Attenuator Frequency Response

Alex52381

Other Parts Discussed in Thread: OPA365

Hello,

I am designing an attenuator and follower stage for voltage measurements. I have designed the voltage divider to do a divide by 20 function. The max amplitude that I will see is 20V. Therefore, this would bring my voltage into the range of 1V for my ADC. During my design process, I read that it is important to introduce a capacitive divider to "even-out" the frequency response of the circuit at higher frequencies.

Figure 1: The circuit concept

Figure 2: The associated frequency response

My issue is that I can't figure out how the capacitor values are related and how they affect the resistors. This frequency response here indicates some kind of obvious RC filtering going on. What I want to do is provide the input with about 25 pF of capacitance, and also obtain a minimum bandwidth of 1MHz.

Is there something that the simulation is not showing me? Or is there some "real-life" constraint that I need to be wary of?

Thanks,

Alex

over 15 years ago

0 Arvind Raj over 15 years ago

TI__Intellectual 1425 points

Hi Alex,

Can you please let me know which op amp you are using for this application?

0 Alex52381 over 15 years ago in reply to Arvind Raj

Intellectual 810 points

Arvind,

Thanks for the quick reply. I don't have an op amp pinned down. I was considering the use of the OPA365 because its low offset voltage would preserve DC accuracy and because of its relatively low cost.

Thanks,

Alex

0 Arvind Raj over 15 years ago in reply to Alex52381

TI__Intellectual 1425 points

Hi Alex,

I dont see any need for the two capacitors and also dont understand why you need to 'even out' the frequency response at higher frequencies. Which application is this going into? Can you also tell me where you are planning to put the 25pF capacitance?

0 Alex52381 over 15 years ago in reply to Arvind Raj

Intellectual 810 points

Arvind,

My customer wants to be able to plug in a standard passive oscilloscope probe and monitor the waveform fed into it. The resistors provide the divide by 20 function, as well as providing 1M ohm input impedance. The capacitors must be designed such that they provide about 25 pF of input capacitance (based on my research on passive probe loading).

I haven't been able to find a good explination as to why the capacitors must be placed in the cirucit in that manner, as well as how to deal with their frequency response. This is the reason I'm posting here. My understanding is that the capacitors act as a voltage divider at higher frequencies, and that they are supposed to compensate for stray capacitance in the circuit.

Alex

0 Alex52381 over 15 years ago in reply to Alex52381

Intellectual 810 points

For all of those who read this looking for closure, after more testing, it looks like the resistive voltage divider was all I really needed. The frequency of my signals didn't go high enough for any other circuitry to matter to me.

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Attenuator Frequency Response