This version is *outdated* by a newer approved version.This version (06 Aug 2017 12:37) is a *draft*.

Approvals: 0/1The Previously approved version (06 Jun 2017 17:45) is available.

Approvals: 0/1The Previously approved version (06 Jun 2017 17:45) is available.

**This is an old revision of the document!**

*Basic ideas behind the classic current mirror are revealed below by using the building approach proposed in the Wikibooks story* How to Reverse Current Direction.

A current mirror is a circuit block which functions to produce a copy of the current in one active device by replicating the current in a second active device. An important feature of the current mirror is that it has a relatively high output resistance which helps to keep the output current constant, regardless of load conditions. Another feature of the current mirror is that it has a relatively low input resistance which helps to keep the input current constant, regardless of drive conditions. The current being 'copied' can be, and often is, a varying signal current. Conceptually, an ideal current mirror is simply an ideal current amplifier with a gain of -1. The current mirror is often used to provide bias currents and active loads in amplifier stages. Given a current source as the input, we convert the current (entering the current mirror) into a voltage and then use this voltage to control a current sink (the current exiting the mirror); as a result, we obtain a current sink (figure 11.1a). Conversely, given a current sink as the input, we convert the input current (exiting the current mirror) into a voltage and then use this voltage to control a current source (figure 11.1b); as a result, now we obtain a current source. We can generalize this basic current mirror structure in a first conclusion:

A current mirror consists of a current-to-voltage converter consecutively connected to a voltage-to-current converter.

Figure 11.1, Current Mirror (a) Sink (b) Source

It should be noted that the two converters may have a linear relationship (for example where V_{OUT} = I_{IN}R and I_{OUT}= V_{IN}/R) like a resistor, but this linear relationship is not required. The converters might be non-linear devices having whatever transfer or I to V characteristics that may even depend on another quantity (such as temperature); the only requirement is the characteristics be the inverse of each other. For example, if the I to V converter implements a function v = f(i) and the other represents the inverse function i = f ^{-1}(v) the whole function is v = f(i) = f(f ^{-1}(v)) . So, we can formulate the second conclusion: a current mirror consists of two consecutively connected converters that have inverse transfer functions.

We need a configuration where our active element of choice, a transistor, serves as the desired current-to-voltage converter. However, the transistor is a unidirectional device, where for the BJT the base emitter voltage controls the collector current or for the FET the gate source voltage controls the drain current. Producing the opposite where the collector current controls the V_{BE} is not possible in the conventional use of the device as a common emitter amplifier. The solution is to incorporate negative feedback. In this case that means making the transistor adjust its base emitter or gate source voltage, V_{BE} or V_{GS}, so that the collector or drain current is I_{IN} = (V_{1}-V_{BE})/R. For this purpose, we simply connect the collector to the base or gate to drain. This results in 100% parallel negative feedback (figure 11.2). As a result, with this reversed transistor, the collector current serves as the input quantity while the base-emitter voltage V_{BE} serves as the output quantity with a logarithmic transfer function. The input part of the simple BJT current mirror is just a bipolar transistor with 100% parallel negative feedback. Similarly, a diode connected enhancement mode MOS FET (gate tied to drain) will serve as a similar current to voltage converter with V_{GS} as the output quantity rather than V_{BE}.

Figure 11.2, Current to Voltage Converter

A bipolar transistor can be driven by a voltage or by a current. If we consider the base emitter voltage, V_{BE}, as the input and the collector current, I_{C}, as the output (figure 11.3), we can think of a transistor as a non-linear voltage-to-current converter having an exponential characteristic. The base can be directly driven by the voltage output of the I-to-V converter we just discussed. The collector provides the output terminal of our simple current mirror: The output V to I converter stage of the simple current mirror is just a transistor acting as a non-linear (exponential for BJT) voltage-to-current converter. Again if a MOS transistor were used for the input stage the output stage would be a MOS transistor with the gate serving as the voltage input and the drain as the current output.

Figure 11.3, Voltage to Current Converter

The final step is to connect the output of the input stage (the base emitter junction of Q_{1}) to the input of the output stage (the base emitter junction of Q_{2}) to build the basic BJT current mirror circuit (figure 11.4). At this point we will concentrate on the issues involved with the BJT current mirror and pick back up with the MOS current mirror in section 11.6.

