V GG -V TO. rout = Saturation Region I OUT. Vout V SAT V GS V OUT. Sheet 1 of 10. Cascode Current Mirror

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1 of 10 Cascode Current Mirror he main property/feature of a current source/sink is that the current though the device is independent of the voltage across it. Figure 1 shows the most basic of current sink. he current Iout is set by the voltage applied across the gate-source of the device, the greater the voltage the larger the current flow through the device. However as you can see from Figure 1 as the current increases then the slope in the saturation increases for an ideal current sink/source we want this region to be flat ie very high resistance. hese saturation slopes extrapolate to a point on the x axis known as the Channel length modulation parameter λ, which is equal 1/-x, typical values are 0.01-0.05. he smaller this value then the smaller the slope in saturation and the better the current source/sink will be. he output resistance r out is given by:- rout 1 λ. I D I OU Saturation Region I OU M SA OU GG - O Figure 1 Simple current sink As the slope in saturation region is determined by the output resistance r out then increasing this will greatly improve the performance of the current source/sink. In addition we would like to reduce sat to allow larger voltage swings across the device.

of 10 One way of increasing the output resistance is to add a resistor between the source and ground as shown in Figure. In reality this method is not to practical because of the voltage drop across the resistor however, we can replace the resistor by an active resistor formed by a CMOS Fet and this forms the basis of the cascode current source/sink. I OU M SA r Rout (gm.rds).r Figure Method for increasing performance of current source/sink by adding a source resistor to increase rout by (gm.rds)r. I OU 1 M SA M1 Rout (gm.rds).rds1 Figure 3 Adding an active load to increase the rout of the current source/sink to improve performance. Figure 3 Shows the addition of an active load to M can increase the output resistance from rds to (gms.rds).rds1.

3 of 10 MOS Diode When the gate and drain terminals are connected together on a CMOS FE the operation is similar to a p-n junction diode. he circuits for the n-type and p-type diode (active resistors) are shown in Figure 4. I DD M1 SA + ON M1 SA + ON N ype Diode P ype Diode I Figure 4 Active resistor/diode configurations. he gates are connected to the drains, and the sources are connected to supplies. As used as an active load the resistance of the diode is 1/gm. I D β ( ) Where β K' W L As in the diode configurat ion the gate is connected to the source then DS and I D β ( ) DS rearrange to get DS DS I β D + In most applications these diodes can be used to generate a fixed bias voltages as shown in Figure 5.

4 of 10 DD M I SA + ON M1 1 SA + ON N ype Diodes Figure 5 wo N-type CMOS diodes giving two fixed bias voltages of SA + ON and A + ON We can use these CMOS diodes to provide bias to the current mirror shown in Figure 3 to form the most popular current source/sink circuits known as the cascode current mirror M1 & M are effectively two diodes in series with a total voltage drop of SA +, which is fairly independent of current I REF. I REF can be set using a resistor or band-gap/resistor network.

5 of 10 DD I REF I OU IN M1 M * + * DSA SA + M3 M4 SA + SS DSA + Figure 6 Cascode current source, showing that the minimum voltage output is SA + above the rail SS (in this case 0). M1 and M3 are wired as diodes and each have a voltage drop of SA + across the drain-source junction. his circuit has higher output impedance than the simple current mirror, but lower output swing due to the extra device. D G - - (1) he bias to M (gs) is SA + Sub into (1) D ( SA + )- SA - R IN 1 gm 1 1 + gm 3 R OU gm go.go 4 gm.r ds

6 of 10 Example Using the circuit of Figure 6, with DD +5 and SS -5. Assuming the following parameters 1, Kp 80E-6, λ 0.0. Calculate gs, and Rout. All FEs are the same (W/L 1). -6 IREF 0E gs + 1+ -6 Kp 80E 1.5 SA - 1.5-1 0.5 R out OU SS - R O4 ( + ) 5 - ( 0.5) + 1-3 + R O SA ( 1+ gm4.r ) O4 gm.id * 0E 1.5 1-6 80E -6 A/ R O4 R O 1 λ.i D 1 0.0 * 0E -6.5MΩ R OU.5E 6 +.5E 6-6 6 ( 1+ 80E..5E ) 505MΩ he above circuit with the data given was simulated in ADS using a DC simulation to verify the calculated results. he simulation setup is shown in Figure 7, for this setup the circuit was analysed and the annotate DC solution selected to add all the node voltage and currents to the circuit.

