類比積體電路設計技術
Current Source and Current Mirror
Hsun-Hsiang Chen
Department of Electronic
Engineering National Changhua
University of Education
Email: chenhh@cc.ncue.edu.tw
Spring 2010
Reference
Behzad Razavi, “Design of Analog CMOS
Integrated Circuits”, 2001
„ Lecture notes from www.rfic.co.uk
„ Microelectronic Circuits, 5th, 2004 by
Sedra/Smith
„
2
Concept of Current Mirror
„
The motivation behind a current mirror is to
sense the current from a “golden current
source” and duplicate this “golden current”
to other locations.
3
MOS Current Mirror
„
The motivation behind a current mirror is to duplicate a (scaled
version of the) “golden current” to other locations.
Current mirror concept
Generation of required VGS
1
⎛W ⎞
2
I REF = μnCox ⎜ ⎟ (VX − VTH )
2
⎝ L ⎠ REF
VX =
2 I REF
+ VTH 1
μ nCox (W / L )1
Current Mirror Circuitry
I copy1 =
1
⎛W ⎞
μ nCox ⎜ ⎟ (VX − VTH )2
2
⎝ L ⎠1
(
W / L )1
I copy1 =
(W / L )REF
I REF
4
MOS Current Mirror – NOT!
„
This is not a current mirror, because the relationship
between VX and IREF is not clearly defined.
„
The only way to clearly define VX with IREF is to use a
diode-connected MOS since it provides square-law IV relationship.
5
Example: Current Scaling
„
MOS current mirrors can be used to scale IREF up or
down
I1 =
0.2mA; I2 = 0.5mA
λ = 0:
6
Effect of Vo on Io
Figure 6.5 Basic MOSFET current mirror.
Figure 6.6 Output characteristic of the current source in Fig. 6.4 and the current mirror of Fig. 6.5 for the case Q2 is matched to Q1.
7
Impact of Channel-Length Modulation
λ≠0
1
⎛W ⎞
μ nCox ⎜ ⎟ (VX − VTH )2 [1 + λ (VDS1 − VD , sat )]
2
⎝ L ⎠1
1
⎛W ⎞
2
= μ nCox ⎜ ⎟ (VX − VTH ) [1 + λ (VDS1 − VGS + VTH )]
2
⎝ L ⎠1
I copy1 =
1
⎛W ⎞
2
I REF = μnCox ⎜ ⎟ (VX − VTH ) [1 + λ (VGS − VD,sat )]
2
⎝ L ⎠ REF
1
⎛W ⎞
2
= μnCox ⎜ ⎟ (VX − VTH ) [1 + λVTH ]
2
⎝ L ⎠ REF
I copy1 =
(W / L )1
(W / L )REF
I REF
(W / L )1 I ⎛⎜1 + λ (VDS1 − VOV 1 ) ⎞⎟
1 + λ (VDS 1 − VGS + VTH )
=
(W / L )REF REF ⎜⎝
1 + λVTH
1 + λVTH ⎟⎠
8
A Current-Steering Circuit
9
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Amplifier Bias Example
Current Mirrors
Ch. 5 # 10
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
例題 5.1
在圖5.6中，如果所有電晶體都位於飽和區時，計算出 M4 之汲極電流。
解：
我們知道 ID2 = IREF[(W/L)2/(W/L)1]，且 |ID3| = |ID2|，而ID4 = ID3 [(W/L)4 /
(W/L)3]。因此，|ID4| =αβIREF，其中α = (W/L)2 / (W/L)1 且β = (W/L)4 /
(W/L)3。適當地選擇α與β可以建立在 ID4 與 IREF 間的大或小的比值。
舉例來說，α = β = 5 時會產生一放大因子25，同理，α =β = 0.2 可以用
來生成一個明確定義之小電流。
Current Mirrors
Ch. 5 # 11
A Cascode Current-Source
Figure 6.43 A cascode current-source.
12
Double Cascoding.
Figure 6.44 Double cascoding.
13
The Folded Cascode
Figure 6.45 The folded cascode.
14
A cascode MOS current mirror.
Figure 6.58 A cascode MOS current mirror.
15
The Wilson MOS mirror
Figure 6.61 The Wilson MOS mirror: (a) circuit; (b) analysis to determine output resistance; and (c) modified circuit.
16
The Widlar current source
Figure 6.63 Circuits for Example 6.14.
17
The wide-swing current mirror
Figure 9.12 (a) Cascode current mirror with the voltages at all nodes indicated. Note that the minimum voltage allowed at the output is Vt + VOV. (b) A
modification of the cascode mirror that results in the reduction of the minimum output voltage to VOV. This is the wide-swing current mirror.
18
Referenced Self-bias Current
Source
„
This circuit
provides an output
current IOUT that is
independent of the
supply voltage, but
instead is
dependant on the
MOS threshold
voltage VT.
19
Referenced Self-bias Current
Source
„
Normally the voltage
required on the gate
of M1 will be:
„ Vgs1 = Vov1 + Vt1
„
2IDL
Vov1 =
knW
„
If we make the ratio
of W/L very large
then
2IDL
Vov1 =
⇒0
knW
„
Therefore,
„ Vgs1 ≈ Vt1
I1 = I 2 = I out = Vt1 / R
20
Referenced Self-bias Current
Source
„
„
„
Example
Design a referenced
current source for Iout of
50uA and a referenced
output voltage of 1.5V.
Also produce a plot of
supply voltage vs output
voltage
Make W/L=100
2IDL
2 × 50 × 10−6 × 1
Vov1 =
=
= 0.09V
−6
knW
110 × 10 × 100
I1 = I 2 = I out = (Vt1 + Vov1 ) / R
R = (Vt1 + Vov1 ) / I out = (0.7 + 0.09) /(50 × 10−6 ) = 15.8kΩ
21
Bias Circuit
Figure 7.42 Bias circuit for the CMOS op amp.
22
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Active Current Mirror
Current Mirrors
Ch. 5 # 23
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Open Loop Constraints
Current Mirrors
Ch. 5 # 24
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Calculation of Gm
ID1 = I D3 = ID 4 = gm1,2 Vin / 2
I D2 = −gm1,2 Vin / 2
Iout = I D2 − I D4 = −gm1,2 Vin ,⇒ Gm = gm1, 2
Current Mirrors
Ch. 5 # 25
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Calculation of Rout
VX
VX
IX = 2
+
2ro1, 2 + 1/ gm 3 ro4
Rout ≈ ro2 || ro4 , (2ro1,2 >> [1/ gm 3 ] || ro3 )
Current Mirrors
Ch. 5 # 26
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Small-Signal Gain
Av ≈ gm 1,2 (ro2 || ro4 )
Current Mirrors
Ch. 5 # 27
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Common Mode Characteristics
ΔVout
ACM =
ΔVin,CM
Current Mirrors
Ch. 5 # 28
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Common Mode (cont.)
ro3,4
1
||
−
2gm3,4
2
gm1,2
−1
ACM ≈
=
1
1 + 2gm1,2 RSS gm 3, 4
+ RSS
2gm1,2
Current Mirrors
Ch. 5 # 29
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Slides
prepared by Travis N. Blalock, University of Virginia.
Common Mode (cont.)
ADM
CMRR =
ACM
gm 3,4 (1 + 2gm1,2 RSS )
= gm1, 2 (ro1,2 || ro3,4 )
gm1,2
= gm 3,4 (ro1, 2 || ro3,4 )(1 + 2gm1, 2 RSS )
Current Mirrors
Ch. 5 # 30