I'm trying to calculate how much power dissipated as heat in an amplifier like LNA. Let's say a HBT device operates at 4.8 V with 80 mA operating current. Gain of the amplifier is 25 dB, OP1dB = 16 dBm. If I feed input power of -20 dBm, output power is +5 dBm.

So, heat dissipation due to DC supply is 4.8*0.08 = 0.384 W. What about heat dissipation due to RF power? How does it contribute? Do we need to consider heat dissipation due to RF like how we consider for power amplifiers taking PAE into consideration?

  • 1
    \$\begingroup\$ But the RF power comes from the DC supply, right? So it's covered by the DC power/dissipation, at least for LNA type devices. \$\endgroup\$
    – SteveSh
    Nov 18, 2023 at 13:07
  • \$\begingroup\$ @SteveSh : so, for small signal devices, DC power dissipation itself is the heat dissipation? \$\endgroup\$
    – Aparna B
    Nov 25, 2023 at 4:23
  • \$\begingroup\$ That's my understanding, though that may change a bit under higher levels of input signals. A lot depends on the architecture and other details of the amplifier. \$\endgroup\$
    – SteveSh
    Nov 25, 2023 at 11:45

1 Answer 1

P (dc)= 4.8*0.08 = 0.384 W. 
Pin= -20 dBm=0.00001 W
Pout= +5 dBm= 0.003 W
P diss= Pdc+Pin-Pout= 0.384+0.00001-0.003=0.38101 W

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