Figure 11.4, BJT Current Mirror

If transistors Q_{1} and Q_{2} in figure 11.4 are identical (that is have the same size emitter and thus equal I_{S}) the input current to output current ratio or gain is ideally 1. There are often occasions when a gain other than one is required. When building circuits from discreet devices only simple integer ratios are possible while in microelectronic integrated circuits it is possible to make transistors with arbitrary emitter areas, A. However, even in integrated circuits the best design practice is to use identical unit size transistors when making current mirrors.

Figure 11.4.1, Current Mirror with non-unary gain ratio

If on the input side of the mirror we connect N identical devices in parallel and connect M devices in parallel on the output side, the gain of the mirror will be M/N. In figure 11.4.1 we see an example where 2 (N=2) devices are connected together on the input and 3 (M=3) devices are connected together on the output. The mirror gain will thus be 3/2 or 1.5. Since all five transistors share the same V_{BE}voltage, their collector currents, I_{C}, will all be equal. The input current I_{IN} splits equally in Q_{1} and Q_{2} such that:

, or

There are three primary error sources that make the simple 2 transistor mirror less than ideal. The first is the mirror gain. Ideally I_{OUT} should exactly equal I_{IN}. There are systematic and random factors that make this not the case. The second is the incremental output resistance, which determines how much the output current varies with the voltage seen at the mirror output. The third limitation is the minimum voltage drop across the output leg of the mirror necessary to maintain the desired output current. This minimum voltage, called output compliance, is determined by the need to keep the output transistor of the mirror in the active region. There are also a number of secondary performance issues with mirrors, for example, temperature stability and frequency response.

An error source in this simple BJT based current mirror is that the transistors Q_{1} and Q_{2} (figure 11.4) each remove a base current I_{B} from the input current I_{IN}.
As a result, the output current is smaller than the input current:

As was already discussed, current mirrors can just as easily be made from MOS FET transistors. The I-to-V and V-to-I functions are different but of course are still the inverse of each other. A significant advantage of the MOS current mirror is the lack of base current induced error that BJT current mirrors suffer. There are methods to correct or compensate for the base current in BJT current mirrors which will be discussed in detail in later sections of this chapter.

It is necessary to keep the output (BJT) transistor out of saturation, V_{CB} = 0 V. Or from another
perspective, not allow the collector base junction to forward bias. That means the lowest output voltage that
results in the correct output current, the compliance voltage, is V_{OUT} = V_{CV} =
V_{BE} under bias conditions with the output transistor at the output current level I_{C} and
with V_{CB} = 0 V or, restating the V_{BE}relation from earlier:

Where V_{T} is the thermal voltage and I_{S} is the reverse saturation current.

The V_{CB} of Q_{1} in the mirror is zero. If V_{CB} is greater than zero in the
output transistor Q_{2}, the collector current in Q_{2} will be somewhat larger than
Q_{1} due to the Early effect. In other words, the mirror has a finite output resistance given by the
r_{O} of the output transistor, namely:

Where:

V_{A} is the Early voltage

V_{CB} is the collector-to-base voltage

As we learned in an earlier chapter, the inclusion of emitter degeneration resistors (R_{E1} and
R_{E2} in figure 11.5) can increase the effective collector impedance seen at the mirror output. In
order for the mirror gain to remain equal to 1, R_{E1} must of course equal R_{E2}. The added
voltage drop across the emitter resistor R_{E2} (I_{OUT} x R_{E2}) adds to the
minimum allowable output voltage (see section 11.5.2).

While resistors could also be added to the sources in an MOS based mirror, it is often more effective to simple increase the channel length, L, of the transistors. The longer the channel the less it is affected by the channel length modulation due to the increasing drain voltage.

Figure 11.5, Emitter degeneration included to boot output resistance.

It is important to note that the inclusion of emitter resistors does not reduce the reduction in the output
current I_{OUT} caused by the finite beta of Q_{1} and Q_{2}. A compensating voltage
can be inserted by including resistor R_{B}, of the correct value, as shown in figure 11.6.

