This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 1) Connect the circuit in figure 1a: Pick some value of Va and then change it so that Va-Vb changes by 1V. Readings: Va Vb Va-Vb ∆Vb rbe rbe=nVt/Ib 8.044V 0.6745V 7.369V---------------- 3.528k 7.040V 0.6713V 6.369V .0032V 3.2 kohms 4.082k rbe=∆Vb/1 micro amps=.0032/1 micro amps=3.2 kilo ohms Ib=(Va-Vb)/1Mohm=6.369/1Mohm=6.369 micro amps rbe=nVt/Ib=26mV/7.369 micro amps=3.528 kilo ohms rbe=nVt/Ib=26mV/6.369 micro amps=4.082 kilo ohms How much does Vb change? It changes .0032V. How does your measured value compare with the conventional formula? 3.2k compared to 4.082k. How does this compare to the diode resistance studied in experiment 3? Rd=26mV/Id which is typically only a few ohms~10. For what theoretical value of Beta would 1 W be dissipated in the transistor? 2) Connect the circuit in figure 1b: Pick some value of Va and then change it so that Va-Vb changes by 1V. Readings: Va Vb Va-Vb ∆Vb rin Rin 8.015V 1.880V 6.134V---------------- 234 k 6.780V 1.645V 5.135V 0.235V 235 k 230 k...
View Full Document
- Fall '10
- Input impedance, Electrical parameters, Output impedance, kilo ohms, kilo ohms Ib=