# Solving Circuits

Discussion in 'Homework Help' started by davidand, Jun 3, 2005.

1. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0
[attachmentid=685][attachmentid=686][attachmentid=687]Can someone point me in the right direction,
I attached a few homework problems I'm looking for some advice maybe someone can pass along or recommend a good book or web site to help me learn more about electronics.

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If you haven't already done so, take a look at the material contained in the tutorials on this website. There is some very good basic information available. Well worth the read.

hgmjr

3. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0

I've been going through the tutorials,
I tried to solve HW 3 and I come up with R=0.564 ohms.
using 11.3/20.
Am I going in the right direction?

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
You are very close. The only thing you overlooked is that the current is 0.020 Amps rather than 20 amps. If you make that adjustment you will be right on the money.

hgmjr

5. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0
Thanks
I'm tryng HW#2 now
so far I have V
B = 7.9V

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I believe you need to take another look at your work in calculating Vb.

HW#2(hint)
The three diodes play a key role in the setting the value of Vb.

One important parameter appears to be absent from the definition for HW#2 problem. I am surprised that the transistor's beta is not part of the set of givens for the problem.

HW#1(observation)
Is it possible that the circuit node where the 430 ohm emitter resistor and the 5V zener diode's anode are connected is also tied to a source of negative voltage that is not shown?

hgmjr

7. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0
Your correct for HW #1 I went online with instructor and he gave me the corrected
HW#1.

For HW#2 I did subtract 0.7 for each diode. I'm researching the chapter on diodes now.

Thanks for you input.

8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Hi davidand,

I think that there may be some information missing on the figure for HW#2. Is it possible that the circuit node associated with the cathode of the bottom diode and the bottom leg of the 15 ohm emitter resistor is attached to a negative power source?

I made the assumption that the node was connected to ground based on the figure. That may be why I did not get the same answer that you did.

Once this question is cleared up I think you are on the right track.

I am guessing that the node in question is tied to -10V. Am I right?

You're doing pretty good so far.

hgmjr

9. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0
Yes it is connected to -10V

10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Now that I have the missing piece of information from HW#2, I can see where you got your answer.

Suggest you take a second look at the sign of your result. Keep in mind that the diodes are referenced to the -10V.

hgmjr

11. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0

thanks hgmij
The book I'm using id a reprint for Bura of Navy,
I'm taking a distance learning course and this is a reference book, it's pretty basic
I'm trying to find a book on solving circuits.
I'm still trying.
again thanks

12. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0
I will get that info to you soon, I left book at work.
I'll post info soon.

13. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0
For HW#2
I find
V
B = -7.9

V
C = 0

V
E =0

I use the following

V
C = 10V - (IR)

It does'nt look right can someone check this and explain it to me?

14. ### hgmjr Moderator

Jan 28, 2005
9,030
214

You need to take a second look at your Vc and Ve calculations.

HINT: The voltage at the emitter will be more negative than the voltage at the base. Keep in mind that you are dealing with an NPN silicon transistor.

Once you have Ve then you will have most of what you need to calculate Vc. Since you have no value for the beta of the transistor we will be forced to assume a value of say 100. That is unless you know the actual value posed by the problem. Or we can assume that Ie is equal to Ic and proceed on that basis.

Hang in there.
hgmjr

15. ### davidand Thread Starter Active Member

Jun 2, 2005
43
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NAVPERS 10087-C
Basic Electronics

I bought it at Barnes +Noble

I'm in Army All the manuals for electronics courses are made advailable for me
but no hard copy I must download them and print it if I want hard copy.
Cheaper to buy the book for \$14.95

16. ### davidand Thread Starter Active Member

Jun 2, 2005
43
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I worked HW#2 some more

V
B = -7.9V

V
E = -8.6V

V
C = 24.4V

17. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Your value for Ve is now correct.

Your value for Vc is off. I think I can understand where you made the error.

HINT: The value of the emitter current is computed by the voltage "drop" across the emitter resistor.

I think what you may have done is used the value of Ve directly in your calculation of the current flowing in the emitter resistor.

HINT: In the transistor stage of HW#2, any computed voltage that is more positive than the positive power supply or more negative than your negative supply should be view with suspicion.

Still you are making progress.

hgmjr

18. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0
I calculate it to be 9.96V now.

19. ### hgmjr Moderator

Jan 28, 2005
9,030
214
That is still not quite it.

What value do you get for the emitter current?

hgmjr

20. ### davidand Thread Starter Active Member

Jun 2, 2005
43
0
-8.6V/15 Ω =0.57mA

10V-(0.57)(60 Ω )=9.49V