What is Current mirrors

Chapter 4.14 Current mirrors

Lessons In Electric Circuits Volume III – Semiconductors Book
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Lessons In Electric Circuits Volume III – Semiconductors Book

  • 266CHAPTER 4. BIPOLAR JUNCTION TRANSISTORSrEE = KT/IEmwhere:K=1.38×10−23 watt-sec/oC, Boltzman’s constantT= temperature in Kelvins ∼=300.IE = emitter currentm = varies from 1 to 2 for SiliconRE ∼= 0.026V/IE = 26mV/IEThus, for the common-emitter circuit Rin isRin =βrEEAs an example the input resistance of a, β = 100, CE configuration biased at 1 mA is:rEE = 26mV/1mA = 26ΩRin =βrEE = 100(26) = 2600ΩMoreover, a more accurate Rin for the common-collector should have included rEERin =β(RE + rEE)This equation (above) is also applicable to a common-emitter configuration with an emitterresistor.Input impedance for the common-base configuration is Rin = rEE.The high input impedance of the common-collector configuration matches high impedancesources. A crystal or ceramic microphone is one such high impedance source. The common-base arrangement is sometimes used in RF (radio frequency) circuits to match a low impedancesource, for example, a 50 Ω coaxial cable feed. For moderate impedance sources, the common-emitter is a good match. An example is a dynamic microphone.The output impedances of the three basic configurations are listed in Figure 276,4.112. Themoderate output impedance of the common-emitter configuration helps make it a popularchoice for general use. The Low output impedance of the common-collector is put to gooduse in impedance matching, for example, tranformerless matching to a 4 Ohm speaker. Theredo not appear to be any simple formulas for the output impedances. However, R. Victor Jonesdevelops expressions for output resistance. 289,[3]• REVIEW:• See Figure 276,4.112.4.14Current mirrorsAn often-used circuit applying the bipolar junction transistor is the so-called current mirror,which serves as a simple current regulator, supplying nearly constant current to a load over awide range of load resistances.We know that in a transistor operating in its active mode, collector current is equal to basecurrent multiplied by the ratio β. We also know that the ratio between collector current andemitter current is called α. Because collector current is equal to base current multiplied by β,and emitter current is the sum of the base and collector currents, α should be mathematicallyderivable from β. If you do the algebra, you’ll find that α = β/(β+1) for any transistor.We’ve seen already how maintaining a constant base current through an active transistorresults in the regulation of collector current, according to the β ratio. Well, the α ratio works