# Envelope detector and DC voltage

I am currently working in the simulation of a physics project. One of the detectors inputs a voltage V1:

V1(t)= D+Acos(wt)

and tries to extract the amplitude. I read that a simple Envelope detector can do the job. The problem is that in my system sometimes D takes really big values and then the detector fails and returns a wrong value. I have very little idea on how to use a circuit in which this doesn't happen, although I guess that a high pass filter before the source could solve the problem.

How do I stop by system from failing to detect and return the wrong value?

• "One of the detectors inputs a voltage current", so it inputs power? Aug 10, 2017 at 13:07
• Eliminate D and use Cs = k*C1 where k is your bandwidth ratio from max to min. Aug 10, 2017 at 13:19
• Put that into an answer @TonyStewart.EEsince'75, it looks like it's better than my answer if you put some meat to the bones. Aug 10, 2017 at 13:20
• What? I dont understand, could you elaborate a little? Aug 10, 2017 at 13:24
• @Victor That's exactly why I want him to turn it into an answer so he can elaborate. Aug 10, 2017 at 13:29

Using a RC high pass filter to get rid of the DC component is the way to go. And that is the easiest solution. Just two components.

Why do you want other solutions? Or rather, what kind of other solutions do you want / expect?

• Thanks! I wanted a circuit that worked with the least number of differential equations, as I only need to simulate the system connected to a bigger project. But i guess this is ok. Aug 10, 2017 at 13:11

you don't need differential equations to solve a DC blocking HPF with a rectifier.

## First the HPF , LPF response.

• The low frequency cutoff affects the step response time Ts and its associated frequency response where ω=1/T. and adding a series cap, Cs, with some known source impedance Rs and parallel load resistance, Rp then
• Ts= (Rs+Rp)Cs
• The high frequency cutoff or low pass filter LPF has the same equation for ω=1/T and parallel load cap Cp, Rp gives
• Tp= (Rs+Rp)Cp

Now if you used matched R source and load eg 50 ohms then midband loss is -6dB and breakpoints occur at -9dB (3dB down) then your bandwidth ratio or its inverse T ratio is Cs/Cp . e.g. for 3 decades of response Cs=1000Cp

If anyone sees a mistake, pls let me know otherwise if Ok pls upvote your concurrence.