# Balanced Amplifier IP3 Improvement

by Thomas Shafer
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## Balanced Amplifier IP3 & P1dB Improvement Due to 90° Hybrid Couplers

A balanced amplifier design is defined by two amplifiers of equal gain, 1dB compression point (P1dB) and Third-Order Intercept (IP3), arranged in the configuration shown to the right. The couplers are 3 dB hybrids, where the input power is split equally between a 0°and a 90° port. The unused ports are terminated in the system impedance – typically 50 Ω. Reflections from the input and output ports of the amplifiers are shunted to the unused port of each coupler, giving the entire arrangement a matched impedance.

For a single, “unbalanced” amplifier, the output third-order intercept point is defined by the device specifications. The third-order intermodulation products (IM3) will appear at approximately 2Δ dB below the signal power of the primary tones at the device output where Δ = IP3in – Pin = IP3out - Pout (in other words, 3×15 dB below the output-referred 3rd order intercept point).

Consider the example of an amplifier with the gain and linearity specs shown above. If the input power of the two primary tones at ƒ1 and ƒ2 are both -30 dBm, then at the output, third-order Intermodulation products (IM3s) will appear at 2ƒ1 - ƒ2 and 2ƒ2 – ƒ1, at a power of:

IM3out = Pout - 2•(IP3out – Pout)
= (Pin + G) - 2•(IP3out – (Pin + G))
= (-30+24) – (2•(29 – (-30+24)))
= -6 – 2•(29-(-6))
= -76

When this amplifier is used in the balanced configuration above, the primary tones are split equally (-3dB) at the outputs of the first hybrid, with one signal phase-shifted 0°, and the other phase-shifted 90°. After the two split signals are amplified to [Pin –3dB + G], they re-combined, with the phase-shift swapped, so that the recombined signal is equal to [Pin + G] 90°.

However, since the input signal to each amplifier has been decreased by 3dB, the IM3s at the output of each amplifier have been decreased by 9dB. Specifically,

IM3out = Pout - 2•(IP3out – Pout)
= ((Pin – 3dB) + G) - 2•(IP3out – ((Pin – 3dB) + G)))
= ([-30dBm – 3dB]+24dB) – (2•(29dBm – ([-30dBm – 3dB]+24dB)))
= -9dBm – 2•(29dBm – (-9dB))
= -85dBm

The IM3s are combined in-phase through the output 3dB hybrid (remember the 90° phase shift is swapped), which is effectively adding 3dB to the number above. Thus the IM3s of the balanced amplifier are -82dBm, which is 6dB lower than the single amplifier.

The effective IP3out can be calculated with the formula

IP3out (dBm) = Pout (dBm) +0.5[Pout (dBm) – IM3out(dBm)]
= (Pin + G) + 0.5[(Pin + G) – IM3out]
= (-30dBm + 24dB)+0.5[(-30dBm + 24dBm) – (-82dBm)]
= +32dBm  3dB higher than the single amplifier                      QED

Here is a slightly different approach that springboards off of Thomas' work, but does not concern itself with the phases of the signals through the splitter and combiner. This solution is also valid using standard 2-way splitters/combiners with individual matched amplifiers that exhibit excellent input and output VSWR†.    --- Kirt Blattenberger

Assume individual device parameters are the same as in Thomas' example, and that Pin for each of the two system input signals is -30 dBm at the system input.

The 3 dB coupler splits the signals equally, so at the amplifier inputs the signals are -33 dBm.

3rd-order products at the output of each device are:

IM3out = Pout - 2•(IP3out – Pout)
= 3Pout - 2IP3
= 3•(-33dBm +24dB) – 2•(29dBm)
= -27dBm - 58dBm
= -85dBm

The IM3out after combining both amplifiers is -85dBm + 3dB = -82dBm

Now, since the gain of the balanced amplifier is the same as the individual devices, we have:

Gain = 24dB
IM3out = -82dBm
Pout = -30dBm + 24dB = -6dBm

Cranking those numbers back into the IP3 formula yields:

IP3 = 1/2 (3Pout - IM3out)
= 1/2 (3•-6 - -82)
= 32dBm   QED (again)

Note: By a similar analysis, it can be shown that the 1 dB compression point (P1dB) also improves by 3 dB.

† Read about the characteristics of a 3 dB hybrid coupler that make it a great tool for building amplifiers
whose individual devices do not match the system impedance, but do match each other's impedance.