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The Basic Current Mirror

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Adding Emitter Resistors

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Adding Emitter Resistors 

If we add an emitter resistor to a single transistor amplifier, we get negative feedback, or "degeneration." When we do the same thing to a current mirror, as shown to the right, we still get degeneration, but we also get a couple of extra effects.
The basic output resistance, r_{O}, is set primarily by the Early voltage, V_{A}, of that transistor. Mathematically, r_{O} ≈ V_{A}/I_{C}. If V_{CE} is very small (as with Q1), very little current will be lost through R_{O}, and this parameter has negligible effect. However, if V_{CE} becomes larger (as with Q2), the current lost through R_{O} becomes significant, and causes the behavior of the transistor to depend on collector voltage.
The presence of the emitter resistors increases the effective output resistance of both transistors. This doesn't matter for Q1, since its V_{CE} is fixed. However, it does help with Q2, by reducing any changes in transistor characteristics as its output voltage changes. The increase in output resistance depends on the value of the resistance and on a transistor parameter called transconductance. Transconductance is given the symbol g_{m}, and is defined mathematically as the change in collector current (I_{C}) divided by the change in emitterbase voltage (V_{BE}) required to produce that change in I_{C}. Then, the extra output resistance provided for Q2 by emitter resistor R2 is g_{m}×R2×r_{O}. Thus the total output resistance for Q2 has now become:
Note: since we aren't dealing with changes due to signals here,
we can calculate g_{m} using the thermal voltage, V_{T},
which amounts to about 26 mV at room temperature. Using this as a
practical value, g_{m} = I_{C}/V_{T}.
The exact value of V_{T} is calculated as:
V_{T}  =  k_{B}T 


q 
Of course, there are a couple of drawbacks to this circuit. The same emitter resistors that increase R_{OUT} also increase the emitter voltages of the two transistors in the current mirror. The base voltages must also increase to maintain the needed V_{BE}. This in turn means that the minimum collector voltage of Q2 is much higher than just V_{BE}, so the available range of output voltages is significantly reduced.
The other drawback is that R1 is now part of the total resistance that determines the reference current, I_{REF}. The circuit will still work, but the choice of resistor values is a bit more complicated since R1 now serves two functions.
If R1 = R2, this circuit still operates as a normal current mirror, with I_{O} very close to I_{REF}. However, if R1 ≠ R2, then the currents through the two transistors will adjust themselves so that the emitter voltages of the two transistors are essentially equal. Thus, I_{E1}R1 = I_{E2}R2. Or, neglecting base currents for a moment, I_{O} = I_{REF}R1/R2. We can still add a third transistor as a base current amplifier if necessary, although the base current of Q1 still flows through R1, but with modern, highgain transistors we can ignore this point.
We can also include multiple mirror transistors as with the basic current mirror, and each transistor can have its own emitter resistor to control its output current in accordance with the above expression. This is one very practical way to provide different but related currents for each of multiple independent circuits.
Something special happens if one resistor is zero, but the other is not. A circuit where R1 = 0 was patented by R. J. Widlar in 1967. We saw that circuit when looking inside the 741 op amp. There, the blue current mirror circuit is a Widlar current mirror. The same circuit is reproduced to the right.
The Widlar current mirror (or current source, as it is sometimes called) is designed to provide very small output currents without requiring very high resistance values or drastically restricting the output voltage range. The mathematical expression for the relationship between I_{REF} and I_{O} includes the fact that Q1's own internal emitterbase characteristics are now significant, and that these characteristics depend partly on temperature:
I_{REF}  = exp  (  I_{O}R_{E}  ) 



I_{O}  V_{T} 
In this expression, V_{T} is the thermal voltage of the transistor, which amounts to about 26 mV at room temperature as we saw above. Unfortunately, the expression cannot be readily solved for I_{O}, since it appears both as itself and as part of the exponential term. It can be solved by trial and error.
The easier way is to determine the desired values of I_{REF} and I_{O}, and then solve for the required value of R_{E}:
R_{E}  =  V_{T}  ln  (  I_{REF}  ) 



I_{O}  I_{O} 
As an example, suppose we want to use I_{REF} = 1 mA, and I_{O} = 50 µA (0.05 mA). Solving the above equation gives us a value of 1557.7808Ω for R_{E}. We can either use a 1.5K resistor and get a slightly higher I_{O}, or a 1.6K resistor and reduce I_{O} slightly. If I_{O} is critical, we can either use one of those fixed values and adjust R to control I_{REF}, or we can use a 1.5K fixed resistor and a 100Ω trimmer potentiometer for R_{E} to precisely set I_{O}. Either way, the low value of I_{O} and the presence of R_{E} will cause R_{OUT} of Q2 to be a high value.


 
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