The emitter is not at -5V since there is an active element (constant current source) between the -5V and the emitter.
<br>Since I = 0.35 mA (at the ...
Solution Summary
This solution calculate the DC voltages at the base and emitter terminals.
... may assume in all cases that the transistor is biased ... Procedure Determine Quiescence or Q point voltages and currents ... and biasing set by the voltage divider at ...
... decreasing the ability of the transistor to amplify AC ... The solution discusses diodes, transistors (including BJT) and ... resistor, a 0.7V fixed voltage source and ...
... below the level of output voltages without affecting ... components (LM317 regulator, any transistors), transformer and ... normal operation and transistor 〖TR〗_1 ...
... 0.65V for the PNP transistor to conduct ... zero input and between input voltages of approximately before either of the transistors start conduction. ...
... The response includes a short analysis of DC current gain impact on voltage output in a simple bipolar junction transistor (BJT) configuration. ...
... i_C=10+50=60mA Transistor forward current gain h_fe. ...Voltage across R_2 In order to estimate the voltage across R_2 we need to sum the voltages at point X ...
... provides a very steady voltage gain even if the transistor's beta gain varies or if the temperature of the transistors rise that cause transistor currents to ...
... inside the diode, and applying sufficient voltage (0.7V ... The transistors are commonly used as amplifiers and ... In a bipolar junction transistor, its 3 electrodes ...
... forward current gain h_i_e = LED current = 10 mA LED voltage drop at 10 mA = V Base/emitter voltage = V Collector emitter voltage when transistor is on = 1 V. ...
... In this case, the NPN transistor acts as a switch that acts like a short when a high enough positive voltage is applied at its base, providing a high current ...
