6.301 Solid-State Circuits - Massachusetts Institute of ...6.301/rec13.pdf6.301 Solid-State Circuits Recitation 13: Current Sources and Current Mirrors Prof. Joel L. Dawson Page 2

6.301 Solid-State CircuitsRecitation 13: Current Sources and Current MirrorsProf. Joel L. Dawson

Many times in complex analog systems, it is useful to have a current source. Consider our “activeloading” concept from the last recitation:

Remember we speculated that if we could make the collector current of Q1match ILOAD , this mightbe a way to get a very high gain stage. Infinite gain, even, were it not for base width modulation.

Now we know that transistors themselves make very good dependent current sources. This suggeststheir use as good independent current sources, provided that we fix the base-emitter voltage in somesensible way.

One sensible way that we looked at as follows:

Large voltage drop here meansVBE variations not important.

ILOAD↓V0

Q1

VB−+

IOUT

−VEE

↓⇒

IOUT↓

VCC


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This is a fine idea when you have a lot of voltage headroom. Today we’re going to look atalternatives, though, that often prove to be more useful.

CLASS EXERCISE

Compute I0 for the following circuit:

(Workspace)

And just like that, we discover the current mirror. It’s a basic analog building block, and is one waythat we have for building a good current source:

Q2

↓ I0

Q1

10k

10.6V

Q2Q1

↓ I0

⇒↓ I0


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Notice how this technique solves a thorny problem for using transistors as a current source. We’vealways avoided doing things like:

Because of the uncertainty in IS . But if Q1and Q2 are matched, we don’t care what the exact value ofIS is. We establish a reference current using the power supply and Q1 , and use the nonlinearity of theI1 −VBE1 relationship to “undo” the nonlinearity of the VBE2 − I2 relationship. In other words, IC1 getsmirrored into the collector of IC2 .

Remember that IS = AE ⋅qDnni

2

WBNB

⎛⎝⎜

⎞⎠⎟

, where AE is the emitter area. Accordingly, we can get some

variety in our current mirrors by sizing the transistors differently:

I0 = IseqVBkT

↓

VB+−

AE2 = 10AE1

I0 = 10 ⋅ IREF↓

AE1

IREF↓


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This idea of using matched devices also comes in MOSFET flavors:

I0 = IREF ⋅

WL

⎛⎝⎜

⎞⎠⎟ 2

WL

⎛⎝⎜

⎞⎠⎟ 1

Matching

Now we agreed that if the devices in a bipolar current mirror are matched, we have close to identicalcurrents in the collectors.

ID =µnCox2

WL

⎛⎝⎜

⎞⎠⎟VGS −VTH( )2

D

S

GWL

⎛⎝⎜

⎞⎠⎟ 1

WL

⎛⎝⎜

⎞⎠⎟ 2

IREF↓I0↓

I2

Q2VBQ1

↓I1IC1 = IS1 exp

qVBkT

⎛⎝⎜

⎞⎠⎟

IC2 = IS2 expqVBkT

⎛⎝⎜

⎞⎠⎟


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We can conclude that for matching, we have:

IC1 − IC2 = IS1 − IS2( )exp qVBkT⎛⎝⎜

⎞⎠⎟

ΔICIC

=ΔISIS

Some common current mirrors:

Simple Current Mirror

We calculate the error according to I0 =IR1+ 2

β

Error = 2β

R0 = r0 [Output impedance]

I0

→

I0β

IR↓

↓I0

KCL : IR = I0 +2I0β

= I0 1+2β

⎛⎝⎜

⎞⎠⎟= I0

β + 2β

⎛⎝⎜

⎞⎠⎟

I0 = IRβ

β + 2⎛⎝⎜

⎞⎠⎟


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Can prevent thermal runaway, boost output impedance, and reduce sensitivity to matching by usingemitter degeneration:

Buffered Current Mirror

IB =2I0

β +1( )β

IR = I0 +2I0

β β +1( )

I0 =IR

1+ 2β β +1( )

Error = 2β β +1( ) ≈

2β 2

R0 = r0

R0 ≈ rπ RE + 1+ gm rπ RE( )( )r0

Error = 2β

I0

RERE

IR ↓↓

I0β

I0β

→←

I0↓

↓I0

IB→

IR↓


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Widlar Current Mirror (How to get very small output currents)

KVL :VBE1 = VBE2 + I2RE

VT lnI1IS

⎛⎝⎜

⎞⎠⎟= VT ln

I2IS

⎛⎝⎜

⎞⎠⎟+ I2RE

VT lnI1I2

⎛⎝⎜

⎞⎠⎟= I2RE

So if, for instance, we wanted I1 = 1mA and I2 = 1µA :

VT ln10−3A10−6A

⎛⎝⎜

⎞⎠⎟= VT ln(1000) = 180mV

I2RE = 180mV

RE =180mV1µA

= 180kΩ

The output impedance of this mirror is

r0 ≈ rπ 2 rE + 1+ gm2 rπ 2 RE( )( )r02

RE

Q2

I2↓↓

Q1

I1

6.301 Solid-State Circuits - Massachusetts Institute of ...6.301/rec13.pdf6.301 Solid-State Circuits Recitation 13: Current Sources and Current Mirrors Prof. Joel L. Dawson Page 2

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