Impedance matching network between a chain of Class C RF amplifiers (?)

[audioguru]

I tried your impedance matching circuit on the last FM transmitter simulation I posted that had an output of 282mW.
The impedance matching made no difference.

 

[audioguru]

Deleted double post.

 

[ccurtis]

- Hi ccurtis,
Which one is best among them? Are there any huge difference?

Best among which? I'm confused. LOL.

 

[Willen]

Best among which? I'm confused. LOL.

Among schematics 'A' and 'B', attached on #20 post.

 

[ccurtis]

Among schematics 'A' and 'B', attached on #20 post.

Oh, okay. There is no performance difference between those two circuits.

 

[Willen]

Hi ccurtis,
again want a simple help from you!

A online calculator preferred by you is confusing me! How to put in these specific box, I tried few terms but result was "NaN"


I filled for Class a amp for a Tx of audioguru (last time you calculated 154nH and 10pf):

Source Resistance: 2K
Source Reactance: 0
Load Resistance: 75
Load Reactance: 0
Desired Q: 0
Frequency: 100MHz

hee hee silly input of mine

 

[MikeMl]

Try it with 3 < Q < 10.

 

[ccurtis]

Hi ccurtis,
again want a simple help from you!

A online calculator preferred by you is confusing me! How to put in these specific box, I tried few terms but result was "NaN"


I filled for Class a amp for a Tx of audioguru (last time you calculated 154nH and 10pf):

Source Resistance: 2K
Source Reactance: 0
Load Resistance: 75
Load Reactance: 0
Desired Q: 0
Frequency: 100MHz

hee hee silly input of mine

Hi. If you go back and look, you'll see I used 200 ohms, not 2K, since that's what produced the most output power for the 23mW circuit. In simulation the 27pf cap alone in series with a 75 ohm load is just as effective a match as the 154nH and 10pf matching network, because the impedance difference looking into both combinations is only about 100 ohms.

 

[Willen]

My actual question was I cannot use calculator. So simply insert the values on calculator and and show a picture (screen shoot) please!

 

[ccurtis]

Okay, here goes. NaN means "not a number"; in other words, no solution was found.

 

[Willen]

Thank you!

But why you used Q=3?
Is it Quality factor of inductor? Then why not 10 or 100? Hahaha
Make me clear please!

 

[ccurtis]

Thank you!

But why you used Q=3?
Is it Quality factor of inductor? Then why not 10 or 100? Hahaha
Make me clear please!

The matching program assumes a perfect inductor. The Q calculated is the Q of the matching network, a measure of the bandwidth of the network. Your matching network attenuates signals at frequencies above and below 100MHz. The higher the Q, the more those signals are attenuated. Q=3 by default. I didn't enter it. For an "L" matching network, the Q "is what it is" based on the what you enter for the other inputs.

 

[Willen]

Hi Willen, here's how I would go about figuring the values for the interstage matching networks:

1. The first stage is supposed to deliver 200 mW. For simplicity, I'm going to ignore the few pf of transistor output capacitance. You can tweak the values of the inductor and capacitor, once built. When you build it there are going to be parasitic L's and C's anyway to deal with. The load resistor for 200 mW is RL=V^2/(2P), or 12^2/(2*200mW)=360 ohms. I'm going to also assume an input impedance for the following transistor of 5-5j, which is typical. Using an impedance matching program (because who wants to calculate all that by hand?), the values for L and C of the first matching (diagram B) network between the first and middle stages are 37pf and 75 nH.....

I am still confused on Load Resistance and Source Resistance to fill in Calculator. Can you show another screen shoot of Quoted values here and explain simply between Load Resistance and Source Resistance please!

And what is the purpose of assuming of 5-5j impedance on calculator?

 

[ccurtis]

I am still confused on Load Resistance and Source Resistance to fill in Calculator. Can you show another screen shoot of Quoted values here and explain simply between Load Resistance and Source Resistance please!

And what is the purpose of assuming of 5-5j impedance on calculator?

I had to assume something for the input impedance (load impedance), so I chose something typical. The actual value would have to be measured or obtained from the datasheet (if given there). Here is the screenshot. The value for the inductor is different from what I originally posted. Maybe I made a typo.

 

[Willen]

You mean you have corrected now (Original post has 75nH, 37 pf for 200mw between first and 2nd stage) Now it's 59nH and 37 pf.

Your 5-5j assume impedance is Load Resistance= 5 on calculator?

If this network has 360 ohms impedance as source (output), then this 360 ohms will be as an input impedance for 2nd network but why not? (because collector of Class C has no change of impedance i think).

 

[ccurtis]

You mean you have corrected now (Original post has 75nH, 37 pf for 200mw between first and 2nd stage) Now it's 59nH and 37 pf.

Your 5-5j assume impedance is Load Resistance= 5 on calculator?

If this network has 360 ohms impedance as source (output), then this 360 ohms will be as an input impedance for 2nd network but why not? (because collector of Class C has no change of impedance i think).

The +5 for the source reactance should be -5, so the inductor value I originally posted is correct afterall.

The collector of the first stage will see a 360 ohm load with the matching network connected and a 5-5j ohms load on the matching network from the next stage transistor. In other words, the matching network transforms the 5-j5 input impedance to 360 ohms for use by the collector of the first stage.

 

[Willen]

OK then I attached final schematics to review from you to be insure that all values are correct. Check once please! I think it works fine

(also, look at C8 and L10, it is opposite network- inductor first than capacitor. Should I have to put capacitor first?)