Figure 11.6, Adding a resistor in the base of Q_{1} tends to compensate for the finite beta of Q_{2}

For V_{CB} close to zero (that is, neglecting base-width modulation errors) the necessary value for
R_{B} is:

For example, using A = 2 ( a mirror gain of 2 ), I_{IN} = 1mA, so r_{E} = 26Ω and
R_{E1} = 500Ω (introducing about 500mV of degeneration) R_{B} should be 1578Ω if β is enough
larger than A so that factor can be ignored, or 1626Ω if it is included and beta is 100. Clearly, this
compensation method becomes unpredictable when β is small and is comparable to A and will never be precise
because β_{1} and β_{2} are not in general equal. However, this technique can be of practical
utility in many cases.

The simple current mirror can, obviously, also be implemented using MOSFET transistors, as shown in figure
11.7. We know that transistor M_{1} is operating in the saturation region because V_{DS} is
greater than or equal to V_{GS}. Transistor M_{2}will also be in saturation so long as the
output voltage is larger than its saturation voltage. In this simple configuration, the output current
I_{OUT} is directly related to I_{IN}.

Figure 11.7 Simple MOS current mirror

The drain current of a MOSFET I_{D} is a function of both the gate to source voltage and the drain to
gate voltage of the MOSFET given by I_{D} = f (V_{GS}, V_{DG}), a relationship
derived from the functionality of the MOSFET device. In the case of transistor M_{1} of the mirror,
I_{D} = I_{IN}. Input current I_{IN} is a known current, and can be provided by a
resistor as shown in the figure, or by a threshold-referenced or self-biased current source to ensure that it
is constant, independent of voltage supply variations.

Using V_{DG}=0 for transistor M_{1}, the drain current in M_{1} is I_{D} = f
(V_{GS},V_{DG}=0), so we find: f (V_{GS}, 0) = I_{IN}, implicitly determining
the value of V_{GS}. Thus I_{IN} sets the value of V_{GS}. The circuit in the diagram
forces the same V_{GS} to apply to transistor M_{2}. If M_{2} also is biased with
zero V_{DG} and provided transistors M_{1} and M_{2} have good matching of their
properties, such as channel length, width, threshold voltage etc., the relationship I_{OUT} = f
(V_{GS},V_{DG}=0 ) applies, thus setting I_{OUT} = I_{IN}; that is, the
output current is the same as the input current when V_{DG}=0 for the output transistor, and both
transistors are matched.

The drain-to-source voltage can be expressed as V_{DS}=V_{DG} + V_{GS}. With this
substitution, the Shichman-Hodges model provides an approximate form for function f(V_{GS},V_{DG}):

Where:

Kp is a technology related constant associated with the transistor,

W/L is the width to length ratio of the transistor,

V_{GS} is the gate-source voltage, V_{th} is the threshold voltage, V_{DS} is the
drain-source voltage

λ is the channel length modulation constant

**Output resistance**

Because of channel-length modulation, the mirror has a finite output resistance given by the *ro* of the
output transistor, namely:

Where:

λ = channel-length modulation parameter

V_{DS} = drain-to-source bias.

**Compliance voltage**

To keep the output transistor resistance high, V_{DG} = 0 V. That means the lowest output voltage
that results in correct mirror behavior, the compliance voltage, is V_{OUT} = V_{CV} =
V_{GS} for the output transistor at the output current level with V_{DG} = 0 V, or using the
inverse of the f- function, f^{-1}:

For the Shichman-Hodges model, f^{-1} is approximately a square-root function.

Figure 11.8 shows a mirror where the simple wire connecting the collector of Q_{1} to its base is replaced by an emitter follower buffer. This improvement to the simple current mirror is referred to as an emitter follower augmented mirror. The current gain (ß_{Q3}) of the emitter follower buffer stage (Q_{3}) greatly reduces the gain error caused by the finite base currents of Q_{1} and Q_{2}.

Figure 11.8 Buffered Feedback current mirror.

