# question about envelope detector filter circuit

Discussion in 'General Electronics Chat' started by count_volta, Nov 29, 2009.

1. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
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P.S. (I reposted this in General Discussions, because nobody looks in the RF forum, and after all this is a circuits related question at the very heart of the matter.

Hey guys, so I was designing an envelope detector in the lab. I also need to do a Pspice simulation of it. The situation is as follows.

Carrier signal: 1Mhz

Message signal: 1khz

This is the circuit I must use.

The input function generator also has a 50Ω internal resistance, so I included that in my simulation.

Here is the circuit I simulated.

R4 is the function generator internal resistance.

So I have a small dilemma here. Let me explain.

The carrier wave, and hence the modulated signal has a frequency of 1Mhz. That means its period is 1μs.

The RC time constant is τ= RC. The resistance R is R4 + R6 in the Thevenin equivalent. Or 60Ω. The capacitor is 3μf. So thus τ = 0.18ms. Times this by 5 = 0.9ms.

So it takes the capacitor 0.9ms to be fully charged. This is much longer than the period of the incoming signal. The capacitor is not being fully charged.

So okay lets say we want to fix this problem. We can't change the 50Ω internal Function Generator resistance. We can alter R6 and the capacitor.

Lets say we make R6 100Ω so the total resistance R is 150Ω and the capacitor 1.3nF.

Now the time constant is 0.2μs. Times this by 5 and we get 1μs. So now the capacitor is fully charged in one period of the incoming modulated signal. That is one problem solved.

But there is another problem This is a low pass filter, whose job it is to recover the message signal at 1khz.

The cutoff frequency of this low pass filter is Fc = 1/2πRC = 1/2π*150Ω*1.3nF = 816,179Hz. Some LOW pass filter right.

So that is the dilemma. I have to either alter the time constant, or alter the center frequency. But if I do one, the other is affected. Believe me I tried many different values of resistances and capacitances and nothing works.

Here is what Pspice gives me. As you can see the envelope detector isnt detecting much. The red signal is the voltage at the output.

Can you guys please help. Just what is more important here, the time constant or the cutoff frequency? Or did I get something else wrong?

2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
I think C1 is way too large. It's function is similar to a power supply's, it shorts the high frequency while letting low frequency through, anything less than 20Khz (actually 10Khz) with AM radio. R6, on the other hand, is way too small.

You are basically simulating the guts of a crystal radio.

You are also not taking the effect of C1 reactance to the input impedance. This circuit will have a complex impedance function that will not be easy flattened.

3. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
Yep, but there are formulas to determine exactly what it should be. As I said before there is the time constant to worry about, and the cutoff frequency of the filter. But if I adjust one paramater, the other parameter suffers.

So the dilemma is, either the cutoff is extremely high, or the time constant is too long. How to balance these out? I think this is what they want us to learn in this experiment. I tried for hours to figure it out. I come to you guys as a last resort.

4. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Try upping R6 to 1KΩ. Your input impedance for 1Mhz will be unchanged, since C1 is a dead short to RF. Also drop your C1 value, say to 0.01µF.

5. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
A 1N914 might be a bit slow, too. Try a change to a 1N4148 if you have to stay with silicon. The best diode for that kind of detector, though, is germanium. The 1N34 is very good for an elementary AM detector.

6. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
Bill I'm going to have to trust your years of experience, but how in heck?

There must be a reason for why those values work. Can you explain please?

When calculating the cutoff frequency Fc = 1/2piRC, which R do you use? The 50Ω R4 or R6?

I remember reading in the ebook that τ= RC only works for series RC circuits and if you have anything other than a series RC, you turn it into one by determining its Thevenin equivalent. I will have to assume the same is true when determining the cutoff frequency.

So then the R we use in the calculations is R4 + R6 in series, if we take the capacitor to be the load.

So now using the values you suggested (which work), R = R4+R6 = 1050Ω

Thus τ= 1050Ω*0.01μF = .0105ms

As I said before the period of the 1mhz input signal is .001ms. Which is actually pretty close. So okay the capacitor is mostly getting charged.

Now the cutoff frequency, Fc = 1/2πRC = 1/2π*1050*0.01μF = 15khz.

So you see, the cutoff frequency is still high as heck, so how is the filter isolating the 1khz message signal here?

Btw, the troughs of the wave are a bit flatter than the original message signal, I wonder if its possible to get them closer to the 1khz message signal?

7. ### Wendy Moderator

Mar 24, 2008
20,772
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You are looking at the difference of experience vs. theory. I was widening the gap between the respective impedance. RF and audio are very different beasts, make the differences in the circuit between the critters so they both go their different ways. I did a quick calculation for the 3µF cap impedance at 1Mhz, but the rest was off the cuff.

