EECE488 Set 5 - Current Mirrors 1SM

EECE488: Analog CMOS Integrated Circuit Design

Set 5

Current Mirrors

Shahriar MirabbasiDepartment of Electrical and Computer Engineering

University of British Columbiashahriar@ece.ubc.ca

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Applications of Current Sources

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Simple Resistive Biasing for Current Source

2

12

2 )(2 THDD

oxnOUT VV

RRR

LWCI −

+≈µ

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Problems

• Output current depends on:

– Supply

– Process

– Temperature

• What if the bias voltage is independent of supply voltage?

• Is there a way of generating reliable currents?

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Basic Idea

Iout is a function of gate-source voltage

Typically we assume that one precisely defined current source isavailable and other current sources copy their current from this precise source.

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Basic Idea

This structure is called current mirror

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Question

• What happens if the two transistors in the basic current mirror have different sizes?

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Example

Assuming all the transistors are in saturation region, find Iout:

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Current Mirrors: Amplifier Bias Example

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Board Notes

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Current Mirrors: Signal Amplification Example

• Find the small signal voltage gain of the following circuit.

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Effect of Channel Length Modulation

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Cascode Current Mirror

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Board Notes

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Cascode Current Mirror

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Cascode Current Mirror

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Cascode Current Mirror

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Board Notes

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Cascode Current Mirror Biasing

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Cascode Current Mirror Biasing

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Current Mirror Biasing

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Basic Circuit to Generate Supply Independent Current

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Supply Independent Current

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Board Notes

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Supply Independent Current

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Start-up Problem

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Start-up Problem

DDGSTHGSDDTHTHTH VVVVandVVVV >++<++ 351351

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Board Notes

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Active Current Mirrors

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Active Current Mirrors in Differential to Single-Ended Amplifiers

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Differential to Single-Ended Amplifiers

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Calculation of Gm

ID1 = ID3 = ID 4 = gm1,2Vin / 2 ID2 = −gm1,2Vin / 2

Iout = ID2 − ID4 = −gm1,2Vin ,⇒ Gm = gm1, 2

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Small-Signal Gain

Av ≈ gm 1,2 (ro2 || ro4 )

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Common Mode Characteristics

ACM =∆Vout

∆Vin ,CM

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Common Mode

ACM ≈−

12gm3,4

||ro3,4

21

2gm1,2

+ RSS

=−1

1 + 2gm1,2 RSS

gm1,2

gm 3, 4

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Common Mode

CMRR =ADM

ACM

= gm1, 2(ro1,2 || ro3,4 )gm 3,4 (1 + 2gm1,2 RSS )

gm1,2

= gm 3,4 (ro1, 2 || ro3,4 )(1 + 2gm1, 2RSS)