MOSFET Current Mirror Current Sources Basic Concepts

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Let us study the most basic Current Mirror circuit, consisting of two matching MOSFET transistors connected back-to-back, such that both have the same Gate-to-Source voltage:

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Given: (*) The two enhancement-type(增强型) NMOS transistors have matching features, as follows: VTH,1 = VTH,2 , kn,1’ = kn,2’, λ1 = λ2 =0 (*) “By structure” we have that VGS,1 = VGS,2 (*) Typically the supply VDD, resistor R and a desired reference current Iref are all given. (*) The ratio (W/L)1 is not necessarily equal to the ratio (W/L)2 (*) “By structure” we have that VGD,1 = 0, and because the transistor is an enhancement-type, this guarantees that transistor 1 is always in Saturation Mode

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Need: Using the transistors geometries (W/L)1 and (W/L)2 as design parameters, we want to create a DC current Io, as long as transistor Q2 is in Saturation Mode

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The Drain of transistor Q2 is connected to a load circuit, not necessarily a resistor. The load circuit typically involves one or more additional MOSFET transistors. Depending on the load, transistor Q2 may be in any of three modes: Saturation, Triode or Cutoff. Of course, only when it is in Saturation it will work as originally planned (a DC current source)

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The current Io always goes away from the load circuit and into Q2. Such a DC current source is said to be a sink.（接收器）

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Design of a Current Mirror DC Sink We shall look first at Q1: I D ,1 = I ref =

1 W = k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) 2 2 L

VDD − VGS ,1 R

So, indeed if Iref is specified and VDD and R are given, then by the right-hand term the needed voltage VGS,1 is specified. Then, using the middle term, need to solve for (W/L)1. 7

Example 1:

Let VDD = 5V, VTH,1 = 1V, kn,1' = 20μA/V2 and R = 1KΩ. What should be (W/L)1 needed for creating Iref = 1mA? 8

Solution of Example 1:

I ref = 1mA = ⇒ I ref = 1mA = ⇒(

VDD − VGS ,1 R

=

5 − VGS ,1 1

⇒ VGS ,1 = 4V

W W 1 1 k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) 2 = ⋅ 20 ⋅10 −3 ( )1 ( 4 − 1) 2 ⇒ L L 2 2

W )1 = 11.11 L

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Let us now focus our attention on the "mirror" transistor Q2:

I D,2

1 W 2 = I o = k n , 2 ' ( ) 2 (VGS , 2 − VTH , 2 ) 2 L

I D ,1 = I ref

1 W = k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) 2 2 L

We now divide the two equations and use all the given matching parameters of the two transistors.

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Io I ref

W ) 2 (VGS , 2 − VTH , 2 ) 2 L = W 0.5k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) 2 L 0.5k n , 2 ' (

Obviously, many terms cancel out and we are left with the following simple design formula:

Io I ref

W ( )2 = L W ( )1 L 11

Example 2 (Follow-up to Example 1) What should be (W/L)2 if we want Io = 7mA? Solution By Example 1 we have that Iref = 1mA and (W/L)1 = 11.11. Therefore:

Io I ref

W W ( )2 ( )2 7 mA W L L = ⇒ = ⇒ ( ) 2 = 77.77 W 1mA 11.11 L ( )1 L

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Question: What do we do if we want to create a source DC current source, rather than a sink ? [A "source" is when the current goes from the current source into the load circuit] Answer: If we build a current mirror current source using PMOS transistors (rather than NMOS) then the output DC current will be "sourced" and not "sunk".

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Question: If we need to generate multiple different DC current sources and sinks, what is the total number of resistors needed for the design? Answer: Just one resistor for the entire circuit!

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Current Steering

The use of a negative DC supply –VSS does not change the fact that, by-structure, both transistors, in every mirror pair, 15 have the same VGS voltage.

Recalculation of Reference Current

I D ,1 = I ref

VDD − (−VSS + VGS ,1 ) W 1 2 = k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) = L R 2

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NMOS Current Mirror Sinks:

I2 I ref

W )2 = L W ( )1 L (

I3 I ref

W )3 = L W ( )1 L (

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Current Steering Mechanism:

The Drain current of the NMOS transistor Q3 comes from the Drain of the PMOS transistor Q4. I4 = I3 Can "steer" a current from NMOS current mirror to PMOS 18 current mirror, or vice versa.

Current Steering Mechanism:

There is no need for the NMOS and PMOS to be matching: Only all the NMOS transistors must match among themselves, and the PMOS transistors must be mutually matching. 19

PMOS Current Mirror

For the PMOS transistors Q4 and Q5, the following parameters must match: VTH,4 = VTH,5 , kp,4’ = kp,5’, λ4 = λ5

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PMOS Current Mirror

W ( )5 I5 L = W I4 ( )4 L 21

We can now relate the sourced current I5 to the reference current Iref:

I5 I ref

W W )3 ( )5 I I I = 3 ⋅ 4 ⋅ 5 = L ⋅1 ⋅ L W W I ref I 3 I 4 ( )1 ( )4 L L (

I5 I ref

W W )3 ( )5 = L ⋅ L W W ( )1 ( ) 4 L L (

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1) We need only one resistor R, no matter how large is the number of DC currents generated. 2) By using current steering we can create sourcing currents from sinking currents or vice versa.

3) The reference current can reside either in a NMOS current mirror or in a PMOS current mirror. 23

Example (a follow-up to the previous example) Given: VDD = 5V, VSS = 0 VTH,1 = VTH,2 = VTH,3 = 1V, kn,1' = kn,2'=kn,3'= 20μA/V2 and R = 1KΩ. VTH,4 = VTH,5 = -1V, kp,4' = kp,5'= 30μA/V2 Need: A reference current of Iref = 1mA, one sink current of 7mA and one source current of 5mA. Question: Using the diagram scheme just discussed, what should be all (W/L) ratios of all five transistors?

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Solution The beginning of the solution is identical to what has been done in the previous examples. Let's quote the results: (W/L)1 = 11.11 and (W/L)2 = 77.77mA. Current I2 is then the desired sink current

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Now:

I5 I ref

W W W W ( )3 ( )5 ( )3 ( )5 5mA = = L ⋅ L = L ⋅ L W W 1mA 11.11 W ( )1 ( ) 4 ( )4 L L L

There are infinitely many choices for the geometric dimensions of transistors Q3, Q4 and Q5. For instance, we may take (W/L)3 = 11.11, (W/L)4 = 10 and (W/L)5 = 50. Current I5 is then the required source current.

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The channel-length modulation effect may be responsible for errors in the operation of a Current Mirror Current Source. For instance, depending on the load of Q2 we may get VDS,2 ≠ VDS,1. As a result the value of I2 will slightly vary, depending on the load. It means that the current source is not ideal - it has a finite output resistance, equal to ro of the respective output transistor.

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