Improving my best diode detector

More complicated than using dropbox.

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Rick C

I'll keep that as a possibility.

Earlier in our discussion there was a question about why have the termination resistor. I turns out to be important if the circuit is capacitively coupled. It doesn't work without a termination. I'm now using an 11K termination after the coupling capacitor, with a 500 ohm source resistance. I don't know how the termination value affects circuit operation.

Mikek

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• Neat,,,just as expected it closes the site. Very useful.

• But i did, tried a number of things, nogo. However, once in a while it works.

>

Actually, i think the bias does not need subtraction; in the "proper" circuit, it helps.

Value should affect the observed linearity, especially if too high.

Not sure what you did. I can make the signup overlay go away and then I see the full page.

What did you do differently to make it work?

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Rick C

If you look at the schematic he is using it is already doing a pretty good job of biasing the diode. He doesn't need the low end, so there is no reason to bias the measurement out of the low end, it's already doing that just by the way it is used.

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Rick C

I can't say I follow the need for a termination resistor after the coupling cap. What does it do for you other than raise the corner frequency of the high pass filter?

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Rick C

I'm putting a cap between the diode. Biasing is an option, but. I'm off the detector for a while, I need to get a flat response out of my amplifier. Here's the amp as wired, measurements are without the diode detector connected.

Up until recently, all my testing has been at 1MHz, yesterday I ran the SigGen up to 10 MHz. At 10Mhz the detector had much more output than at 1MHz. I decided the 27PF need to be increased.

about 35% higher at 10MHz than at 1MHZ. I have seen circuits where they use an inductor where the termination resistor is. (68 ohm in the article)> Maybe this would block the high frequency and equalize the response.

I have another problem, my amplifier is down about 17% at 10MHz. So, I'd like to make that flat before I get involved working on the detector. I just swapped in a 2N3866 for the BC547b, it didn't help.

the increase of the detector. Mikek

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I'll get back to you on that, When I removed it earlier, the detector output dropped to zero. I have disassembled the detector, when I put it back together, I'll verify it. Btw, I have tried several emails to you, they all bounced. I have other gmail accounts I send to, they are fine, it is only your address that

Mikek

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I'm curious, what is the AC voltage on the emitter of Q3? It should be about the same as the base voltage. If that is correct, I don't get why the gain is 4 rather than 500/220. Oh, I think it would instead be

500/(220/2) because the other 220 ohm resistor is AC coupled and in parallel with the resistor in the emitter leg. Then the gain should be over 4, about 4.5. What is the AC voltage on the point between the 220 ohm resistor and the 3.47 kohm resistor? Certainly 10 uF should be nearly invisible to 1 MHz.

How about the detector input? Is your amp level between 1 and 10 MHz?

The use of the inductor is to DC ground that point. I don't think you need to do that.

I can't follow what you are talking about, which parts are which. I would try to make the amp flat and then figure out how to deal with the detector.

First, which stage of the amp is not flat? Have you measured the output of each stage to see where the response problem is? What are the AC readings at the base and emitter of Q2 and Q3 at 1 and 10 MHz?

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Rick C

Also check for AC signal on your power rail. Why is there no cap larger than 100 nF?

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Rick C

The problem is not gmail, it is my arius.com server rejecting your emails because they are on spam lists. You need to deal with the spam listing problem. I've given you info about that and it is in the bounce reply you get. btw, you shouldn't be emailing me at gmail. I gave you an arius.com address to use.

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Rick C

Yes, you have that correct, the 10uf cap AC couples the other 220 into the circuit.

What is the AC voltage on the point between the 220

with 14Vpp input that point is 3.15Vpp.

My amplifier is down about 17% at 10MHz.

T1 and Q2 output is flat. Q3 rolls off between 1MHz and 10MHz.

Yes, Q3.

The problem lies with Q3, the input is flat the output rolls off.

Mikek

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I tried both. I looked at what you posted, I had 5 items from 2008 and before. Do you think that is the problem?

Here's what the page gave me.

Summary information for me. Note: Times shown are for the latest entry only! Found 4 network entries and 0 host/domain entries.

Problem Entries, (listings will cause email problems.)

5 "Escalated" entries [16:30:03 21 Nov 2008 GMT+00].

Usage classification (only important if you run your own mailserver.)

1 "DUHL" entries [13:59:07 19 Sep 2004 GMT+00]. 1 "exDUHL" entries [14:05:20 19 Sep 2004 GMT+00]. Note: Active "exDUHL" entries mean that the IP/Network has been unblocked for some or all IPs from the DUHL. Mikek

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That shows your AC coupling is not very good. This is the bootstrap to reduce the input capacitance of this stage. You need to up the value of either the capacitor or the resistor or both. You didn't say what the AC reading is on the emitter, but it should be pretty close to 14 Vpp and both sides of the caps should be about the same AC voltage for the bootstrap to work properly.

The 220 ohm resistor on the base side of the caps can be increased if you reduce the 3.47 kohm resistor by the same amount. The limiting case is two resistors of 1.75 kohms giving an AC load to the caps of 8.75 kohm which is about 40 times the 220 they now see. This also increases the input impedance and makes the capacitance reduction work properly which may be part of your frequency response problem.

I don't know what "my amplifier" includes. Is that from the 14.14 volt input to the collector of Q3? Ok, I see later you say the first two stages are flat to 10 MHz. Where do you measure the response of Q2? Is this while connected to Q3?

Try the changes to the Q3 bootstrap and see if that helps the frequency response. Otherwise try proportionally increasing the values of all the resistors. If you double them all and the response increases, you are fighting something fundamental with the circuit.

To isolate the circuits, try driving the Q3 input directly without the rest of the amp connected. The Q3 bootstrap can impact the loading on the Q2 stage output.

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Rick C

I don't know, but doesn't the page indicate a way to get taken off the blacklist? Either way you should complain to your hosting provider. You are sharing a server and someone is using it to send spam, although that might be long ago.

This is what I see when I check the email you forwarded to me...

Problem Entries, (listings will cause email problems.)

85 "Spam" entries [02:08:44 20 Apr 2017 GMT+00]. 1 "Virus" entries [20:38:42 21 Nov 2016 GMT+00].

You ask for your IP to be delisted, but if the spam continues it will be put back on.

Clearly there is an ongoing problem with your shared server hosting. Open a ticket with your hosting support and send them the bounce messages.

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Rick C

I might have confused you, I should have wrote, with 14Vpp on the input to the circuit (before T1) I have 3.15Vpp on the 220 ohm 3.47k junction. The AC voltage is the same on both sides of the 10uf cap,

At the emitter of Q2.

Yes.

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That is not my IP, that is the example they use on their site.

You're using info from the wrong IP.

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