Cascode and wide-swing mirrors

A cascode on the mirror

E4.1's simple mirror has r_out = r_o: 0.80 MΩ at 1.2 V for W/L = 8/1 µm. To raise it, stack a cascode device on the output (E3 lesson "The cascode") and a matching diode on the reference side to bias it.

O2 holds O1's drain near R1's, so the copy is accurate: 19.99 µA at 0.9 V, 20.00 µA at 1.7 V. Seen from the output, O2 multiplies O1's r_o by its own gain: r_out ≈ (g_m2 + g_mb2)·r_o2·r_o1. The lab computes that from the device parameters at 1.2 V: 100.8 MΩ against 102.2 MΩ measured, 128 times the simple mirror.

The cost is headroom. R2's V_GS is 0.853 V, larger than R1's, because its source sits 0.714 V above its body (the body effect). O2 stays saturated only while V_out > V_x + V_Dsat ≈ 0.708 + 0.14 = 0.85 V (V_Dsat: the smallest V_DS that keeps a device saturated). Below that, r_out collapses: 15.5 MΩ at 0.9 V, 5.39 MΩ at 0.85 V, 2.15 MΩ at 0.8 V.

The current falls slowly, because O1 then acts as a simple mirror. So the 1 % test gives 0.621 V, below the 0.85 V where the cascode stops working.

The wide-swing mirror and the W/4 rule

The stacked mirror wastes headroom: O1 sits at 0.708 V but needs only about 0.14 V. A wide-swing (low-voltage) cascode mirror biases the cascode gate from a separate diode device, so O1 sits just above V_Dsat and the output can fall to about two V_Dsat.

The textbook sizes that bias device from the square law. At the same current, a device of width W/4 needs twice the overdrive V_ov = V_GS − V_th, so its gate sits at V_th + 2V_ov. The cascode, of width W, needs V_th + V_ov, and its source lands at V_x = V_ov, exactly the square law's edge of saturation.

The lab's W/4 mirror does that: V_x = 0.137 V, at O1's simulated V_Dsat of 0.134 V. Two things go wrong. At 20 µA these devices are in moderate inversion (between weak and strong), where saturation has no sharp edge. And the reference device sits at 0.714 V while O1 sits at 0.137 V, so E4.1's V_DS error returns, enlarged.

A smaller bias device moves x up: at W/8, V_x = 0.249 V and the copy is 18.96 µA, still 5.2 % low. Sizing alone cannot fix a V_DS mismatch.

Match the reference side, then size by measurement

Put a cascode in the reference branch too, on the same bias line, and tie the bottom gates to the top of that branch. Both bottom devices then sit at nearly the same low V_DS.

With the reference matched, the bias width sets only where the bottom devices sit, and so r_out and the minimum output voltage.

Read the matched rows. At W/4 the bottom devices sit at 0.129 V, in the soft knee: the copy is right, but r_out is only 5.7 MΩ. At W/8 they sit at 0.246 V: 44.8 MΩ and the lowest minimum output voltage, 0.297 V. The rule of thumb "two V_Dsat" gives 0.274 V. At W/12, r_out reaches 77.9 MΩ for a minimum of 0.340 V.

The stacked cascode gives the most r_out where headroom is plentiful; the matched wide-swing mirror keeps most of it down to about 0.3 V.

Try it

Run the bench: seven mirrors, each copying 20 µA, on one output node. Each prints its current at 0.9 V, its r_out at 1.2 V, its minimum output voltage and, for the wide-swing ones, V_x. Plots shows every current and r_out against VO.

Then size the bias yourself. .param wb=1 sets the bias device of mirror m8; try 2, 1.333, 0.8 and 0.5, and for each read vx09_m8, r12_m8 and vmin_m8. Then delete m8's reference cascode (tie R1's drain to its gate, as in w8) and watch the copy fall back.

Check yourself

* a matched wide-swing mirror with its output held at the reference branch's top node g by EO
.lib sky130 tt
VDD vdd 0 1.8
IREF vdd g 20u
XR2 g b y 0 sky130_fd_pr__nfet_01v8 W=8 L=1
XR1 y g 0 0 sky130_fd_pr__nfet_01v8 W=8 L=1
IB vdd b 20u
XB b b 0 0 sky130_fd_pr__nfet_01v8 W=1 L=1
XO2 o b x 0 sky130_fd_pr__nfet_01v8 W=8 L=1
XO1 x g 0 0 sky130_fd_pr__nfet_01v8 W=8 L=1
* EO copies v(g) onto the output without loading g
EO o 0 g 0 1
.control
dc IREF 19u 21u 1u
let iout=-i(EO)
meas dc iout FIND iout AT=20u
.endc
.end