# BJT transistor circuit

1. Sep 2, 2010

### Loke

1. The problem statement, all variables and given/known data

Find IB, IC, IE, VB and VC in Figure 7, where Bdc is 40.

3. The attempt at a solution

VCC-ICRC-VCE=0
IC=BdcIB
VBB-IBRB-VBE=0
IE=IB+IC
VBE=VB
VCE=VC

#### Attached Files:

• ###### BJT transistor circuit.jpg
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Last edited: Sep 2, 2010
2. Sep 2, 2010

### stevenb

Why don't you try to apply your equations? For example, why can't you solve for Ib using your third equation? Let's start there. Make an attempt, and if you can't figure that out, let's find out why.

Once you know Ib you can progress further in the problem.

3. Sep 2, 2010

### Loke

i tried ady...but i still cant find it....3rd equation got 2 unknown Ib and Vbe...i just dont know how to solve it without the type of transistor given...can you help me to figure it out?

4. Sep 2, 2010

### stevenb

Technically you are correct that you don't know Vbe, but there is an approximation you can make in this type of circuit. Assume Vbe is 0.6 V and you will get a decent answer.

The alternative is to use the Schokley diode law, but then you need more information about temperature and transistor characteristics. Given those uncertainties, assuming Vbe=0.6 should be good enough.

5. Sep 2, 2010

### Loke

ermmm....can you solve it with kirchoff's law because i haven study schokley diode law.... i cannot apply it on my work yet.

6. Sep 2, 2010

### stevenb

ermmm ... you missed my main point. Assume Vbe is 0.6V and you can solve.

7. Sep 3, 2010

### Loke

Teacher only taught me silicon transistor,Vbe=0.7 and germanium transistor,Vbe=0.3 ....but i dont know can assume one ^^? ......and in this question both type of transistor is not given that's why i dont know how to proceed..:(

8. Sep 3, 2010

### stevenb

Well, you are making good points that technically the problem does not specify the transistor. But, this approach is not going to let you solve the problam.

Traditionally, if the transistor is not specified, one can assume it is a silicon device, and 0.6 or 0.7 V can be used for Vbe.

Try it. You need to think like an engineer on this one.

9. Sep 3, 2010

### Loke

ohh... thanks...i ask u 1 more question....if let say you dont assume Vbe=0.7....can you actually find an answer? i'm wondering.

10. Sep 4, 2010

### stevenb

Yes, you can find an approximate answer based on assumptions, but it amounts to the same thing. Any reasonable assumptions will put Vbe in the 0.6 V to 0.7 V range and the answer is not very sensitive to changes to Vbe when Vbb is 3 V or greater.

There is going to be sensitivity to changes in temperature and transistor characteristics and part of the challenge to designing transistor circuits is making circuits that are less sensitive. You will be learning about this soon.

11. Sep 5, 2010

### Loke

As long as the question does not provide the type of transistor,then we just assume it to be Vbe=0.7 ...am i right? i meant other than this question also?....

Last edited: Sep 5, 2010
12. Sep 5, 2010

### stevenb

Typically, yes, but there are always exceptions to any rule. The basic idea in most circuit design is to make robust circuits that are not sensitive to temperature and transistor characteristics. The end result of this is that usually changes in Vbe are not important and the above approximation gives reasonable answers.

One example of an exception is the well known current mirror circuit.

13. Sep 5, 2010

### Redbelly98

Staff Emeritus
stevenb gave a good answer. Also, you can think of the base-to-emitter path as similar to a diode. As long as there is something providing voltage between the base and emitter, it will (usually) be about 0.7V.

14. Sep 5, 2010

### Loke

Determine IB, IC, IE, VCE and VCB in Figure 8 for the following values:
RB = 5 k ohm, RE = 500 ohm , VBB = 3.0 V, VCC = 20 V and Bdc = 80

problems:
-If i assume the Vbe in this question also = 0.7...i still cant solve it right away...can u help me figure it out?
-can i assume IC=IE....so IE=BdcIB ...any idea?

attempt at solution:
VBB-IBRB-VBE-IERE=0
VCC-VCE-IERE=0
VBE=0.7
IC=BdcIB

#### Attached Files:

• ###### transistor.jpg
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Views:
56
15. Sep 6, 2010

### stevenb

Remember that Ie=(beta+1)*Ib

16. Sep 6, 2010

### Loke

oh...should use Ie=(beta+1)*Ib instead of Ie=beta*Ib right?... THANKS alot ^^ !