One thing to note that is different in this mirror configuration vs. the simple two transistor mirror is that
the Collector-Base voltage, V_{CB}, of Q_{1} is no longer zero. It is equal to the
V_{BE} of Q_{3}. Given the effect of the finite output resistance (Early effect ) the output
current I_{OUT} in Q_{2} will most closely match I_{IN} when the collector voltage of
Q_{2} is the same as that of Q_{1} which is 2XV_{BE} above the common voltage. Also
note that when driven by a resistor, like R_{1}, I_{IN} will now be
(V_{1}-V_{BE1}-V_{BE2})/R_{1}.

Another consequence of adding the emitter follower buffer is, in general, a loss in the frequency response of
the mirror. Transistor Q_{3} is potentially operating at a very small current of 2I_{B}. If
there were to be a significant capacitance to ground at the base connection common to Q_{1} and
Q_{2} the current available to discharge this current will also be small equal to 2I_{B}. But
the current available to charge this node is potentially equal to ß_{Q3}I_{IN} which is very
much larger than 2I_{B}. This asymmetry in the charging vs. discharging current available for this
node in the current mirror can lead to very undesirable response to fast changes to I_{IN}.

A Wilson current mirror or Wilson current source, named after George Wilson, is an improved mirror circuit configuration designed to provide a more constant current source or sink. It provides a much more accurate input to output current gain. The structure is shown in figure 11.9.

Figure 11.9 The Wilson Current Mirror

We will be making the following two assumptions. First, all transistors have the same current gain ß. Second,
Q_{1} and Q_{2}are matched, so their collector currents are equal. Therefore, I_{C1}
= I_{C2} (= I_{C}) and I_{B1} = I_{B2} (= I_{B}) .

The base current of Q_{3} is given by,

The emitter of Q_{3} current by,

Looking at figure 11.9, it can be seen that _{IE3} = I_{C2} + I_{B1} + I_{B2}
Substituting for I_{C2}, I_{B1} and I_{B2}, I_{E3} = I_{C} + 2I_{B}

so,

Substituting for I_{E3}

rearranging,

The current through R_{1} is given by, I_{R1} = I_{C1} + I_{B3}

But, I_{C1} = I_{C2} = I_{C}

Substituting for I_{C} and since
we get,

Therefore,

And finally,

From the above equation we can see that if

And the output current (assuming the base-emitter voltage of all transistors to be 0.7 V) is calculated as,

Note that the output current is equal to the input current I_{R1} which in turn is dependent on
V_{1} and R_{1}. If V_{1} is not stable, the output current will not be stable. Thus
the circuit does not act as a regulated constant current source.

In order for it to work as a constant current source, R_{1} must be replaced with a constant current
source.

**Advantages over other configurations**

This circuit has the advantage of virtually eliminating the base current mismatch of the conventional BJT
current mirror thereby ensuring that the output current I_{C3} is almost equal to the reference or
input current I_{R1}. It also has a very high output impedance due to the negative feedback through
Q_{1} back to the base of Q_{3}.

Adding a fourth transistor to the simple Wilson current mirror in figure 11.10, we have the modified or
improved Wilson mirror. The improved input to output current accuracy is accomplished by equalizing the
collector voltages of Q_{1} and Q_{2} at 1 V_{BE}. This leaves the finite ß and
voltage differences of each of Q_{1} and Q_{2} as the remaining unbalancing influences in the
mirror.

Figure 11.10 The improved Wilson Current Mirror

A Widlar current source is a modification of the basic two-transistor current mirror that incorporates an emitter degeneration resistor for only the output transistor, enabling the current source to generate low currents using only moderate resistor values. This circuit is named for its inventor, Robert Widlar, and was patented in 1967.

The Widlar circuit may be used with bipolar transistors or MOS transistors. An example application is in the now famous uA741 operational amplifier, and Widlar used the circuit in many of his designs.

Figure 11.11 A version of the Widlar current source using bipolar transistors.

Figure 11.11 is an example Widlar current source using bipolar transistors, where the emitter resistor
R_{2} is connected in series with the emitter of output transistor Q_{2}, and has the effect
of reducing the current in Q_{2} relative to Q_{1}. The key to this circuit is that the
voltage drop across the resistor R_{2} subtracts from the base-emitter voltage of transistor
Q_{2}, thereby reducing the collector current compared to transistor Q_{1}. A simulation plot
showing this reduction in I_{C2} is presented in figure 11.12.

