Peak Performance

The problem

Build a common-source amplifier with a resistive load and a series inductor: the drain “out” connects through an inductor L to a resistor Rd, whose other end is on VDD. The gate is driven by an Input source “in”. The test bench loads “out” with 150 fF. Choose Rd and L so that the −3 dB bandwidth is at least 380 MHz (about twice what the resistor alone gives) with no more than 1.5 dB of peaking.

A plain resistive load makes a pole at 1/(2π·Rd·CL). The inductor’s impedance rises with frequency and keeps the load impedance high where the capacitor would pull it down: it adds a zero at R/L that extends the bandwidth by up to 1.7 times. Too much inductance makes a peak in the response; the optimum, about L = 0.4·Rd²·CL, gives a flat response.

The test bench

• Supply: VDD = 1.8 V.
• The test bench hangs a 150 fF load capacitor from “out” to ground and applies the AC stimulus to “in”, sweeping from 100 MHz to 10 GHz.

Numbers to hit

• ◆Gain at 100 MHz at least 8 dB
• ◆Bandwidth (−3 dB) at least 380 MHz
• ◆Peaking above the low-frequency gain at most 1.5 dB
• ◆Output DC level between 0.6 V and 1.7 V
• ◆Every transistor in saturation