Another BJT Transistor Example-FE/EIT Exam

From the CBT NCEES practice exam...finding Collector Current, DC bias circuit with Capacitors...

[@kbz0n2007]

Your assumption isn't that beta equals 1. You actually assumed that beta is infinite (which is a good approximation in most cases), because you didn't consider the base current when calculating the base voltage divider. A transistor with a beta of 1 would be useless because it can't amplify anything. Also, because of Kirchhoff's law, the condition Ic=Ib=Ie is only true if the three currents are zero.

[@manishbhatt7653]

Ic=Ie? You assume the transistor is in saturation that is why alpha = 1 because beta = inf. If Vc>Vb>Ve isn't true than your assumption will be wrong and you have to switch to another mode. Ic != Ib. if they were equal, Ie = Ic/2. -_-

[@voiceofmayor649]

Thanks and thanks to you. I just did my FE exam and I passed. Though the exam was tough but I passed it. I used the NCEES prep questions, Michael R. Lindeburg, Wasim Asghar and your video. Your video broke down most of the topics. thanks to you once again.

[@aminul248]

I think there is mistake in calculation , you did not pay attention on the 2k and 1 k resistor , because of these two resistor control the base current , as I did to find a base current (30-,07/2k)-.7/1k=IB, when you see the current comes from 30 volts , it is divided in 1 k and in transistor. IB=13.95 mA

[@DK_427]

thank you so much ! I know exactly where this question is coming from and the official answer is the same as yours. I much pretty your way of doing it. thanks !

[@shirhad]

Nicely explained, Thanks

[@kkinki7774]

The video content is so excellent, congratulations

[@RaiyaAcademy]

Thank you so much, appreciate the support.

[@michaelstaublin1817]

Thanks!!

[@seankendrick1231]

I had a similar circuit and to solve for IC, the solution recommended IC = (VCC)/(Rac + Rdc) Rac = resistors under AC condition, Rdc = resistors under DC condition. I calculated this problem on the video and I got 17.6 mA

[@Bxlwto]

i have a question is the voltage around 1K resistor Vb if it is then current flowing is ib and by using Ie=Ib+Ic equation ,vb by divider Vb/Ik=Ib and Vb-.7=Ve,Ve/500=Ie... but the answer we get is 8.6.I dont't think so current around 1k is ib which is why i will follow your method please reply if you can.

[@Barreleye]

Thanks a lot. Even though you solve it in a different way, Ic gave me 0.018Amps so it is far good that approach too.

[@brimmed]

huh I_c = I_e = I_b? how does that work. the equation in the handbook is i_e = i_c + i_b. so if i_c = i_e since we're assuming beta = 1, wouldn't i_b = 0?

[@YahyaAlbdairy]

Ib here =0 because no impedance on Ib branch , and Ic=Ie

[@hoovie3000]

KVL: -30V + Ic*600 + Vce + Ie*500 = 0 where Ie is as explained in the video, Ie = 18.6mA and Vce = Vcc - Ieq(Rc+Re) where Ieq = ( Vb-Vbe ) / ( (Rb/(Beta+1)) + Re ); Beta is infinite so the Rb term goes away. Ieq = Ie = 18.6mA and Vce = 30-18.6mA(600+500) = 9.54V Solve for Ic in the first equation; Ic = 18.6mA

[@magdymousa81]

thank you , very smart solution any easy

[@ThePositiev3x]

If beta is equal to 1 then that means Ic = Ib . How can you say they are equal to Ie? I thought Ie = Ic only if beta is too big that we could neglect the base current Ib. But they are equal...(Ic=Ib)

[@kebo04]

KVL: -Vb + Vbe + Ve = 0 -Vb + 0.7 + Ie(500) = 0 Voltage Divider: Vb = 30*(1000/(1000+2000)) = 10 Plug into KVL and solve for Ie. Ie = 18.6mA

[@shakshaelsafty]

I wonder if we could make an assumption while we could solve it without making any assumptions. By having Ic= Ie - Ib so Ic= (Ve/500) - (Vb/1K)..We could get Vb by doing Voltage Divider..then get Ve from Vbe=0.7=Vb-Ve (and we have Vb)..then we can get Ic...Thanks a lot for your great effort.

[@shakshaelsafty]

By solving it this way, i got Ic= 8.6 mA not 18.6 mA

[@milestopping123]

I think the circuit diagram for voltage divider to determine Vb is not correct. Wikipedia has the one resister in series and 1 resister in parallel with the voltage source. Rp/Rs+Rp. en.wikipedia.org/wiki/Voltage_divider (love the vids!)

[@p2ryantwothousand3]

the assumption is that beta is a very large number making (beta)/(beta +1) roughly equal to 1

[@WisdomVendor1]

the emitter current is NOT = to the base current .

[@RaiyaAcademy]

I made an oversimplification/assumption for the purpose of this particular problem.

[@WisdomVendor1]

Raiya Energy Your simplification assumes the emitter current and collector current are approximately equal. The base current is substantially much smaller.

[@chenxiaomaoify]

Raiya Energy That is totally wrong that you said Ic=Ie=Ib. Ib is much much smaller!

[@chenxiaomaoify]

Raiya Energy And there is a relation between the currents that Ib+Ic=Ie. So with the approximation that Ic=Ie(belta is big enough that alpha is approximately equal to 1), Ib should be approximately to be 0!

[@milestopping123]

lu chen I agree. She should not have written Ie=Ib because it adds confusion. She probably did not catch it because she never used Ib to get the solution. (good catch!)

[@cyrusIIIII]

why did you delete the source? you can only delete the capacitors not the dc source. the source is still creating voltage difference!!!