[SOLVED] 2 Power amplifiers in parallel

Two related questions:
1.
if i split an RF signal into 2 lines and amplify it with 2 same PAs and then combine it again into one RF signal, how do i get ~3dB of power gain? from where i win 3 db ? i split amplify and combine... isn't it energy from nothing?

2. if for e.g. in a specific operational point i input RF signal with qpsk digital modulation and measure ACPR of -30dBc at a single Power amplifier, if input this signal into two parallel PAs as in question 1 , will i see same ACPR of -30dBc or somehow it can become -27dBc?

Hi,

I'm no expert in RF...thus there may be better answers..

isn't it energy from nothing?

Click to expand...

The energy comes from the power supplies, not from nothing...

You gain 3dB of power and you do not...
* You get the same voltage at the output, but you may drive twice the current.
Thus you can get 3dB of power, only when you lower the load impedance....thus could be when you drive two 50 Ohm loads.
* But I assume you don't want to change your load, thus I said you don't gain 3dB. In this case you gain nothing.

Klaus

you are right i got it from the power supply i drive two PA,
regarding your load note, if i combine the signals from the two PAs and drive same 50 ohm load , are you sure i not get a 3db of power gain? as you mentioned i now drive two PAs and not single one...

do you have and assumptions regarding the ACPR or linearity question?

You do not get more power gain, if each amplifier has a gain of 10dB the output will, assuming no losses be 10dB higher than the input.
What you will get, again assuming no losses is 3dB more power output; if each amplifier has 1W output then the combined outptut will be 2W.
Combining amplifiers is a way of getting more output power not more gain.

As was stated, if you split the signal and combine again, at the output of the PA you gain nothing IF the input power is the same.
But IF you increase the input power with 3dB, you will get 3dB more output power (double the power in Watts).
This is the reason that all the parallel or push-pull PAs need (at least) 3dB more input power than single ones.
I mentioned "at least 3dB" because most of the time needs more than this, to compensate for transformer/splitter losses.

Regarding ACPR, generally parallel or push-pull configuration, gives better PA linearity. Somewhere between 3dB to 6dB better IP3 than in case of single active device. There are also many other factors that affect linearity and ACPR in this case.

thanks a lot for clarification,
regarding the second answer, are there some equoations or place where i can see why theoreticly there is linearity improvment ?

i understand that if you use hybrid 90deg splitter and combiner you achieve better match for input and output yet why it improves linearity?

i'm confused because when you for e.g. try to calculate noise floor so you wants to have at least a 10dB difference for example between the thermal noise to the quantization noise, or between the thermal noise and the nonliniarity noise , so here if both PAs have same acpr and you combine them together so you dont have at leaset 10dB but zero db, i think i miss here something very basic (maybe because they are parallel and not cascade? maybe becasue the input signal is coharent?)

What you're missing is that you don't seem to understand decibels. If you take a 100watt amplifier and combine it with another 100watt amplifier, you get 200watts out (assuming all losses), or you have DOUBLED YOUR POWER!!. If you double your power, the power gain is 3dB. The output power goes from 40dBm to 43dBm.

SLK001 said:

What you're missing is that you don't seem to understand decibels. If you take a 100watt amplifier and combine it with another 100watt amplifier, you get 200watts out (assuming all losses), or you have DOUBLED YOUR POWER!!. If you double your power, the power gain is 3dB. The output power goes from 40dBm to 43dBm.

Click to expand...

Hey its clear to me, what i asked is about what happens to the acpr of the combined output

regarding the second answer, are there some equations or place where i can see why theoretically there is linearity improvement ?

Click to expand...

I think the operation conditions aren't yet clearly defined, If you utilize the parallel circuit to double the output power (you need additional 3 dB of power gain in front of the input splitter to do so, as explained), then each amplifier is operating at the same output power and has roughly the same non-linearity and e.g. intermodulation caused ACPR as in the single amplifier setup. The ACPR would stay at previous -30 dBc.

ACPR (which is made mainly due to harmonics of the amplifier) is definitely better when use parallel or push-pull devices.
A push–pull amplifier produces less distortion than a single-ended one, because symmetrical construction means that even-order harmonics are cancelled, which will reduce distortion.

