# Transistor Problem

Discussion in 'General Electronics Chat' started by james88, Aug 12, 2013.

1. ### james88 Thread Starter New Member

Aug 7, 2013
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0
Hi All,

First off, I've tried to search for an answer to this, but I've not had any luck.. If this is by any chance one of those questions that keeps being asked then I do apologise.

My issue is that I am trying to make logic gates using transistors (BC547's I think) and the voltage from my base connection is flowing straight through and out of the emitter ....

Can anyone see where I may be going wrong?

Thanks
James

2. ### ScottWang Moderator

Aug 23, 2012
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777
What kinds of the logic gates that you are trying to make?
Do you have any circuit, if you have one then post it?

3. ### james88 Thread Starter New Member

Aug 7, 2013
6
0
My circuit is like this:

The problem is that the voltage is able to flow directly from the base of the second transistor to the emitter.

4. ### ScottWang Moderator

Aug 23, 2012
4,926
777
The circuit as I posted, you have to raising the value of Re to 330Ω, that value is for the TTL, you also need to adding Rb1,Rb2 to protecting the Base of bjt.

The Vce1 is around 0.2V, and The vce1 is around 0.9V.

You can testing and adjust the value of resistors by yourself.

• ###### TwoInputANDGate_james88.gif
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Last edited: Aug 12, 2013
5. ### james88 Thread Starter New Member

Aug 7, 2013
6
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Thank you for your help but I still don't understand sorry.

James

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I would like to help but I don't understand your question/problem.

7. ### ScottWang Moderator

Aug 23, 2012
4,926
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1. What's the purpose of this AND gate?
2. What's the value of Vbat? (5V,6V,?)
3. Does the 100Ω is the Load?
4. If you want to using the bjt to be a switch, then the Vce of bjt should be in a saturation status, the vce will be around 0.2V and C of Bjt had load in it, Vbe around 0.7V, if C of bjt there is no load and connecting to Vcc then Vce will be around Vbe+0.2V=0.9V.

If you didn't have any experience of bjt in logic circuit, you better adjust the Rb1, Rb2 from 1K~10K and measuring the Vce1, Vce2 that I added, it will help you to know more about the relation with Ib(b current) and Vce.

Last edited: Aug 12, 2013
8. ### Ron H AAC Fanatic!

Apr 14, 2005
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A A two-transistor AND gate is generally a bad idea, as you have discovered. Even if you add base resistors, the resulting logic levels are dependent on transistor beta. You will get much better results by making a NAND gate, and then, if you need AND, add an inverter. See attachment.

If you need to drive a load, post as many details about your application as you can.

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9. ### BobTPH Active Member

Jun 5, 2013
806
121
The two transistors making up the NAND gate should be connected in parallel, not series. I.e. both collectors should connect to the pullup and both emitters to ground.

Bob

10. ### Ron H AAC Fanatic!

Apr 14, 2005
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That will make a NOR gate. Either input high will make the output low.

11. ### ScottWang Moderator

Aug 23, 2012
4,926
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OR gate : Only one input is Hi then the output will be Hi.
NOR gate : OR gate + Inverter

AND gate : Only one input is low then the output will be low.
NAND gate : AND gate + Inverter.

12. ### LDC3 Active Member

Apr 27, 2013
920
160
AND gate: so it will be high when both are low?
I think you mean if, not only.

13. ### ScottWang Moderator

Aug 23, 2012
4,926
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Thanks for your mentioned.
I used the wrong word to described, I should say anyone of inputs.

OR gate : Anyone of inputs is Hi then the output will be Hi.
NOR gate : OR gate + Inverter

AND gate : Anyone of inputs is low then the output will be low.
NAND gate : AND gate + Inverter.

* the temperature are too high and drive me going to crazy, the room now is 95℉(35℃).

Last edited: Aug 13, 2013
14. ### LDC3 Active Member

Apr 27, 2013
920
160
Anyone refers to a person; any would refer to an object.

15. ### ScottWang Moderator

Aug 23, 2012
4,926
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Oh, thanks.
OR gate : Any one of inputs is Hi then the output will be Hi.
NOR gate : OR gate + Inverter

AND gate : Any one of inputs is low then the output will be low.
NAND gate : AND gate + Inverter.

16. ### LDC3 Active Member

Apr 27, 2013
920
160
That would get your ideas expressed. Better would be

OR Gate: If any input is HIGH, then the output is HIGH.

XOR Gate: If both inputs are the same, output is LOW.

17. ### BobTPH Active Member

Jun 5, 2013
806
121
You are right, I had realized my mistake right after posting and thought I had deleted the post, but I guess not.

Bob

18. ### ScottWang Moderator

Aug 23, 2012
4,926
777
When you using "then the output is HIGH", the "is" seems like a static statement, but I used will be, it looks like a active changing status, maybe using the following sentence is more close to the meaning that I want.
- If any input is HIGH, then the output will go HIGH.

19. ### james88 Thread Starter New Member

Aug 7, 2013
6
0
That was just what I was after, thanks Ron, great help!!

Thanks Everyone

20. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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Why are you using discrete components? Why don't you just buy a quad AND gate?