PNP junction transistor
The PNP junction transistor pictorial presentation is:
VEE forward-biases emitter-base pn junction, electrons in base flow to emitter. Emitter base junction property is low impedance and low voltage drop but collector base junction is reverse-biased with high impedance. Current in the emitter circuit is emitter current while current in the collector circuit is collector current. Collector current comprised of emitter current and collector base current, ICBO. The emitter current has constant of proportionality α or αIE.
Both currents αIE and ICBO . This is written as IC = αIE + ICBO . Apply KCL to this transistor yields
IE = IB + IC
From IE = IB + IC , IB = IE - IC. But IC = αIE + ICBO rearrange α we get
The ratio α/(1-α) represents β.
Emitter Base Junction
Apply KVL to emitter base loop the emitter current is IE = (VEE - VEB)/ Re . Simplify circuit that voltage collector dependence is negligible and dependence of emitter current on collector base voltage is postponed then emitter-base circuit is exactly same as the analysis of diode circuits.
Voltage threshold or VEBQ is about 0.7 for silicon transistor. When VEE = 4 volts and Re = 1 kΩ. Note that VBE varies between 0.7 and 0.9 volts depends on base current.
DC with oscillator
Insert oscillator vi in series with dc source VEE. The ac signal is vi = Vim cosωt . The presentation picture is
Collector Base Junction
Collector Base junction is bring into consideration after plotting the common-base output characteristics. Such a figure consider below:
For values VCB less than 0.5 volts the curve plotted above is a family of straight line equation given by
IC = αIE + ICBO
Convert this equation into its equivalent yields:
Active elements like transistor has emitter current transmitted to collector current thus αIE is a dependant source. To avoid the nonlinear region to the left of iC axis it is necessary that VCB less than 0.5 volts at all times.
Large signal model
In the circuit above α = 1, ICBO = 0, VEE = 2 volts, Re = 1 kΩ, VCC = 50 volts, RC = 20 kΩ and a 1 volt peak sinusoidal source is connected in series with VEE.
For emitter-base circuit, the junction is forward biased as long as Vim < 1.3 volts. The transistor amplifies input ac voltage and resulting voltage gain is AO = Vcbm/Vim = 20/1 = 20.
A change in emitter current in pnp transistor produces a change of approximately the same amount of collector current iC and smaller change in base current a factor of 1 - α. To achieve current amplification the change is initiated in the base current not in the emitter current. It assumes that β = α/(1-α) does not vary with base current. Neglecting ICBO, we have iC = βiB . In a typical range of 1 to 100 mA, hfe is approximately equal to β, independent of any changes in collector current.
Current amplification circuit
Current amplification circuit is made so that base-emitter junction is forward-biased and collector-base junction reverse-biased. The picture presentation is
The derivation involved in this circuit is made based on Kirchoff's Voltage Law