Improving my best diode detector

Yes, you could label the axes.

NT

I believe you care much less about the low end of the scale than the high end. So I would replace the two 4.7 kohm resistors with a 10 kohm pot and tweak it up a bit to improve the middle section of the curve. You can use a small pot with a slightly smaller than 18.13 kohm resistor to set the full scale point, then adjust the 10 kohm resisotr to get an accurate value at midscale.

What is the purpose of the 68 ohm resistor? As the circuit is shown it does nothing but draw more current from the voltage source. I expect your driver will have some output resistance which will make the pair act as a voltage divider which isn't especially useful. The non-linearity of the diodes comes from the low voltage end of their characteristic. If you run with a lower input voltage it just accentuates the non-linearity.

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Rick C

Your right I expect to run the large majority of tests in the upper

3/5ths of the meter range. I am not there yet I expect the graph to run up to 5Vrms, but I don't have a generator that will go that high. I assume it will not deviate from the linear line at the high end. Any agreement there? OK, I'll try the 10k pot. and run some more graphs.

My guess would be to match the source, The author was working at

900MHz and 1.8GHz. I also note it is a return path for the first diode, some circuits replace the Resistor with an inductor for a dc return path but blocking the AC. I would like to run it from a higher impedance source, say 500 or even 1000 ohms, but I don't know how it would affect the output.

If I changed it to a 500 ohm resistor, I would have less current flow through the first diode and as you say accentuate the non-linearity. Mikek

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I changed the termination from 68 to 500 ohms. I made a few iterations until R2 is 3.915k and R1 is 5.85k and the line is really good. One thing I didn't do, I need to change the source to 500 ohms, this will match the amp I want to drive this with. Here's the graph.

Thanks for the suggestions, Mikek

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I have no idea why you need a special "return" path for the diode. The diode passes current one way, that is the point. There is no current the other way. The driving circuit won't care if no current flows the other way. The return path is through the meter and the other diode back to the other side of the voltage source.

BTW, you talk about R numbers in your write ups, but the diagram has no R numbers to match up.

Did you ever figure out what the impedance is of your amp?

Don't think so. The current through the diode comes from the voltage source, not the parallel resistor. If anything, the current will be higher without the 68 ohm resistor. Check out "Thevenin equivalent". The 68 ohm resistor creates a voltage divider with the source impedance and lowers the drive voltage more than it lowers the source impedance and so lowers the current through the rest of the circuit.

That looks pretty durn good. But you need to see how it works with your amp driving it.

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Rick C

The 68 was a simple termination resistor, I have dropped it, I also found that driving it from a 500 ohm source didn't have much effect on linearity. So that will worh with my existing amp impedance. It's all coming together now.

The first resistor after the diode is R2 and the second is R1. Refer to this paper,

Yes, it equals the collector resistor. Loading it with an equal resistance drops the output by half.

Yep, dropped the termination, raised the driving impedance.

Yes, I very happy with how good it looks. Baby steps for me, so far I've only drove it with 450 ohm resistor in series with my 50 ohm generator. Thanks, Mikek

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Possibly. The thing to do to make life easier is subtract off the least squares best straight line and show the residuals x10 on the same scale.

You perhaps want to concentrate on the fit to 0.7v-3v being most linear.

Bat62 looks pretty good to me if you allow it to cross the y axis with a DC offset and ignore the region

The Range to optimize is 3Vrms to 5Vrms. I couldn't run my graph out to 5Vrms, I didn't have the capability, but I'm getting there. My RF millivolt meter only goes to 3volts, I have 100 to 1 adapter, but it seem to cause a 10% error. When using the Q meter the drive can be adjusted to keep the meter in that 3V to 5V range. I also think that will be the most linear range.

Thanks, Mikek

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Why are you so concerned about linearity of your diode detector if you are merely using the detector to look for a resonance peak?

All you require of its amplitude detection is that it is monotonically increasing with input voltage and not too frequency dependent.

I think you really should be checking how it behaves at constant input amplitude over the range of frequencies that you expect to use...

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Regards, 
Martin Brown

Can't get/see that schematic; in any case, for low(er) level sensitivity and "linearity", try adding a small forward DC bias to move the average OP up the curve.

Still works for me.

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The bias would then need to be subtracted out again after rectification.

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Rick C

That's not the purpose of the circuit.

Amplitude *measurement*.

You are not up to date.

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Rick C

He's has a point there, I already know T1 and Q2 start to roll off at 12 or 13 MHz. That is well above where my interest lies, but, I'll try to improve that later, I need to see something work before I lose interest. I have the first section which is mostly an impedance changer (from very high to low) with a gain of 0.96. I have a working detector I'm happy with, it has a better frequency response than my impedance hanger. Now all I need is a amp with a gain of 4.5 and a 500 ohm collector resistor. I have the circuit just need to get the gain and bias right and it should be ready for testing.

On a different note, I have had the AD8307 log amplifier recommended. It would save several parts, but I don't understand it's output in relation to it's input. It says 25mV/db. I want to drive an analog meter, will that work? Or is is scaled in db, am I confused?

Mikek

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The gain of the circuit can be modified by adding an emitter resistor where G = Rc/Re. The gain to Re is always about 1 and the current in the two resistors is nearly equal, so the gain at the collector will be proportional by Rc/Re.

The voltage on the output goes up by 25 mV with each dB of increase on the input. dB is a scale based on logarithms : dB = 20 log(V/Vref) where V is the input voltage and Vref is the input voltage for a zero dB output.

That means the Q will be on a logarithmic scale which you would have to mark on the meter face. You may need to add an op amp circuit to adjust the zero point of the meter or you might be able to use a voltage source and a pot on the ground leg of the meter like the original Boonton circuit, but with a much lower voltage.

This would crowd the numbers on the high end of the meter and space them further apart on the low end. Maybe this is good, maybe not.

BTW, if you are only interested in a given voltage range at the output of the diode circuit, you can connect the meter ground leg to the adjustable voltage source (a pot) and make the zero current point on the meter correspond to a 3 Vrms input and the max reading on the meter face correspond to a 5 Vrms input. Then you have the entire range of the meter to measure the Q more accurately. There is no reason why the low end has to correspond to a Q of 1.

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Rick C

Then it doesn't work with my plans for an analog meter.

If biasing the diode detector up by 3 volts will make it more linear on the bottom end, That's good. I can bias the other end of my meter up by three volts to remove the 3 volts. The Q meter already is setup with an adjustable bias on the negative side of the meter. I'll keep this in mind. I would just as soon keep the zero to 250 scale. Thanks, Mikek

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What I am talking about doesn't affect the diode circuit, it changes the range of your meter range from 0-5 to 3-5 or whatever numbers you want. Because you can change your power input level to bring the output of the circuit to whatever level you want there is no reason to work with the low end of the output range, so no need to linearize the entire output range. Besides, your circuit is pretty durn linear as it is.

If you want to keep the 0 to 250 scale, consider using an adder (+250 say) before you use the multiplier. That would make the true Q scale

250 to 500.

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Rick C

• Nope; get: "Download and save to your Dropbox by creating a free account, or sign in" I do not have a dropbox...

• No schematic..

Too complicated??