# Frequency Multiplier - Return losses vs Conversion Loss

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#### MobiNaz

##### Full Member level 3
Hello all,

I am working on single diode frequency multiplier. Could anyone please tell (with reference) what is the relationship between return losses of the multiplier and conversion loss.

If the input is matched to the diode impedance, more power will be delivered to it, this implies less return loss at the input and less conversion loss. If output impedance is also matched then overall the return loss at the output should be low.

However, I am getting high return losses and low conversion loss. Isn't it weird? I have used transmission lines for matching.

Can this be due to the insertion loss as stated in Chapter 1: Understanding Key RF Switch Specifications - Developer Zone - National Instruments

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#### volker_muehlhaus

##### Guest
Low return loss means that you are able to get the input power into the diode.

However, this does not tell us anything about the harmonics generated by the diode (= nonlinear behaviour).

#### tony_lth

More power input to diode is better, you can have more power output. So that's better return loss meaning.

Freq Multiplier's Conversion Loss is different from other conversion loss. This conversion loss should be all output harmonic waves power divide input power. Such as the input power is 15dBm, and conversion loss is 20dB, so the 30 tooth harmonic total power should be -5dBm. If all the tooth are equeal power, so the desired power output is (-5dBm-15dB)=-20dBm. But if you match specially for one harmonic, you can have greater power. The conversion loss relate with the nth harmonic.

MobiNaz

### MobiNaz

points: 2

#### vfone

What is your conversion loss that you are complaining?
Just as an example, a 20dB conversion loss is a normal loss for a fifth harmonic frequency multiplier using diodes, having tuned impedance matching network at the output.

MobiNaz

### MobiNaz

points: 2

#### MobiNaz

##### Full Member level 3
First of all thanks for replying. Well the thing is it is actually a frequency doubler, using just a schottky diode with Pin = -40dBm. When I have no matching network my conv loss is around 14 dB and as soon as I match the input and output (for 2nd harmonic) the return loss improves but conv loss deteriorates to 141 dB. The question is why?it should improve as the diode is now getting more power?

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#### biff44

-40 dBm is a very low power level to be expecting a simple schottky diode to be doing much multiplying. These things work well with -10 to +20 dBm driving them. At the very least, you should be forward biasing the diode with a DC current of maybe a microamp to even attempt such a thing.

MobiNaz

### MobiNaz

points: 2

#### MobiNaz

##### Full Member level 3
I am actually forward biasing it, this reduces the conv loss to 97dB only. My question still remains..

#### biff44

You are going to have a hard time forward biasing the diode with a short circuited stub on either side of it.

MobiNaz

### MobiNaz

points: 2

#### biff44

Assuming your schematic is correct, all the current would go thru the stub into the ground plane, instead of thru the diode.

MobiNaz

### MobiNaz

points: 2
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#### volker_muehlhaus

##### Guest
Assuming your schematic is correct, all the current would go thru the stub into the ground plane, instead of thru the diode.
I agree to that. With the two shorted stubs, you force 0V DC across the diode.
To avoid that, you can make the stubs longer with open end (add 1/4 wavelength in length).

#### biff44

I agree to that. With the two shorted stubs, you force 0V DC across the diode.
To avoid that, you can make the stubs longer with open end (add 1/4 wavelength in length).
You can do all sorts of stuff, but he did not and wonders why it is not working.

MobiNaz

### MobiNaz

points: 2

#### MobiNaz

##### Full Member level 3
I agree to that. With the two shorted stubs, you force 0V DC across the diode.
To avoid that, you can make the stubs longer with open end (add 1/4 wavelength in length).
But if I open the stub and add a quarter wavelength, won't my stub become a short stub? If you mean to say that I should replace short stub by open one, then it would be that I increase the length by quarter wavelength, isn't it?

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#### volker_muehlhaus

##### Guest
But if I open the stub and add a quarter wavelength, won't my stub become a short stub?
Your shorted line TL4 is so short (21µm) that it is effectively a short circuit at all frequencies, including DC and the 5GHz fundamental frequency. If you replace this with a lambda/4 open stub at 5GHz, you will get a short at 5GHz but open at DC and 10GHz. That was the trivial idea to avoid the DC short across the diode.

But biff44 is right, there are many possible solutions. I now payed more attention to your dimensions, and noticed really strange values like 125nm and 21nm line width and 10µm line length. I think that your matching does not work this way and your input power might actually go the the line losses rather than the diode.

#### MobiNaz

##### Full Member level 3
The thing is I need really small dimensions, thats why instead of using 50 Ohm Transmission line I have altered the length and width both to get impedance matching.

Further to above reply, I do not want the line to be a short at 5GHz and Dc both, cause they both are required for proper functioning of the diode, that is for better conversion losses.

I know now why the design is not working, so what do you all suggest I should do?

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#### volker_muehlhaus

##### Guest
The thing is I need really small dimensions
But your dimensions are impossible to realize. They just make no sense at all.

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