Figure 11.12 Plot of the collector current of Q_{1} and Q_{2}
R_{1} = 15KΩ, R_{2} = 300

This observation is expressed by using KVL around the base emitter loop of the circuit in Figure 11.11 as:

Where ß_{2} is the beta of the output transistor, which may not be the same as that of the input
transistor, in part because the currents in the two transistors are very different. The variable
I_{B2} is the base current of the output transistor, V_{BE} refers to base-emitter voltage.
If we neglect the effect of finite ß and use the V_{BE} equation we can obtain a useful formula for
the output current:

where V_{T} is the thermal voltage, I_{IN} = I_{C1} and I_{OUT} = I_{C2}.

Suppose we want to create a 100uA output current from a 300uA input current as in the simulation plot of
figure 11.12. V_{T} is 26mV times ln(3) is 28.5mV. 28.5mV divided by 100uA is 285 ohms.
This equation makes the approximation that the currents are both much larger than the saturation currents
I_{S1}, I_{S2}, an approximation valid except for very low current levels. In the following
the distinction between the two scale currents is dropped, although the difference can be important, for
example, if the two transistors are designed with different emitter areas.

Figure 11.13 Small-signal circuit for finding output resistance of the Widlar source shown in figure 11.11.

A test current I/_{x} is applied at the output, and the output resistance is then R_{O} =
V_{x} / I_{x}.

An important property of a current source is its small signal incremental output impedance, which should
ideally be infinite. The emitter degeneration resistance introduces local current feedback for transistor
Q_{2}. Any increase in the current in Q_{2} increases the voltage drop across R_{2},
reducing the V_{BE} for Q_{2}, thereby countering the increase in current. This feedback
means the output impedance of the circuit is increased, because the feedback involving R_{2} forces
use of a larger voltage to drive a given current.

Output resistance is found using a small-signal model for the circuit, shown in Figure 11.13. The transistor
Q_{1} is replaced by its small-signal emitter resistance r_{E} because it is diode connected.
In a diode-connected transistor the collector is short-circuited to the base, so the transistor
collector-base junction has no time-varying voltage across it. As a result, the transistor behaves like the
base-emitter diode, which at low frequencies has a small-signal circuit that is simply the resistor
r_{E} = V_{T} / I_{E}, with I_{E} the DC Q-point emitter current. The
transistor Q_{2} is replaced with its hybrid-pi model. A test current I_{x} is attached at
the output.

Using the figure, the output resistance is determined using Kirchhoff's laws. Using Kirchhoff's voltage law
from the ground on the left to the ground connection of R_{2}:

Rearranging:

Using Kirchhoff's voltage law from the ground connection of R_{2} to the ground of the test current:

or, substituting for I_{b}:

Eq. 4

According to Eq. 4, the output resistance of the Widlar current source is increased over that of the output
transistor itself (which is r_{O}) so long as R_{2} is large enough compared to the
r_{p} of the output transistor. (Large resistances R_{2} make the factor multiplying
r_{O} approach the value (ß +1).) The output transistor carries a low current, making r_{p}
large, and increase in R_{2} tends to reduce this current further, causing a correlated increase in
r_{p}. Therefore, a goal of R_{2} » r_{p} can be unrealistic, and further discussion
is provided below. The resistance R_{1} / r_{E} usually is small because the emitter
resistance r_{E} usually is only a few ohms.

**ADALM1000 Lab Activity 6, BJT Current Mirror **

**ADALM1000 Lab Activity 6M, MOS Current Mirror **

**ADALM2000 Lab Activity 6, BJT Current Mirror **

**ADALM2000 Lab Activity 6M, MOS Current Mirror **

When designing a circuit it is important to take into account the wide variation in certain device values from one to another. A central objective of the designer is to desensitize the circuit to these variations to produce a circuit which meets the specifications across all possible conditions. One aspect of design which is common to nearly all circuits is the establishment of stable bias or operating point levels. This seemingly minor portion of a design can provide the most challenging and interesting circuit problems. Many bias generators are centered around the generation of currents to operate the core of the circuit. Current generated from simple resistors and diodes or diode connected transistors connected across the power supply will vary approximately proportional to the variation of the supply voltage. This variation in the resulting bias current is frequently undesirable.

