This is a demonstration of an PNP transistor. The emitter is at +2V, and the base and collector voltages can be controlled using the sliders at right. Move the mouse over the transistor to see labels for the three terminals. Compare it to the NPN example.
The emitter-base junction acts like a diode. Unlike an NPN transistor, current flows out of the base, not into it. Little current flows out of the base unless it is below about 1.4V (0.6V below the emitter). Assuming the collector is at a lower voltage than the base, the emitter-collector current is 100 times the base current. So, this transistor has a beta (current gain) of 100. Moving the collector voltage higher or lower won't have any effect as long as it's lower than the base voltage. This is forward active mode.
A transistor is often considered to be in saturation mode when the collector is higher than the base. But it still acts like forward active mode unless the voltage difference, Vcb, is on the order of a diode drop (.6 V). If the base is at 1.3V and the collector is raised to about 1.86V or higher, the base current will go up and the collector current will go down, so it will no longer be 100 times the base current. This is saturation, where the transistor acts like a low-resistance switch, with a small voltage drop from the emitter to the collector. -- Credits: Mr. Paul Falstad.
|Navdeep Singh Channa said: (May 27, 2013)|
|How did you calculate the Current value at base and collector ?
I've worked with NPN Transistors, but never with PNP. The query I have is that, I gave 3V on base through a 100E Resistor, and 12V on Emitter through a 100 E resistor.
How is the Voltage across resistor (at base) 81 mV. And hence 3.08V on base?
I mean other values can be derived from that only.
|Raj Kumar said: (Nov 11, 2015)|
|Flow of current of PNP transistor.|
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