imirrors - Current Mirrors Basic Current Mirror Current...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Current Mirrors Basic Current Mirror Current mirrors are basic building blocks of analog design. Fig. 1(a) shows the basic npn current mirror. For its analysis, we assume identical transistors and neglect the Early e f ect, i.e. we assume V A →∞ . This makes the saturation current I S and current gain β independent of the collector- base voltage V CB . 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. Figure 1: (a) Basic mirror. (b) Mirror with base current compensation. 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. Thus, we can write the equation I REF = I O + 2 I O β Solution for I O yields I O = I REF 1+2 / β Because the Early e f ect has been neglected in solving for I O , the output resistance is in nite. If we include the Early e f ect and assume that it has negligible e f ect in the solution for I O , the output resistance is given by r out = r 02 = V CB 2 + V A I O For a more accurate analysis, we can include the Early e f ect in calculating the output current. If the transistors have the same parameters, we can write I C 1 = I S 0 exp V BE V T I B 1 = I C 1 β 0 I O = I S 0 1+ V CB 2 V A exp V BE V T I B 2 = I O β 0 (1 + V CB 2 /V A ) By taking the ratio of I O to I C 1 , we obtain I O = 1+ V CB 2 V A I C 1 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Summing currents at the node where I REF enters the circuit yields I REF = I C 1 + I C 1 β 0 + I O β 0 (1 + V CB 2 /V A ) = I C 1 + 2 I C 1 β 0 Thus I C 1 is given by I C 1 = I REF 1+2 / β 0 It follows that I O is given by I O = 1+ V CB 2 V A I C 1 = I REF (1 + V CB 2 /V A ) 1+2 / β 0 The output resistance is given above. Note that the e f ect of a nite β is to reduce I O but the e f ect of the Early e f ect is to increase it. Because of the Early e f ect, the output current is commonly greater than the input current. One way of obtaining a better match between the input and output currents is to use series emitter
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

imirrors - Current Mirrors Basic Current Mirror Current...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online