Current Mirrors. Basic BJT Current Mirror. Current mirrors are basic building blocks of analog design. Figure shows the basic NPN current mirror.

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Albert Thornton
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1 Current Mirrors Basic BJT Current Mirror Current mirrors are basic building blocks of analog design. Figure shows the basic NPN current mirror. For its analysis, we assume identical transistors and neglect the Early Effect, i.e. we assume V A. This makes the saturation current IS and current gain β independent of the collector base voltage V CE. The input current to the mirror is labeled I REF. This current might come from a resistor connected to the positive rail or a current source realized with a transistor or another current mirror. The emitters of the two transistors are shown connected to ground. These can be connected to a dc voltage, e.g. the negative supply rail. The simplest way to solve for the output current is to sum the currents at the node where I REF enters the mirror. Because the two transistors have their base-emitter junctions in parallel, it follows that both must have the same currents.

2 Thus, we can write the equation Solving for I O, we will get Note that this equation predicts that I O < I REF unless β. Because the Early Effect has been neglected in solving for I O, the output resistance is infinite. If we include the Early effect and assume that it has negligible effect in the solution for I O, the output resistance is given by For a more accurate analysis, we can include the Early Effect in calculating the output current. Let us consider the circuit as shown in figure. If the transistors have the same parameters, we can write

3 Summing currents at the node where I REF enters the circuit yields Thus I C1 is given by It follows that I O is given by Note that this equation predicts that I O can be larger that I REF. The output resistance is given above. Note that the effect of a finite β is to reduce I O but the effect of the Early Effect is to increase it. Because of the Early Effect, the output current can be greater than the input current. One way of obtaining a better match between the input and output current is to use series emitter resistors on the transistors. If the current in one transistor increases, it causes the voltage across its emitter resistor to increase, which causes a decrease in its base-emitter voltage.

4 This causes the current to decrease, thus causing the two transistors to have more equal currents. A typical value for the emitter resistors is 100 Ω. With these resistors, R te2 is no longer zero so that the output resistance is increased. It is given by r out = r ic2 which can be much greater than r 02. BJT Mirror with Base Current Compensation The figure shows the basic current mirror with a third transistor added. The collector of Q 3 must be connected to a positive reference voltage, e.g. the positive supply rail, which biases it in the active 2 mode. If we neglect the Early effect and assume all transistors are identical, we can write Solution for I O yields

5 For a non-infinite Early voltage and V CB1 = V BE3 << VA, it can be shown that the output current is given by where BJT Wilson Mirror A Wilson current mirror is shown in figure. We neglect the Early Effect in the analysis and assume the transistors to have identical parameters. The emitter current in Q 3 is I O /α. This current is the input to a basic current mirror consisting of Q 1 and Q 2. This current is mirrored into the collector of Q 1 by dividing by (1 + 2/β). At the node where I REF enters the mirror, we can write

6 Solution for I O yields The advantage of the Wilson mirror over the current mirrors examined above is that it has a much higher output resistance. This is caused by two positive feedback effects. To see how this occurs, suppose a test current source is connected between the mirror output and ground. If the source delivers current to the output node, the voltage increases. This causes a current to flow through r 03, causing the emitter voltage of Q 3 and the base voltage of Q 1 to increase. The increase in voltage at the emitter of Q 3 causes its collector voltage to increase because Q 3 is a commonbase stage for an emitter input. Because Q 1 is a common-emitter stage for a base input, the increase in voltage at its base causes the collector voltage of Q1 and the base voltage of Q3 to decrease. Because Q3 is a common-emitter stage for a base input, the decrease in voltage at its base causes its collector voltage to increase. Thus there are two positive feedback effects which cause the collector voltage of Q3 to increase to a larger value. Because rout is the ratio of the collector voltage of Q 3 to the current in the test source, it follows that the output resistance is increased.

7 BJT Low-Level Mirror The circuit shown in figure is a low-level current mirror. It can be used when it is desired to have a much lower output current than input current. For the analysis, we neglect the Early Effect, assume identical transistors, and assume that β. We can write By taking ratios, we obtain This equation cannot be solved for I O. If I REF and I O are specified, it can be solved for R E to obtain

8 FET Current-Mirror Common-Source Amplifier The figure shows a common-source amplifier. The active device is M 1. Its load consists of a current mirror active load consisting of M 2 and M 3. The current source I REF sets the drain current in M 3 which is mirrored into the drain of M 2. Because the source-to-drain voltage of M 2 is larger than that of M 3, the Early Effect causes the dc drain current in M 2 to be slightly larger than I REF. The voltage V G is the dc component of the gate input to M 1. It is a dc bias voltage which sets the dc drain current in M 1. This current must be equal to the drain current in M 2 in order for the dc output voltage to be stable. In applications, V G would usually be set by feedback. It will be assumed that r 01 = in calculating i o(sc) but not in calculating r out.

9 Note that the body effect must be accounted for in M 1. Assume that I D1 = I D2 = I D3 = I REF. (a) Use the simple T model to calculate the shortcircuit output current. Assume that r 01 =. where (b) Calculate the output resistance. Assume that r 01 <. where (c) Calculate the output voltage and the voltage gain

10 Common-Gate Amplifier The figure shows a common-gate amplifier. The active device is M 1. Its load consists of a current mirror active load consisting of M 2 and M 3. The current source I REF sets the drain current in M 3. This current is mirrored into the drain of M 2. As with the common-source amplifier, the Early Effect makes the drain current in M 2 slightly larger than that in M 3. The dc voltage V G is a dc bias voltage which sets the drain current in M 1. This must be equal to the drain current in M 2 in order for the dc output voltage to be stable. In applications, V G would usually be set by feedback. It will be assumed that r 01 = in calculating i o(sc) but not in calculating r out. Note that the body effect must be accounted for in M 1. Assume that I D1 = I D2 = I D3 = I REF.

11 (a) Use the simple T model for M 1 to calculate shortcircuit output current. Assume that r 01 =. (b) Calculate the output resistance. Assume that r 01 <. (c) Calculate the output voltage and the voltage gain. (d) Use the simple T model for M 1 to calculate the input resistance. Assume that r 01 =.\

12 Common-Drain Amplifier The figure shows a common-drain amplifier. The active device is M 1. Its load consists of a current mirror active load consisting of M 2 and M 3. The current source I REF sets the drain current in M 3 which is mirrored into the drain of M 2. As with the common-source amplifier, the Early Effect makes the drain current in M 2 slightly larger than that in M 3. The dc voltage V G is a bias voltage which sets the dc output voltage. Note that the body effect must be accounted for in M1. Assume that I D1 = I D2 = I D3 = I REF. (a) Use the pi model for M 1 to calculate i o(sc). Note that the body effect is not present because v bs1 = 0 when v o = 0. where

13 (b) Use the simple T model with body effect to calculate the output resistance. (c) Calculate the output voltage and the voltage gain.

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