1 Transistor Characteristics and A simple BJT Current Mirror
2 Current-oltage (I-) Characteristics Device Under Test DUT i v T T 1 R X R X T for test Independent variable on horizontal axis Could force current instead; same info. Resistor must include the origin (why?)
3 Current Source Characteristic i v T T 0 i X Slope = 0 I X Constant current; no dependence on voltage across terminals We desire to implement an approximation of an ideal current source
4 Characteristic of Thevenin Eq. R X v X v R X X v X i T Slope still 1/R v T R X v X Can use axis intercepts to find slope Can use calculus to find slope i v T T 1 R X
5 Characteristic of Norton Eqiv. r o I r C0 o I C0 I C0 v X Norton Equivalent Parameters Parameter names used here are from BJT analysis All capital notation for voltages and currents (bias) An ideal current source has r o = (flat line) Can you find I C0 and r o from and data? What if must be > 0?
6 BJT Large-Signal Model Forward-Active Region B I B IB BE E IS BE exp T C0 C0 I B r o IC IB IS T r I I C exp BE CE I I B expx e x o I C I So what is r o? I I I 1 CE CE C C0 C0 ro IC0ro C C0 1 CE r o A I A C0 where I A C r I A o C0 6 th ed. P. 374 Figure th ed. P. 328 Figure 6.19 Text uses I C instead of I C0
7 Transistor as a Current Source I C I C0 Slope = 1/r o v CE r o BE IC0 IB IS exp T I I I CE CE C C01 C0 A A v CE r o I I I The curve is taken at a fixed BE A A CE A C0 C C Classically, BJT curves are at a fixed I B Right side of graph looks like current source
8 Different BE alues See Figure 6.17 page 372 in 6 th ed. Or Figure 6.18 page 327 I C02 I C01 I C Slope = 1/r o2 Slope = 1/r o1 BE BE2 BE BE1 A Different BE implies different curve v CE Both I C0 and r o are different for different BE Identify saturation region and forward active
9 Simple BJT Current Mirror
10 Use for a Current Source CC Desire gain close to 1 v s R S R E Q3 C big For CC circuit above: v out v RE RL g out m RE RL v r R R R 1 ( R r ) g R R R L High I C gives high g m But need small R E for high I C and small R E decreases gain Replace R E with a current source, R E but there is bias current s S E L S m E L
11 BJT Current Source Try I C BE I C0 Slope = 1/r o vce ro exp 1 I I I I C S BE T CE A C S BE T 0 exp v CE Apply BE to get characteristic above Do we know I S? Depends on base-emitter area Do we know T? Depends on temperature We do not know the value of BE to apply! We don t know CE either, but the curve is fairly flat so setting I C0 is really the primary concern.
12 Problem Statement To complete our BJT current source, we need a way to find the BE that when applied to a transistor, will cause that transistor to have a collector current of a desired value. We can exploit the matching property of integrated transistors. Two transistors on the same substrate are at the same temperature and will be fabricated with nearly the same area.
13 The Diode-Connected Transistor I I I I REF C1 B1 C I B I I 1 C REF I REF IC1 I B 1 Q1 BE1 Current-voltage characteristic of this element looks like diode with the I S of the transistor Collector current is nearly equal to I REF Resulting BE1 is the value for that transistor to cause I C1 = I REF Apply BE1 to a matched transistor to cause it to have a collector current nearly equal to I REF
14 Do Not Overlook the Feedback I REF IC1 I B 1 The basecollector connection is important! I C1 I REF Q1 CE1 Q1 BE1 BE1 I I I REF C1 B1 If I REF >>I C1, I B1 will increase If I B1 increases, I C1 increases Then I REF I C1 decreases If I REF >>I C1, and there is no mechanism to increase I C, CE1 will rise until the current source ceases to function or until the transistor melts.
15 Finally A Current Mirror I C1 Q1 I REF I C2 CE2 Q2 Relate I C1 to I REF : I I I I REF C1 B1 B2 C I I B1 1 I I I IREF IC1 1 2 B2 C2 C1 Relate I C2 to I C1 : In general, I I exp 1 C S BE T CE A Matched transistors have the same I S and the same T. And Q1 and Q2 have the same BE (why?) Hence: I 1 C 2 CE2 I 1 A C1 CE1 A I C CE2 A I 1 2 REF 1 CE1 A
16 Current Mirror Summary I REF I C I C1 I C2 CE2 I C0 Slope = 1/r o vce ro Q1 Q2 v CE If we sweep v CE2, we get the same result as above. The output resistance of this current mirror is r o2. The previous page finds the current transfer ratio is equal to I CE 2 C A I REF CE1 A
17 Example +2 I 1.3 1k 1.3mA REF 5k mA I C 2 1.3mA v s v out g m3 1.3mA 25m r m 1.3mA 1.9k I REF 1k I C2 600 ro mA 77k k I C1 Q1 Q2 vout gmre v 1 R r g R s S m E -2