The cascode

How the top transistor holds the bottom one still

A cascode stacks a common-gate transistor, M2, on the drain of a common-source transistor, M1. M1 turns its gate voltage into a current, and M2 passes that current up to the output while it shields M1 from the output voltage.

A single transistor's current rises with its V_DS, and that slope is its r_o (E2.4). In the cascode the output swing lands on M2 instead, and M1's V_DS moves 88 times less than the output.

To find R_out, push a test current i into the output. It flows through M1, whose gate is still, so node x rises by i·r_o1. M2 now has a resistor r_o1 under its source: the degenerated stage of E3.4. Its output resistance is r_o2 + r_o1 + (g_m2 + g_mb2)·r_o2·r_o1.

The lab prints r_o1 = 276 kΩ, r_o2 = 505 kΩ and g_m2 + g_mb2 = 284.8 µS. The formula gives 40.57 MΩ and the sweep measures 40.76 MΩ. The factor (g_m2 + g_mb2)·r_o2 = 144 is M2's own intrinsic gain: the cascode multiplies r_o1 by it.

Why only 71 times the single device, then? The single device sits at V_DS = 1.2 V, where its r_o is 572 kΩ. Inside the cascode M1 keeps only 0.38 V, and at that V_DS its r_o1 is 276 kΩ, half as much (E2.4).

XS d0 in 0 0 sky130_fd_pr__nfet_01v8 W=4 L=0.5

XM1 x in 0 0 sky130_fd_pr__nfet_01v8 W=4 L=0.5
XM2 d12 g12 x 0 sky130_fd_pr__nfet_01v8 W=4 L=0.5

What the cascode costs

Both transistors must stay saturated. M1 needs V_x above its saturation voltage V_Dsat1 = 126 mV, and M2 needs the output above V_x + V_Dsat2 = 0.380 + 0.126 = 0.51 V. A single device needs only its own V_Dsat, about the same 0.13 V.

The cascode gate voltage V_B decides where x sits: one V_GS of M2 below V_B. The plot sweeps the output for the single device and for V_B = 0.9, 1.2 and 1.5 V. The y axis is logarithmic.

Too low a V_B starves M1. At 0.9 V, x sits at 0.137 V, barely above V_Dsat1: M1 slips toward triode, the current drops to 16.2 µA, and R_out never reaches 10 MΩ. Too high a V_B wastes swing, because at 1.5 V the 10 MΩ point moves up to 0.83 V. A good V_B puts x a little above V_Dsat1, so V_B ≈ V_GS2 + V_Dsat1 plus some margin.

A cascode amplifier

Now use each sink as an amplifier. An ideal 20 µA source loads each drain, with 1 pF across it, and a sweep finds the input that puts the output at 0.9 V. With an ideal load the low-frequency gain is the input device's g_m times the stage's R_out.

The single device gives g_mr_o = 126.8, its intrinsic gain (E3.3). The cascode gives g_m1·R_out: 6445 by the formula, 6359 measured. That is a product of two intrinsic gains, 50 times one device's. It falls short of 126.8 squared because the two devices now share the 0.9 V output: M1 keeps 0.37 V and M2 0.53 V, and at a lower V_DS each r_o is lower (E2.4). M1's own g_mr_o drops to 61.

The cascode also shields M1 from the output. At 1 MHz the single device's drain swings 38 times the input, so its gate-drain capacitance C_gd acts many times larger (the Miller effect, E6). Node x swings only 0.89 times the input there. At low frequency x swings 60.8 times, because the ideal load makes M2's input resistance high, as in E3.6.

Try it

One sweep of VOUT drives all four sinks; each has its own ammeter, and deriv turns each current into R_out. The op first reads M1 and M2 of the 1.2 V cascode at V_out = 1.2 V for the formula.

Predict, then run. Make both cascode devices longer (L = 1): r_o rises in both, so R_out rises roughly as the square. Lower VIN to 0.7 V: less current, a larger r_o, and V_x rises a little. Tie M2's gate to its drain instead of VG12 and watch R_out collapse.

Check yourself

* the cascode's top device on an ideal 20 uA sink
.lib sky130 tt
VB vb 0 1.2
VOUT out 0 1.2
XM2 out vb x 0 sky130_fd_pr__nfet_01v8 W=4 L=0.5
I1 x 0 20u
.control
dc VOUT 0 1.8 0.005
let iout=-i(VOUT)
meas dc iout_12 FIND iout AT=1.2
.endc
.end