ohm.

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