Calculate the value of IB if IE is 1 milli amps, IC is 0.95 milli amps in a common base transistor configuration a.0.04b.-0.07c.0.07d.0.05e.-0.04
Question
Calculate the value of IB if IE is 1 milli amps, IC is 0.95 milli amps in a common base transistor configuration
- a. 0.04
- b. -0.07
- c. 0.07
- d. 0.05
- e. -0.04
Solution
In a common base transistor configuration, the emitter current (IE) is the sum of the base current (IB) and the collector current (IC). This relationship can be expressed by the equation:
IE = IB + IC
We can rearrange this equation to solve for IB:
IB = IE - IC
Given that IE is 1 milliamp (mA) and IC is 0.95 mA, we can substitute these values into the equation:
IB = 1 mA - 0.95 mA = 0.05 mA
So, the value of IB is 0.05 mA. Therefore, the correct answer is d. 0.05.
Similar Questions
In common emitter BJT (npn) configuration if IB = 1mA , β = 100 . find emitter current
Which of the following is the correct relationship between base and emitter current of a BJT?(1 Point)IE = (β + 1) IBIB = β IEIB = (β + 1) IEIB = I
Calculate IE2 . RB = 2.4 MΩ RE = 510 Ω β1 = 60 β2 = 100 VB1E1=VB2E2=0.8V 17.6 A 15.8 mA 17.6 mA 15.8 A
Input voltage and input current of a common emitter configuration of BJT circuit isa.VBE and IBb. VCE and ICc. VBE and ICd. VCE and IB
In a given transistor amplifier, 𝑅𝐶 = 2.2𝑘Ω 𝑎𝑛𝑑 𝑟𝑒′ = 20Ω, 𝑡ℎ𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑔𝑎𝑖𝑛 𝑖𝑠(a) 2.2 (b) 110 (c) 20 (d) 44
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.