7 of 10 DC DC DC1 Sweepar Start0 Stop5 Step.01 ar Eqn AR AR W1 DD5 L1 LAMBDA0.0 MOSFE_NMO MOSFE3 ModelMOSFEM LengthL um WidthW um 0 ua I_DC SRC1 Idc0 ua 0 ua -40.0 ua _DC SRC dcdd -.98 0.0 ua vd4 R R1 R400000-0.0 ua 0.0 ua I_Probe I1 LEEL1_Mode MOSFEM NMOSyes to1 Kp80e-6 LambdaLAMBD -5-5.11 pa -5-1.71 pa -5 MOSFE_NMO MOSFE1 ModelMOSFEM LengthL um WidthW um -0 ua 0 ua -1.61 vgs3 0 A -3.30 vgs 0 A -5-0 ua 0 A -3.63 pa -0.0 ua -3.41 vd1 0.0 ua -5 0 A -1.60 pa -5 40.0 ua -5-5 -0.0 ua _DC SRC3 dc-dd MOSFE_NMO MOSFE4 ModelMOSFEM LengthL um -5 WidthW um MOSFE_NMO MOSFE ModelMOSFEM LengthL um WidthW um Figure 7 ADS DC simulation of the cascode current source example. he resistor load has been calculated assuming a current of 0uA and a minimum of 3. NOE that the bulk connections have been connected to the lowest circuit potential ie DD or 5. he simulation has given slightly different results from the simplified hand calculations as you would expect.

8 of 10 Decreasing OU A cascode current sink can be used in the tail in a long-tail pair/differential amplifier to improve the common mode rejection. However, because we are using two stages the current mirror will reduce the rail voltage available by OU SA +, which could be typically ~ less than the supply rail. Other cascode circuits have been modified to reduce OU as much as possible. One such circuit is shown in Figure 8. ON + Sorearranging ON - is also SA K'W Id L K'W L ( - ) wecansubin to get Id ( ) ON And ON.Id.L K'W If we set the W/L ratio of M1 to 0.5 then.id.4w 4 K'L ON. ON herefore, gs1 is + ON D G - (1) o bias to M (gs1) is SA + Sub into (1) D ( SA + ) SA

9 of 10 DD DD I REF1 I REF W/L 1/4 I OU M1 M SA SA + M3 M4 SA SA + SS Figure 8 Method for decreasing the voltage drop across the two output devices to give a greater available voltage swing. Although this circuit has a low output voltage drop (ON) it suffers in that M1 connected to M are not matched devices (M1 has a W/L ratio of 0.5). As a result I OU will not track I REF over temperature. o eliminate this problem the circuit shown in Figure 9 is used. Each pair of FES in the cascode are fed with the same bias. M3 supplied SA + to the top pair FES M4 and M5.

10 of 10 DC DC1 DC LEEL1_Mode MOSFEM1 NMOSyes to1 Kp80e-6 LambdaLAMBD 5 0 ua I_DC SRC1 Idc0 ua 0 ua MOSFE_NMOS MOSFE3 ModelMOSFEM Length4*L um -.39 pa 0 A WidthW um -60.0 ua 5 _DC SRC 5 dcdd 0 ua -.40 pa -.6 vgs4-3.73 0.0 ua vd4-0 ar ua AR -0 ua -0.0 ua Eqn AR -5-4.3 W1 Rload1 0 ua -4.33 0.0 ua -5 vd1 DD5 L1-3.30 LAMBDA0.0-5 vgs -5-69 -5 fa 0 A 0 A -681 fa -5-5 -5-5 -5 60.0 ua -5-5 MOSFE_NMOS -5-0 ua MOSFE6 _DC -0.0 ua ModelMOSFEM SRC3 LengthL um dc-dd WidthW um 5 I_DC SRC4 Idc0 ua MOSFE_NMOS MOSFE5 ModelMOSFEM LengthL um 0 ua WidthW um 0 A R R1 R437 kohm -0.0 ua 0 A -1.96 pa 0.0 ua I_Probe I1 MOSFE_NMOS MOSFE4 ModelMOSFEM LengthL um -5 WidthW um MOSFE_NMOS MOSFE ModelMOSFEM LengthL um WidthW um Figure 9 ADS ransient simulation of the improved high-voltage swing version of the cascode current source. he in this case is 3.73 giving voltage across the two output FEs of 1.7 (giving a ON for each device of 0.635).

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