In an npn power transistor the collector current is 20mA. If 98 percent of
the electrons injected into base region reach collector then the base
current in mA is nearly ?
Here, in this problem is it okay to take \$\beta=I_c/I_b=98/2=49\$ ?
Electronic – Is it okay to take \$\beta=I_c/I_b=98/2\$ in this problem
homeworknpntransistors
Related Solutions
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If you look at the Ebers-Moll model, you can see that each junction is desctibed by a diode and a voltage controlled source. So the collector current will be, as in the first equation, dependent on Vbc (not Vbe) with the Shockley equation, but also on Ib, like in the second equation.
The second factor will be of some meaning only if Vb > Vc, which happens in the saturation region, where the B-C junction is forward biased and current can flow from the base to the collector. In the forward-active (also called linear) region, the B-C junction is reverse biased and the collector current is basically only given by Beta.
Answering questions:
@LvW Just out of curiosity: What if VCE = VBE? C-B pn junction won't be reverse-biased then, so it won't attract electrons in base region. Thus, IC will be zero, and IE will be equal to IB?
But the C-B diode is not forward biased. This is an application where the BJT is used as a diode and no "classical" amplification is possible (transition region between saturation and amplifying region).
IC and IE are controlled and only controlled by VBE; IB is just a side product; Once VCE is greater than VBE, its specific value does not matter, because E-B junction is reverse-biased. Am I right?
It does not matter too much - on the other hand: Look at the Ic=f(VCE) curves. Ic slowly rises with VCE because of the Early-effect.
Given VBE, IE is fixed, and as a result, the sum of IB and IC is fixed. When VCE < VBE, what IB and IC are depend on VCE. The greater VCE is, the greater IC/IB is. However, the value of IC/IB is capped by "beta", which is reached when VCE = VBE. " Is this right?
In this case (VCE < VBE) the C-B diode is open and there is a small current Ic which has a direction opposite to the "normal" Ic direction. Example: For VCE=0 we have a current Ic which is negative (The Ic=f(VCE) curves do NOT cross the origin!).
Best Answer
It sounds like they mean 98% of the emitter current, not the base current. That means you're given \$\alpha = 0.98\$. You can use that to calculate \$\beta\$ and find the base current.