Widlar current source

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.^[1]^[2]^[3]

The Widlar circuit may be used with bipolar transistors, MOS transistors, and even vacuum tubes. An example application is the 741 operational amplifier,^[4] and Widlar used the circuit as a part in many designs.^[5]

This circuit is named after its inventor, Bob Widlar, and was patented in 1967.^[6]^[7]

1 Analysis
2 Design procedure with specified currents
3 Finding the current with given resistor values
- 3.1 Exact solution
4 Output impedance
- 4.1 Current dependence of output resistance
5 References
6 External links
7 See also
8 Further reading

Analysis

Figure 1 is an example Widlar current source using bipolar transistors, where the emitter resistor R₂ is connected to the output transistor Q₂, and has the effect of reducing the current in Q₂ relative to Q₁. The key to this circuit is that the voltage drop across the resistor R₂ subtracts from the base-emitter voltage of transistor Q₂, thereby turning this transistor off compared to transistor Q₁. This observation is expressed by equating the base voltage expressions found on either side of the circuit in Figure 1 as:

$V_B = V_{BE1} = V_{BE2}%2B(\beta_2%2B1)I_{B2}R_2 \ ,$

where β₂ is the beta-value of the output transistor, which is not the same as that of the input transistor, in part because the currents in the two transistors are very different.^[8] The variable I_B2 is the base current of the output transistor, V_BE refers to base-emitter voltage. This equation implies (using the Shockley diode law):

Eq. 1

$(\beta_2%2B1)I_{B2} =\left(1 %2B 1/\beta_2 \right) I_{C2} = \frac{V_{BE1} - V_{BE2}}{R_2} = \frac{V_T}{R_2} \ln \left(\frac {I_{C1}I_{S2}}{I_{C2}I_{S1}}\right)\ ,$

where V_T is the thermal voltage.

This equation makes the approximation that the currents are both much larger than the scale currents I_S1, I_S2, an approximation valid except for current levels near cut off. 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 chosen with different areas.

Design procedure with specified currents

To design the mirror, the output current must be related to the two resistor values R₁ and R₂. A basic observation is that the output transistor is in active mode only so long as its collector-base voltage is non-zero. Thus, the simplest bias condition for design of the mirror sets the applied voltage V_A to equal the base voltage V_B. This minimum useful value of V_A is called the compliance voltage of the current source. With that bias condition, the Early effect plays no role in the design.^[9]

These considerations suggest the following design procedure:

Select the desired output current, I_O = I_C2.
Select the reference current, I_R1, assumed to be larger than the output current, probably considerably larger (that is the purpose of the circuit).
Determine the input collector current of Q₁, I_C1:

$I_{C1} = \frac{\beta_1}{\beta_1%2B1} \left( I_{R1}-I_{C2}/\beta_2 \right)\ .$

Determine the base voltage V_BE1 using the Shockley diode law

$V_{BE1} = V_T \ln \left(\frac{I_{C1}} {I_S} \right) = V_A\ .$

where I_S is a device parameter sometimes called the scale current.

The value of base voltage also sets the compliance voltage V_A = V_BE1. This voltage is the lowest voltage for which the mirror works properly.

Determine R₁:

$R_1 = \frac {V_{CC} - V_A}{I_{R1}}\ .$

* Determine the emitter leg resistance R₂ using Eq. 1 (to reduce clutter, the scale currents are chosen equal):

$R_2 = \frac{V_T}{\left(1 %2B 1/\beta_2 \right) I_{C2}} \ln \left(\frac {I_{C1}}{I_{C2}}\right)\ .$

Finding the current with given resistor values

The inverse of the design problem is finding the current when the resistor values are known. An iterative method is described next. Assume the current source is biased so the collector-base voltage of the output transistor Q₂ is zero. The current through R₁ is the input or reference current given as,

$I_{R1} = I_{C1} %2B I_{B1} %2B I_{B2} \$

$= I_{C1} %2B \frac{I_{C1}}{\beta_1} %2B \frac{I_{C2}}{\beta_2}$

$= \frac{V_{CC}-V_{BE1}}{R_1} \ .$

Rearranging, I_C1 is found as:

Eq. 2

$I_{C1} =\frac{\beta_1}{\beta_1%2B1} \left( \frac {V_{CC}-V_{BE1}}{R_1}-\frac{I_{C2}}{\beta_2} \right)$

The diode equation provides:

Eq. 3

$V_{BE1} = V_T \ln \left( \frac{I_{C1}}{I_{S1}}\right) \ .$

Eq.1 provides:

$I_{C2} = \frac{V_T}{\left(1 %2B 1/\beta_2 \right) R_2} \ln \left(\frac {I_{C1}}{I_{C2}}\right)\ .$

These three relations are a nonlinear, implicit determination for the currents that can be solved by iteration.