Did you change the diode as suggested? I didn't think of it, but I suspect it will make a difference too. If you have a shottky diode in your library you might try that too.

8. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
I can't really use another diode because that diode is part of the assignment. In fact we simulated this actual circuit in the lab. We got a pretty good result also but I forgot which values we used.

Okay, but as important as experience is, I really need to understand why this works in terms of time constants and filter cutoff frequency because that is what my professor wants me to understand in the assignment.

And besides, I want to know myself. If this is a low pass filter, why do we get a good result with a cutoff frequency of 15khz when the message its supposed to filter out is 1khz?

9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Going through your match (and simplifying everything) the roll off of a RC low pass filter using 3µF and 50Ω is 1Khz, not 15Khz. The thing you have to remember about all filters, the roll off frequency is ½ the power, it is already well attenuated.

I picked 20Khz because this circuit strongly resembles an AM radio detector.

10. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
It is an AM radio detector.

How do you calculate the cutoff frequency correctly?

11. ### Wendy Moderator

Mar 24, 2008
20,772
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I didn't. My calculation was 3µF for 1Mhz was 0.05Ω. I figured 100 times smaller. Actually just went through the math again, I would have used 0.1µF cap if I had calculated it (1.6Ω). As for the 1KΩ, I knew the value had to be bigger, and I like 1KΩ.

12. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
Bill how do you get a real number for the capacitor impedance? We were taught that its Zc = 1/jωC, meaning j is always in the impedance.

Also, again experience is awesome, but my professor won't like it if I say here is the result of the envelope detector based on the experience of another engineer. LOL.

13. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Like I said, I over simplified. Zc = 1/(6.28fC) (same equation, just expressed differently, and calculator friendly).

I didn't bother to figure the parallel impedance.

BTW, you're not seeing an engineer's approach here, but a technicians.

14. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
So can someone please explain to me how to calculate the capacitor and resistor values which Bill got by intuition. I will have an exam on this soon, and I need to be able to do this. My professor didn't explain. I'm supposed to figure this out on my own, and I tried. You can see what I tried in the first post. It didn't work.

I want to be able to calculate the values that give me this. http://img121.imageshack.us/img121/3806/yjjk.jpg

I know what the values are. R = 1kΩ and C=0.01µF. But I gotta obtain these values myself from analysis of the circuit.

15. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Just curious, have you done the math for 0.01µF and 1KΩ? I calculate 15.9Khz roll off. Given that 3db is not necessarily a good freq response point it matches nicely with 10Khz freq response.

I explained how I got the 0.01µF, it was basically 1/10 of the 50Ω and severely rounded off. The 1KΩ was a lucky guess, but it does work out to the roll off we were talking about.

The thing is, you want just enough capacitance to do the job and no more, goes with separating those critters I was talking about. 0.01µF is about right for the RF, any less it wouldn't do the job, any more it would be major over kill.

I think you could justify those numbers pretty easily in your assignment based on the rationals I've just given. BTW, I keep a simple scientific next to my keyboard just for these kind of problems. The design problem is pretty open ended, you could use the my assumptions, tweak the values using the assumptions, and it still work as well.

So to maintain input impedance C1 needs to be around 10% or less of what the input impedance is. If 10% seems to big make it 1%. Then calculate R according to roll off.

Instead of using image shack, which has lots of ad-ware and isn't a clean site to use, why not use the local albums that are part of AAC? Less advertisement (none actually), and easy to use as part of your existing AAC account.

16. ### Audioguru New Member

Dec 20, 2007
9,411
896
An AM radio has a cutoff frequency of about 3.5kHz, not 20kHz.

17. ### count_volta Thread Starter Active Member

Feb 4, 2009
435
24
Okay my professor explained it. I totally get it now.

What people never explain is that the diode in the envelope detector affects the frequency. You don't usually think of a diode as something that affects frequency but it does. It introduces the frequency of the original message signal back into the spectrum of the modulated signal.

And then you use the low pass filter to filter that frequency out. And you should use Fc = 1/2piRC for the cutoff frequency. Then for my particular signal Fc should be somewhere in between the message signal frequency and the carrier frequency, once you know that you solve for R and C.

18. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
295
Using an excessive RC time constant in an envelope detector results in a particularly obnoxious form of distortion known as diagonal clipping...sometimes referred to as "failure to follow" distortion.

The optimum RC time constant is dependent on the modulation bandwidth as well as the modulation percentage.

In general, you want the RC time constant to be no longer than twice the reciprocal of the highest modulating frequency. (rule of thumb...not chiseled in marble!)

eric

19. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Actually it is 10Khz, you are confusing it with telephone, which is 3.4Khz. Maybe Canada uses different specs.