This is to introduce another kind of current mirror, actually a stabilized current source, which has an output which had been desensitized to variation in input current. To understand this configuration, it is helpful to examine the behavior of a zero gain amplifier. A NMOS version is shown in figure 11.14 but PMOS, NPN or PNP transistors will just as well function in this configuration

Figure 11.14 NMOS Zero Gain Amplifier

Remembering back to the previous explanation of the common emitter/source amplifier (figure 11.14(a)), the
gain is a function of the drain (or collector) current and the load resistor. For a given drain current, if
the drain resistor R_{L} is set equal to r_{s} then the gain A will be minus 1.

Alternatively. if the gate is held fixed at the same DC bias level which produces an identical drain current
as in (a) and an input signal is applied to the top of load resistor R_{L} (figure 11.14(b)), then
the gain will be plus 1. That is if the drain/source output impedance of the transistor is neglected. If we
now connect the gate to the top of resistor R_{L} as in figure 11.14©, the net gain superimposing
both paths will be 1 - 1 = 0.

In figure 11.15 we have an NPN transistor biased into conduction with a collector voltage V_{C} which
is less than the base voltage V_{BE} by the thermal voltage V_{T}= kT/q, (equal to Ic times
R_{L}) and is essentially constant with input voltage changes applied from V_{IN}. The
voltages seen at V_{BE} and V_{C} are plotted vs. the applied voltage at V_{IN} in
figure 11.14. As we can see while V_{BE} continues to rise, V_{C} remains much more constant
and actually decreases above a certain level of V_{IN}.

Figure 11.15 NPN Zero Gain Amplifier

Figure 11.16 Plot of V_{BE} and V_{C} vs V_{IN} for R1 = 10KΩ and
R_{L}=75Ω

A zero gain amplifier made using an enhancement mode NMOS 2N7000 transistor was simulated where the small signal AC gain and phase was calculated as the drain current was swept. As can be seen in figure 11.17 there is a sharp null or dip in the gain curve at around 345uA. This also occurs at the point where the phase makes a sharp transition from 0 degrees to 180 degrees.

Figure 11.17 Small signal gain/phase plot of zero gain amplifier vs. I_{D}

**ADALM1000 Lab Activity 7, BJT Zero Gain Amplifier **

**ADALM1000 Lab Activity 7m, NMOS Zero Gain Amplifier **

**ADALM2000 Lab Activity 7, BJT Zero Gain Amplifier **

**ADALM2000 Lab Activity 7m, NMOS Zero Gain Amplifier **

Now that we understand the concept of the zero gain amplifier, the objective is to investigate its use to
produce an output current which is stabilized (less sensitive) to variations of the input current level. This
current source configuration, figure 11.17, is also sometimes called a Peaking Current Source or
g_{m}-compensated mirror. Because the collector voltage V_{C} of transistor Q_{1} is
now more constant with changes in the input supply voltage as represented by V_{IN}, it can be used
as the base voltage of Q_{2} to produce a much more constant collector current in transistor
Q_{2}.

Figure 11.18 Stabilized (Peaking) current source (g_{m}-compensated mirror)

Figure 11.19 Plot of the collector current of Q_{1} and Q_{2}
R_{1} = R_{2} = 10KΩ, R_{L} = 75Ω

Figure 11.20 PTAT current plot of Peaking Current Source

Figure 11.21 A combination of the Widlar mirror and peaking current source provides further improvement in the regulation of a variable input current.

Figure 11.22 Simulation of circuit in figures 11.18 and 11.21 with R_{1} = 20KΩ,
R_{L}= 250Ω and R_{2}= 1.2KΩ

Figure 11.23 cascade of NPN and PNP peaking current sources

**ADALM1000 Lab Activity 8, BJT Stabilized current source**

**ADALM1000 Lab Activity 8m, NMOS Stabilized current source**

**ADALM2000 Lab Activity 8, BJT Stabilized current source**

**ADALM2000 Lab Activity 8m, MOS Stabilized current source**

**Return to Previous Chapter**

**Go to Next Chapter**

**Return to Table of Contents**

university/courses/electronics/text/chapter-11.1502015875.txt.gz · Last modified: 06 Aug 2017 12:37 by Circuit_dreamer