We guess starting values for I_C1 and I_C2.
We find a value for V_BE1:

$V_{BE1} = V_T \ln \left( \frac{I_{C1}}{I_{S1}}\right) \ .$

We find a new value for I_C1:

$I_{C1} =\frac{\beta_1}{\beta_1%2B1} \left( \frac {V_{CC}-V_{BE1}}{R_1}-\frac{I_{C2}}{\beta_2} \right)$

We find a new value for I_C2:

$I_{C2} = \frac{V_T}{\left(1 %2B 1/\beta_2 \right) R_2} \ln \left(\frac {I_{C1}}{I_{C2}}\right)\ .$

This procedure is repeated to convergence, and is set up conveniently in a spreadsheet. One simply uses a macro to copy the new values into the spreadsheet cells holding the initial values to obtain the solution in short order.

Note that with the circuit as shown, if V_CC changes, the output current will change. Hence, to keep the output current constant despite fluctuations in V_CC, the circuit should be driven by a constant current source rather than using the resistor R₁.

Exact solution

The transcendental equations above can be solved exactly in terms of the Lambert W function.

Output impedance

An important property of a current source is its small signal incremental output impedance, which should ideally be infinite. The Widlar circuit introduces local current feedback for transistor $Q_{2}$ . Any increase in the current in Q₂ increases the voltage drop across R₂, reducing the V_BE for Q₂, thereby countering the increase in current. This feedback means the output impedance of the circuit is increased, because the feedback involving R₂ 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 2. Transistor Q₁ is replaced by its small-signal emitter resistance r_E because it is diode connected.^[10] Transistor Q₂ 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 R₂:

$I_b \left( ( R_1 \parallel r_E ) %2B r_{\pi} \right) %2B (I_x %2B I_b ) R_2 = 0 \ .$

Rearranging:

$I_b = -I_x \frac{R_2}{( R_1 \parallel r_E ) %2B r_{\pi} %2BR_2} \ .$

Using Kirchhoff's voltage law from the ground connection of R₂ to the ground of the test current:

$V_x=I_x ( r_O %2BR_2) %2BI_b (R_2-\beta r_O)\ ,$

or, substituting for I_b:

Eq. 4

$R_O = \frac {V_x} {I_x} = r_O \left( 1%2B \frac { \beta R_2} {( R_1 \parallel r_E ) %2B r_{\pi} %2BR_2} \right)$ $%2B \ R_2 \left( \frac { ( R_1 \parallel r_E ) %2B r_{\pi}} {( R_1 \parallel r_E ) %2B r_{\pi} %2B R_2 } \right) \ .$

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 R₂ is large enough compared to the r_π of the output transistor (large resistances R₂ make the factor multiplying r_O approach the value (β +1)). The output transistor carries a low current, making r_π large, and increase in R₂ tends to reduce this current further, causing a correlated increase in r_π. Therefore, a goal of R₂ >> r_π can be unrealistic, and further discussion is provided below. The resistance R₁//r_E usually is small because the emitter resistance r_E usually is only a few ohms.

Current dependence of output resistance

The current dependence of the resistances r_π and r_O is discussed in the article hybrid-pi model. The current dependence of the resistor values is:

$r_{\pi} = \frac{v_{be}}{i_{b}}\Bigg |_{v_{ce}=0}$ $=\ \frac{V_\mathrm{T}}{I_\mathrm{B2}} = \beta_2\frac{V_\mathrm{T}}{I_\mathrm{C2}}\ ,$ in ohms, and

$r_O = \frac{v_{ce}}{i_{c}}\Bigg |_{v_{be}=0}$ $=\ \frac {V_A}{I_{C2}}$ is the output resistance due to the Early effect when V_CB = 0 V (device parameter V_A is the Early voltage).

