I got the answer! Ic = 12-2-0.2 / 820 = 11.95mA
Use B = 20. Ib = 598μA.

However there is something I dont understand. When we want the minimum Ib, I thought that Ib should be the lowest at 239μA when B=50??

I am also able to calculate Rb2 as 7107Ω. Is this case how are we supposed to know if we want a lower Ib or a greater Ib?

Thanks again

 

Good work hitmen!

Yeah, the wording they have is a little confusing, but they basically want to know the current that guarantees reliable operation for any beta in the range of 20-50. If you use the lower current value, and then beta turns out to be 20, then the transitor won't saturate. However, if you use the higher current value, then the transisor will saturate for any value of beta in the specified range.

Steve

 

Ummm... for the question in part B , which resistors does the current flow through?

 

I'm not sure what you're asking, but the Collector current and Base current are separate, so you have 12ma in the collector (per Steve's calculation) and whatever you calculated for the base current.

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Do you mean 6b?

 

I meant that for part 6c where does the current flow. Through which resistor then I can calculate the appropriate values for the resistor. I dont exactly know how to get the answer

 

Hitmen,

I'm not sure if I'm understanding you correctly, but if you are talking about 6c, the current is the base current Ib which flows through the base resistor Rb2.

Your original work shows you do know how to get the answer. Your original answer was Rb=7.1K which they apparently round down to a standard 5% tolerance value of 6.8K. They round down in order to guarantee that the transistor will saturate under worst case conditions.

As I said before, I feel that the 5% tolerance is becoming outdated, but it's clear that that's how they do it.

Steve

 

No . I dunno how to do question 6c as I am not sure how the current flows when Q2 is attached.

Ok. When Q1 does not saturates, what is the voltage at the node below Rc1 in fig 6(Vce)? When Q1 does saturate, is the voltage there 0.2V? I am confused by what happens to the collector current when the base current saturates.

For Q6c, I am not sure of the specific resistance for Rc1 and Rb1. Can anyone give me a hint for I am not sure of the loading effect.

Thanks again.

 

Hitmen,

One way to approach this problem is to work backward from the input of the second stage. You know you need 5V out of the first stage to drive the second stage. This means that Rc1 will drop 7V. Now the issue is finding the current in Ic1 to determine Rc1. Well, you know you need to be able to drive 598 uA into the second stage, so that is the minimum current that must flow. The assumption here is that Q1 is cutoff and is not affecting the second stage yet.

Rc1=7V/598uA = 11.7K.

Next you round down to 10K. This is because rounding up will leave you short on current.

So basically, when the Q1 is cutoff, Q2 is guaranteed to saturate.

Now, to cutoff Q2, you need to divert the current in Rc1 to ground by saturating Q1. This means that Ic1 will equal 11.8V/10K=1.2 mA. Now divide by the low-end beta of 20 to get Ib1=60uA. So Rb1=4.3V/60uA=72Kohms

Now round down to the standard value of 68K.

I hope this helps.

 

Hi Steveb, I finally got the answers. However, I seem to be taught a different way in my lectures. I was taught that when Q1 is cut off, the collector current flows from Rc1 thru Rb2 into the BJT.

so Rc1 + Rb2 = (12 - 0.75) / 598μA

Since Rb2 = 6.8kΩ, we solve to get Rc1 = 12kΩ, which is equilvalent to your answer as above.

Once again, thanks. My lecture notes did not explain it as thoroughly as your post above.