Current Mirrors and Ac0ve Loads

Transcription

1 Current Mirrors and Ac0ve Loads

2 Simple Bipolar Current Mirror The two V BE s are equal. The currents are equal to a first approxima0on. However There is a systema0c error due to base current. The two V CE s are not necessarily equal. Thus, there is an error depending on r 0. The two transistors cannot be iden0cal. Thus, there is an error due to mismatch. V o,min = V CE,sat

3 Simple MOS Current Mirror The two VGS values are the same. Hence, the currents are the same to a first approxima0on. However, The VDS values need not necessarily be the same. The two transistors are not iden0cal. V o,min = V ov

4 Simple Current Mirror for 10 µa ua -i(vload) current sweep V Minimum Length of 0.35 µm

5 Simple Current Mirror for 10µA Increase W and L together to 3 0mes. ua -i(vload) current sweep V

6 Simple Current Mirror with β Helper

7 Cascode Current Mirror

8 Cascode Current Mirror V IN =2V T + 2V ov V OUT(min) =V T + 2V ov Normally, overdrive voltage is not chosen too small. Hence, not suited to low voltage opera0on.

9 Cascode Current Mirror V v(1) v(2) voltage sweep V

10 Cascode Current Mirror ua -i(vload) current sweep V

11 Low Voltage Current Mirror

12 Low Voltage Current Mirror ua -i(vload) current sweep V

13 Another Low Voltage Current Mirror The reference voltage can be set as the overdrive voltage which is around 0.2V. The minimum input and output voltages will be in this range.

14 Output with V REF = 0.25V ua -i(vload) current sweep V

15 Lowest Voltage Current Mirror The output voltage for this one can go down to 50mV because transistor M2 can enter the resis0ve region.

16 Lowest Voltage Current Mirror ua -i(vload) 10 8 current sweep V

17 High Swing Cascode Current Mirror

18 High Swing Cascode Current Mirror V G3 = V G6 = V G1 = V T + V ov V G4 = V G5 = 2V T + 3V ov V G2 = V T + 2V ov V D1 = V ov V IN = 2V T + 3V ov V OUT(min) = 2V ov The V DS values of M1 and M3 are different, systema0c error present.

19 Sooch Cascode Current Mirror

20 Sooch Cascode Current Mirror I in = I in = k ʹ 2 k ʹ 2 W L W L 6 5 ( V GS 6 V T ) 2 2 2( V GS 5 V T )V DS 5 V DS5 [ ] V DS5 = V ov V GS6 = V T +V ov V GS2 = V GS 4 = V T + 2V ov W = 1 W L 3 L 5 6 Note that M4 is included so that M3 and M1 have the same drain voltages.

21 Modified Sooch Current Mirror The previous circuit s0ll has V IN = 2V T + 3V ov Although high swing at the output, low voltage opera0on not possible due to high input voltage. Why not split the input branch into two?

22 Modified Sooch Current Mirror

23 Modified Sooch Current Mirror The W/L of M5 is s0ll 1/3 of the others. V G2 = V T + 2V ov V G1 = V T + V ov V IN(max) =V T + 2V ov

24 Wilson Current Mirror

25 Wilson Current Mirror The difference between the input current and I C3 flows into the base of Q2. This base current is mul0plied by β+1 and flows into Q1. The same current as Q1 flows through Q3. Thus, I C3 is forced to be nearly equal to the input current.

26 Wilson Current Mirror If the output voltage (V C2 ) increases, I C2 increases slightly (Early effect). Thus, I C1 increases by the same amount. Thus, I C3 increases by the same amount. This causes a decrease in I B2, bringing the I C2 back. R out βr 0 2

27 Wilson Current Mirror Wilson current mirror idea can be used in MOSFET s as well. Using an extra transistor M4 balances the two branches. R out ( 2 + g m 2 r 03 )r 02 V OUT (min) = V T + 2V ov V IN = 2V T + 2V ov

28 Ac0ve Loads Current Mirror Load Deple0on Load Diode Connected Load Inverter Type Circuit

29 CMOS Inverter The inverter is an amplifier in the transi0on region of its characteris0cs. Assume V in is biased at V DD /2. The DC value of the output will also be V DD /2. For the same I DS and V GS - V T, g mn = g mp = g m If V An L n = V Ap L p = V A, r 0p = r 0n = r 0 A v = 2g m 2g ds = 2V A V DD 2 V T

30 BW and GBW If the load capacitance is dominant, A v = 2g m R out R out = r 0 2 BW = Also, poles/zeros due to R sig C gs and C gd. We know how to calculate them. 1 2πR out C L GBW = 2g m 2πC L

31 Current Mirror Load Very common type of amplifier. Good output swing, V sat,n V out V DD V sat,p A v = g ( m1 r 01 r ) 02

32 Deple0on Load Need a deple0on transistor 1 A v = g m1 r 01 r 02 g mb2 g m1 = 1 g mb 2 χ ( W L) 1 W L ( ) 2 If M2 can be built in an isolated well, A v = g ( m1 r 01 r ) 02

33 Diode Connected MOS Transistor Will have a small output resistance, R out = 1 g m r 0 BW = g m ( ) f T 2π C gs + C ds

34 Diode Connected Load This will form a wideband amplifier. A v = g m1 1 g m2 1+ g mb g m2 g m 2 r 01 1 g m2 r 02 g m1 g m2 = ( W L ) 1 W L = V V GS 2 T ( ) 2 V GS1 V T However, note that the DC supply is connected to the output via V GS2. Also, body effect on M2.

35 Another Version of Diode Connected Load Folded diode connected load.

36 Folded Diode Load Like a current mirror. Same V OUTDC as V INDC No body effect Good PSRR Double power consump0on