The combining is the key. In parallel with the same input
signal and gain and output termination, all you do is
spread the heat.

Look to Doherty PAs for what an N-element PA array
does to get parallel PAs to sum power efficiently.

Thanks to all,
i ment to increase the input power and thus get additional 3dB at output, for example if single is 40 dBm so each one will provide 40dBm and both summed power will be 43dbm
i also wonder if there is some condition that the sumed ACPR will be worser, for example if the PAs them selves be out of phase due to their production process / tolerance

levnu said:

Thanks to all,
i ment to increase the input power and thus get additional 3dB at output, for example if single is 40 dBm so each one will provide 40dBm and both summed power will be 43dbm
i also wonder if there is some condition that the sumed ACPR will be worser, for example if the PAs them selves be out of phase due to their production process / tolerance

Click to expand...

The ACPR will be the same since the equivalent OIP3 is 3dB higher as well. Magnitude/phase imbalance will main affect the even-order harmonics. OIP3 is less sensitive to magnitude/phase imbalance.

its interesting, can you please explain why its only even order harmaonics? is there any theoretical information regarding this question?

It's just about the fact that amplifiers in push-pull configuration (180° combiner) ideally cancel out even harmonics, as mentioned before. Although it has been claimed that the push-pull configuration could also improve ACPR, I don't yet see how.

If we assume that the modulating base band signal perfectly keeps the spectrum mask, spurious signals in the adjacent channel can be generated by distorting the base band signal in the modulator and due to a nonlinear PA characteristic.

- - - Updated - - -

O.K., if we assume that the push-pull configuration improves the magnitude linearity, it would also increase ACPR.

In fact improving linearity means DECREASE of ACPR. People when say "better ACPR" refer to lower ACPR.

The increase of ACPR (or ACLR, or whatever you named) is mainly due to increased adjacent channel occupancy by 3rd and 5th order inter-modulation components. But not only them contribute to the increase of ACPR. If you are able to reduce all the inter-modulation components, you get even better ACPR.

In the article below from RFMD you will find how that using push-pull configuration (against single-device) reduce the 2dn order distortions by reducing the 2nd harmonic. This improves the P1dB (which is a single tone test) by 2dB.
Read at page 4: "Figure 9 illustrates that there is a 2dB improvement in P1dB when comparing a push-pull configuration to a single device".
https://www.qorvo.com/products/d/da001686

Improving the compression point P1dB of an amplifier (by any existing methods) it will improve the linearity of the amplifier, which means better ACPR.

levnu said:

its interesting, can you please explain why its only even order harmaonics? is there any theoretical information regarding this question?

Click to expand...

s.
Amplifier can be modeled as: y = C0 + C1*x + C2*x^2 + C3 * x^3 + C4 * x^4 + ..., where C0, C1, C2, ... are constant coefficient. C0 repuresents the DC level and usually can be omitted.
Then if you have push-pull structure. x(t) is the input to one channel, -x(t) is the input to another one. Subtract the output of one channel by the other one, you will see clearly the odd-number harmonics will be cancelled perfectly, while the even-number harmonics will be added together since they have the samaae phase.

If there is little magnitude/phase imbalance, the even-number harmonics will be greatly affected. For example the second-order harmonics level will be
C2*(1+delta)*exp(j*theta) - C2 ~= C2 * (1 + delta) * (1 +theta) - C2 ~= C2 * (delta + theta) , where delta is the magnitude imbalance and theta is the phase imbalance. We also assume delta and theta are rather small. We see that imbalance will greatly affect the level in terms of dB.

For the odd-number harmonics, for example the third-order harmonics, the level will be:
C3*(1+delta)*exp(j*theta) + C3 ~= C3*(2 + delta + theta).

Since delta and theta are small, we see that the level of odd-number harmonics is about the same in terms of dB.

thanks to all answers ,
did i get it right push pull is 180 deg delta at input and balanced is 90 degree at the input and 90 degrees back in the output , i just want to try and be clear with the difference, i also now read more on the web try to be sure what is the difference ... if there is a difference i guess the push pull is more for second harmanic removal and balanced is just to increase the output power by 3dB (driving input with more 3dB) with improvment of input return loss and output return loss

great comparison, balanced vs. pull push
**broken link removed**
**broken link removed**