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How is VBC calculated in transistor?
one part of that stack is the >base-emitter junction (Vbe). the other part of that stack >is the base-collector junction (Vbc). Add them together and >you have the total stack voltage (Vce), or Vce=Vbe+Vbc. >
What are the influencing factors for BJT’s performance?
Mainly three parameters are used to define BJT transistor performance. Current Amplification factor, Base Transport Factor, Emitter Injection Efficiency parameters shows the performance of NPN transistor and PNP transistor.
Why is the equation VBC = VBE-VCE not the equation?
Rearranging Vce = Vbe + Vbc gives Vbc = Vce – Vbe not Vbc = Vbe – Vce The equation in question is a reference to setting up a four-resistor biasing network for a BJT common emitter circuit. See: http://www.csus.edu/indiv/o/oldenburgj/EEE102/Chapter5/BJTs.ppt slides 20-22, where the equation is derived from a mesh analysis.
When is an approximation of a BJT valid?
The approximation is valid when V b e is such that the exponential term dominates, which is the case even for even a minuscule value of V b e, i.e 75mV. A BJT is obviously more complicated than your equation (s) provide. But those equations are often good enough when just considering the forward active region.
How to calculate the current of a BJT?
Take a few measurements of your BJT, where you set V C E = V B E (without actually shorting the collector to the base!) Then just take out some graph paper and plot the logarithm of the collector current at these several measurement points, against the voltage V B E = V C E. Draw a line with a ruler back to the y-axis.
Which is the correct equation for rearranging VCE and VBE?
Good answer, but misses the subtlety picked up by Andrew. Rearranging Vce = Vbe + Vbc gives Vbc = Vce – Vbe not Vbc = Vbe – Vce The equation in question is a reference to setting up a four-resistor biasing network for a BJT common emitter circuit.