Equations and constants
Beta = I_C/I_B ( current gain )
Gamma = V_C/V_B ( voltage gain )
Use electron band theory to distinguish between insulators and conductors and to explain what happens when the temperature rises in a semiconductor and what is the function of impurities in semiconductors
Insulators have a filled valance band and a band gap typically 5-10eV to the conduction band. Conductors have a partially filled high energy band. Semiconductor's resistivity decreases with temperature increase because of the smaller energy gap to jump from valence band to conduction band. The function of impurities in semiconductors is to encourage electron transfer/current. That's called doping.
Explain what is meant by the following: pn junction, forward bias, reversed bias, diode, rectifier, half wave rectifier, full wave rectifier, transistor.
a pn junction is a diode . One part is negative holes and the other part is positive holes. In forward bias, a pn junction allows conventional current to flow, + to -. In reverse bias, no current is allowed to flow because of a separation of charge due to the terminals of the supply. A rectifier is a device (diode) that changes ac to dc. A half wave rectifier makes the signal not really DC but unidirectional. A full wave rectifier uses two or more diodes to make the output in one direction. A transistor is a device that is composed of a crystal of one type doped semiconductor sandwiched between two crystals of the opposite type.
Describe the structure and use of a pnp and npn transistor.
A pnp transistor has a negative base in the middle sandwiched between a collector and emitter that are positive. A npn transistor has a positive base in the middle sandwiched between a collector and emitter that are negative. They are used to amplify signals.
A silicon diode whose current-voltage characteristics are given in the figure is connected in series with a battery and a 660ohm resistor. What battery voltage is needed to produce a 12*10^(-3)A current?
Use Ohm's law: V = IR, 12*10^(-3)*660 => 7.92V. According to the figure, about 0.7V is in the forward bias, so 7.92V + 0.70V => ~8.62V.
Suppose that the current gain of the transistor in the given figure is B = I_C /I_B = 80. If R_C = 3.3*10^3ohms, calculate the output voltage for a time-varying input current of 2.0*10^(-6)A.
Voltage is directly proportional to current according to Ohm's law, V = IR. With the input current as 2.0*10^(-6)A, the output current is then 80x that, 0.00016A. Now, using V O = I_C R_C , the output voltage is then 0.528V
If the current gain of the transistor amplifier in the figure is B = I_C /I_B = 100, what value must R_C have if a 1.0*10^(-6) base current is used to produce an output voltage of 0.40V?
The input current is a hundredth of the base current. V O = I_C R_C , V O /I_C = R_C . V O /(I_B *100) = R_C => 4000ohms.
A transistor whose current gain B = I_C /I_B = 90, is connected in the figure with R_B as 2.2*10^3ohms and R_C as 6.2*10^3ohms. Calculate a) the voltage gain and b) the power amplification
a) The voltage gain is the ratio V_C /V_B which can be expressed as I_C /I_B * R_C /R_B => 253.64V_B) the power amplification is P C /P B => V_C /V_B * I_C /I_B => 228327.273.
An amplifier has a voltage gain of 70 and a 14*10^3 load (output resistance). What is the peak output current through the load resistor if the input voltage is an ac signal with a peak of 0.080V?
a) The voltage gain is the ratio V_C /V_B . We are looking for I_C , the output current. With an input voltage of 0.080V, the output voltage is 5.60V. Given the output resistance, we can use Ohm's law to compute the peak output current. I_C = V_C /R_C . 0.0004A.