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Linear RF Power amplifier

Chinku_420

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I have to design a Linear RF Power (class A). My frequency is 2.4 GHz. I need 25 dB gain. If I choose Pin= -20 dBm and the Pout should be Pout= 5dBm. Right?
If Pout id 5dBm= 3.162 mW. So, Rload will be 9/(2* 3.162m) = 1.43 kOhm. (Vdd is 3V & process is TSMC 180nm). So, should I transform this 1.43 k to 50 Ohm? Is this Rl and Ropt same?
I need your help. Thanks
 
You are apparently assuming an ideal class A output stage with output inductor, achieving 2*Vdd swing. I think that's completely unrealistic and no suitable way to design microwave amplifiers. Also 1.4k:50 ohm impedance matching is no good design idea.
 
You are apparently assuming an ideal class A output stage with output inductor, achieving 2*Vdd swing. I think that's completely unrealistic and no suitable way to design microwave amplifiers. Also 1.4k:50 ohm impedance matching is no good design idea.
Thanks for the reply.

At this point, I am only doing calculations while assuming everything ideal. I know that the the drain inductor has very low Q factor. But, my question is how can I chose the load resistance with the same specs.

I know about the load-pull analysis and conjugate match concept. But at this point, I want to do the simple calculations.
Thanks
 
Your theoretical load resistance equation is fine. Rload=(Vcc^2)/(2*Pout). The reason that you get such high load resistance is because the output power is low (3mW).
For the given Vcc, if you increase the output power to about 100mW you get a load resistance close to 50 ohms.
The high load resistance in very low power amplifiers should not be an issue.
Problems in PA design starts to appear in high power amplifiers, when the load resistance is very low (under 1 ohm).
 
Your theoretical load resistance equation is fine. Rload=(Vcc^2)/(2*Pout).
Now I understand the numbers ... but NO, this is not the correct math for a class A amplifier.
Even if you have ideal amplitude (impossible for a class A amplifier) you have to calculate with
* a sine of 3Vpp
.. which is +/-1.5Vp
.. which is about 1.06V RMS

Although meaningless ... this 1.06V RMS, combined with 3.16mW would give 355 Ohms.
And if one wants calculate the pull up resistor for a single transistor output stage .. it again is different.

****
An RF semiconductor amplifier neither physically nor mathematically is a power amplifier. It usually is a voltage amplifier .. in some cases a current amplifier.
For sure it "amplifies" power ... but the power directly depends on the load.
Whereas a voltage amplifier amplifies the voltage and only partly (source impedance) depends on load.
Same for current amplifiers.

So back to the thread:
If the input is given by -20dBm ... it misses the reference impedance. The characteristic impedance. 50 Ohms for example.

So in case of -20dBm on 50 Ohms the voltage can be calculated:
-20dBm = -20dB referenced to 1mW = 10uW (@50 Ohms)
10uW referenced to the 50 Ohms. --> 22.3mV RMS
This 10uW does not go "to the amplifier" .. but most of it gets dissipated by the termination resistor.

If it was a power amplifier ... and you remove the device (termination resistor) that dissipates the 10uW ... the the "input power gets reduced".
And if it was a power amplifer .. it also needs to reduce the ouput power..

But this is not how it works.
If you remove the resistor ... the voltage usually rises a bit (and it causes echoes and distorted frequency flatness) ... and thus the ouput power also rises.
(In ideal case the input voltage doubles, and thus the output power becomes 4x)
But all this is not reliable, maybe not realistic ... not the way to design an RF amplifier ....

****

--> I recommend to first specify the impedance you design your amplifier for
--> then decide the amplifier topology
--> in most cases: then do the calculations with voltages referenced to your impedance
--> in some cases you find out you will need a transformer to adjust impedances. (to your voltage levels)

Klaus
 
Now I understand the numbers ... but NO, this is not the correct math for a class A amplifier.
Even if you have ideal amplitude (impossible for a class A amplifier) you have to calculate with
* a sine of 3Vpp
.. which is +/-1.5Vp
.. which is about 1.06V RMS

Although meaningless ... this 1.06V RMS, combined with 3.16mW would give 355 Ohms.
And if one wants calculate the pull up resistor for a single transistor output stage .. it again is different.

****
An RF semiconductor amplifier neither physically nor mathematically is a power amplifier. It usually is a voltage amplifier .. in some cases a current amplifier.
For sure it "amplifies" power ... but the power directly depends on the load.
Whereas a voltage amplifier amplifies the voltage and only partly (source impedance) depends on load.
Same for current amplifiers.

So back to the thread:
If the input is given by -20dBm ... it misses the reference impedance. The characteristic impedance. 50 Ohms for example.

So in case of -20dBm on 50 Ohms the voltage can be calculated:
-20dBm = -20dB referenced to 1mW = 10uW (@50 Ohms)
10uW referenced to the 50 Ohms. --> 22.3mV RMS
This 10uW does not go "to the amplifier" .. but most of it gets dissipated by the termination resistor.

If it was a power amplifier ... and you remove the device (termination resistor) that dissipates the 10uW ... the the "input power gets reduced".
And if it was a power amplifer .. it also needs to reduce the ouput power..

But this is not how it works.
If you remove the resistor ... the voltage usually rises a bit (and it causes echoes and distorted frequency flatness) ... and thus the ouput power also rises.
(In ideal case the input voltage doubles, and thus the output power becomes 4x)
But all this is not reliable, maybe not realistic ... not the way to design an RF amplifier ....

****

--> I recommend to first specify the impedance you design your amplifier for
--> then decide the amplifier topology
--> in most cases: then do the calculations with voltages referenced to your impedance
--> in some cases you find out you will need a transformer to adjust impedances. (to your voltage levels)

Klaus
I do understand your Point. Thanks for the detailed explanation.

Can you please suggest any good resources to study practical design? I have read the Razavi Book, and Cripps books for Load-pull analysis. But none of them mentioned the points you highlighted. Highly Appreciated.
 

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