Question

Ammeter lout 2N5192G Q1 32K2 SRscale = 20 VDC Rload Point A 01 47k 5.1 volts Fig. 1 Procedure 1. Calculate and record the values for load, road,"out, and out-op-camp in Table 1. typical would be in the range of 50 to 100, depending on the pass transistor used. Rioad Rscale Vload lout Iload out-op-amp 100k 2 | 47k 2 100k 100k 2 1002 47k 2 1002 100k 2. Note that the op-amp is acting as a series-parallel feedback amplifier where the load voltage is equal to Vener (1+ Rscaie/47k). Changes to the load resistor should cause no change in the load voltage. Also, note that love is equal to the transistor's B times the output current of the op-amp. The transistor is, therefore, being used as a current booster, and because it is located inside of the feedback loop, it should not affect the load voltage.

Answer 1

Here Veload (load voltage) = Vzener (1+ Rscale) 47k/ - Voi (1+ Rscale) Here Voi=5.1 V A 5.10 Therefore Veoad = 5.1 (1+ Rscale) 47k and Iload & Vload Rload Rload looksu Recale 47ka Vload 10.2V Iload 0.102 mA looka looka 15.95 V 0.15 95 mA 10or 47 kr 10.2V 0.102 A 0.15 95 A 100r Iwokr 15.95 V Tout = ß (I out op-amp) Transistor a, 2N5192G Silicon NPN Power Transistor

Jout + - Vad Y I load • Rload 3 Lov 5.17 25.1V 4476

The purpose of this circuit is to control the voltage across the load Rload, which requires more current than can be provided by the output of the op-amp alone. An NPN power transistor is used to provide the required current gain. The op-amp is used in non-inverting configuration to provide very high input resistance for the control voltage von Jout = IE - Ic Jaut opamp = IB where IB is base current of 81 It is emitter current of St. It is collector current of a - Ic = ß Jout opamp = Jout Ic = Is exp UBE - I out VT Head Now I = 5.1 mA 47 Since no current enters op-amp I = 0.1085 mA => [Tout = It I load)

Pload 100kr Rscale 47 kr I doad 0. 102mA Iout 0.2105 mA 0.260 mA looke 100 kr 0.1595mA loor 4 7 kr 0.102A 0.1595A 102, 1085 mA 159.6085mA 1oor looka