From earlier in this article (setting the scale currents equal for convenience):

Eq. 5

$R_2 = \frac{V_T}{\left(1 %2B 1/\beta_2 \right) I_{C2}} \ln \left(\frac {I_{C1}}{I_{C2}}\right)\ .$

Consequently, for the usual case of small r_E, and neglecting the second term in R_O with the expectation that the leading term involving r_O is much larger:

Eq. 6

$R_O \approx r_O \left( 1%2B \frac { \beta_2 R_2} { r_{\pi} %2BR_2} \right) \$ $=r_O \left( 1%2B \frac { \beta_2\ln \left(\frac {I_{C1}}{I_{C2}}\right)} {\left(\beta_2 %2B 1 \right) %2B \ln \left(\frac {I_{C1}}{I_{C2}}\right)} \right) \ ,$

where the last form is found by substituting Eq. 5 for R₂. Eq. 6 shows that a value of output resistance much larger than r_O of the output transistor results only for designs with I_C1 >> I_C2. Figure 3 shows that the circuit output resistance R_O is not determined so much by feedback as by the current dependence of the resistance r_O of the output transistor (the output resistance in Figure 3 varies four orders of magnitude, while the feedback factor varies only by one order of magnitude).

Increase of I_C1 to increase the feedback factor also results in increased compliance voltage, not a good thing as that means the current source operates over a more restricted voltage range. So, for example, with a goal for compliance voltage set, placing an upper limit upon I_C1, and with a goal for output resistance to be met, the maximum value of output current I_C2 is limited.

The center panel in Figure 3 shows the design trade-off between emitter leg resistance and the output current: a lower output current requires a larger leg resistor, and hence a larger area for the design. An upper bound on area therefore sets a lower bound on the output current and an upper bound on the circuit output resistance.

Eq. 6 for R_O depends upon selecting a value of R₂ according to Eq. 5. That means Eq. 6 is not a circuit behavior formula, but a design value equation. Once R₂ is selected for a particular design objective using Eq. 5, thereafter its value is fixed. If circuit operation causes currents, voltages or temperatures to deviate from the designed-for values; then to predict changes in R_O caused by such deviations, Eq. 4 should be used, not Eq. 6.

References

^ PR Gray, PJ Hurst, SH Lewis & RG Meyer (2001). Analysis and design of analog integrated circuits (4rth Edition ed.). John Wiley and Sons. pp. §4.4.1.1 pp. 299–303. ISBN 0-471-32168-0. http://www.worldcat.org/search?q=0471321680&qt=owc_search.
^ AS Sedra and KC Smith (2004). Microelectronic circuits (5th Edition ed.). Oxford University Press. Example 6.14, pp. 654–655. ISBN 0-19-514251-9. http://www.worldcat.org/search?q=0195142519&qt=owc_search.
^ MH Rashid (1999). Microelectronic circuits: analysis and design. PWS Publishing Co.. pp. 661–665. ISBN 0-534-95174-0. http://www.worldcat.org/search?q=0534951740&qt=owc_search.
^ AS Sedra and KC Smith (2004). §9.4.2, p. 899 (5th Edition ed.). ISBN 0-19-514251-9. http://www.worldcat.org/search?q=0195142519&qt=owc_search.
^ See, for example, Figure 2 in IC voltage regulators.
^ RJ Widlar: US Patent Number 03320439; Filed May 26, 1965; Granted May 16, 1967: Low-value current source for integrated circuits
^ See Widlar: Some circuit design techniques for linear integrated circuits and Design techniques for monolithic operational amplifiers
^ PR Gray, PJ Hurst, SH Lewis & RG Meyer (2001). Figure 2.38, p. 115. ISBN 0-471-32168-0. http://www.worldcat.org/search?q=0471321680&qt=owc_search.
^ Of course, one might imagine a design where the output resistance of the mirror is a major consideration. Then a different approach is necessary.
^ 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. See diode small-signal circuit.

External links

Comparison of MOSFET cascode and BJT Widlar current source (PDF)