LNA with degenerated inductor - Where does the power go?
时间：04-04
整理：3721RD
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Hi,
From reviewing the classical input impedance of an LNA with inductor degeneration (assuming CMOS), it is known that by adding the degeneration inductor, the input impedance looking into the transistor has a real component Re{Z_in} = (g_m * L_s) / C_gs. So if the input of the LNA is matched to the source, there is a net power delivery into the circuit, but the full input circuit consists of a resonance circuit in which the only physical resistance (ideally speaking) is the R_s of the source since the other L's and C's cancel out.
From the equation above, it seems that the g_m of the transistor is actually transferred to the input but I still don't get the full picture where is the input power actually being burnt? Could anyone help to clarify this? Or cite some interesting source about it?
Thanks!
MK
The expression that you mentioned is derived under the condition that the output is isolated from the input (i.e the drain is connected to an AC ground) and Cgd does not exist.
Note that across the transistor (i.e between source and drain) you will have a voltage difference (AC) = Vx and a current Ix. Because the transistor itself is a lossy element, the loss in power happens across the transistor in this case.
From reviewing the classical input impedance of an LNA with inductor degeneration (assuming CMOS), it is known that by adding the degeneration inductor, the input impedance looking into the transistor has a real component Re{Z_in} = (g_m * L_s) / C_gs. So if the input of the LNA is matched to the source, there is a net power delivery into the circuit, but the full input circuit consists of a resonance circuit in which the only physical resistance (ideally speaking) is the R_s of the source since the other L's and C's cancel out.
From the equation above, it seems that the g_m of the transistor is actually transferred to the input but I still don't get the full picture where is the input power actually being burnt? Could anyone help to clarify this? Or cite some interesting source about it?
Thanks!
MK
I maybe wrong but this is how I would interpret this result.
The following is an excerpt from Razavi's RF microelectronics:
The expression that you mentioned is derived under the condition that the output is isolated from the input (i.e the drain is connected to an AC ground) and Cgd does not exist.
Note that across the transistor (i.e between source and drain) you will have a voltage difference (AC) = Vx and a current Ix. Because the transistor itself is a lossy element, the loss in power happens across the transistor in this case.