# AND-gate diagram

1. Jun 10, 2010

### TsAmE

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

Draw a circuit diagram (using only transistors and resistors) showing how a 2-input AND-gate can be realised.

Component values are not required.

2. Relevant equations

None.

3. The attempt at a solution

(Refer to attachment for correct answer).

I am not too sure how the current would flow (from the beginning to the end of the circuit). At what point does the current originate (i.e. start)?

Why also is the output taken at the collector?

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2. Jun 10, 2010

### Staff: Mentor

Write a truth table for the circuit operation, including the state of the intermediate outputs (at the collectors of the first two input transistors).

Do you understand how the common-emitter (CE) transistor stage works? What does the transistor do when its base is pulled high? How about when its base is pulled low?

3. Jun 10, 2010

### TsAmE

Sorry but I dont understand. How can its base be pulled high or low?

4. Jun 10, 2010

### Staff: Mentor

http://en.wikipedia.org/wiki/Resistor–transistor_logic

The input bases are "pulled" high or low by the input signals. When the NPN transistor's base is driven high through a resistor, that causes the transistor to pull down (pull the voltage down) on its collector. And when you drive the base low, that turns off the transistor, and the pullup resistor on the collector pulls the collector high. So a single stage (like the two input stages in your schematic) acts like a NOT gate. Do you see that part?

5. Jun 11, 2010

### TsAmE

So the base will be pulled high to V+ (positive rail) if the input signal (Vin) > 0.7V or pulled low to 0V if Vin < 0.7V?

The NPN's base (Vin) is driven high by the resistor, since by Ohms Law, adding Rb increases Vin right?

Why does the transistor pull the voltage down on its collector?

Im also not sure why the pullup resistor on the collector pulls the collector (voltage?) high.

Sorry if I am bombarding you with questions.

6. Jun 11, 2010

### Staff: Mentor

It sounds like you need to learn a bit about basic Bipolar Junction Transistor operation:

http://en.wikipedia.org/wiki/Bjt

RTL operates the transistors in the saturated and cutoff regions.

7. Jun 11, 2010

### vk6kro

In the following diagram:

[PLAIN]http://dl.dropbox.com/u/4222062/Transistor%20voltages.PNG [Broken]

In circuit A you can see that the voltages in a series circuit divide so that the voltages are proportional to the resistances and always add up to the supply voltage. This is because the resistors both have the same current flowing in them and V= I * R.
Don't go on until you understand this and why it happens.

In circuit B the bottom resistor has been replaced by a transistor, but the transistor has no base current so it isn't conducting. It has a very high resistance compared with the 1 K resistor so the voltage across it is almost 12 volts.

In the last circuit, circuit C, the transistor now has a lot of base current and the transistor now behaves like a small resistance resistor. Can you see that the voltage across the transistor will now be small compared with the 1 K resistor, so most of the supply voltage will appear across the 1 K resistor?

If you disconnected the top of the 10 K resistor and connected this to 0 volts (the negative side of the battery) then to +12 volts, you can see that the collector will go from 12 volts to 0 volts. So it works as an inverter.

Last edited by a moderator: May 4, 2017
8. Jun 12, 2010

### TsAmE

A: I understand but dont know how you worked out the voltages across the resistors, as it looks to me as too little info is given.

B: Why does the transistor have 12V across it, if there is no current flowing through the circuit?

C: Why isnt there a voltage drop across the 1k? VCE = 0V as the transistor is saturated and has almost no resistance right?

I thought the collector (output) would be on (1), if the transistor was switched on, and off (0) if the transistor was switched off.

9. Jun 12, 2010

### vk6kro

A: I understand but dont know how you worked out the voltages across the resistors, as it looks to me as too little info is given.

Not really. Assume a gain of more than 20 and the voltages would be pretty close to those given. Almost any transistor would have this much current gain.

B: Why does the transistor have 12V across it, if there is no current flowing through the circuit?

Have a look at the first diagram. If the top resistor had 50 times as much resistance as the bottom one, it will have 50/51 of 12 volts across it. The bottom one would have 1/50 of 12 volts across it.
If the top resistor had infinite resistance, all of the 12 volts would appear across it.
The transistor with an open circuited base would be like an infinite resistance, ignoring leakage effects.

C: Why isnt there a voltage drop across the 1k? VCE = 0V as the transistor is saturated and has almost no resistance right?

There is. The diagram shows 12 volts across it.

I thought the collector (output) would be on (1), if the transistor was switched on, and off (0) if the transistor was switched off.

No, if the transistor is conducting, the output would be zero volts (well, a small voltage). If it was open circuit, it would have all the supply voltage across it.

Incidentally, your AND gate can be made with just one resistor and two diodes.

10. Jun 13, 2010

### TsAmE

Howcome there is a current gain in A if there is no transistor? I ment how did you work out the voltages across the resistor, the unknown R?

Otherwise I understand the rest of the stuff you said, thanks a lot.

11. Jun 13, 2010

### vk6kro

You don't have to know the resistor values. The voltage in a voltage divider always splits in the ratio of the resistors.

Try an example.

Supply = 12 volts......Top resistor is 30 ohms......Bottom resistor is 10 ohms. Ratio 30 / 10 = 3

Total Resistance = 40 ohms
Current = 12 volts / 40 ohms = 0.3 amps

Voltage across 10 ohm resistor = I * R = 0.3 * 10 = 3 volts
Voltage across 30 ohm resistor = I * R = 0.3 * 30 = 9 volts

ratio 9 / 3 = 3

12. Jun 14, 2010

### TsAmE

Oh I see. Another thing I want to know is in diagram C, when the transistor is switched on does the current flow in the following way:

*It starts at positive terminal of battery, then flows through the 10k resistor, into the base, then into the emitter then earth. This is the base current right?
*After this has happened, another current (which is amplified by the base current) starts at the top of the 1k and flows through it to earth. This is the current of the 1k right?

13. Jun 14, 2010

### vk6kro

Yes, that is exactly right.

When a transistor is used as an amplifier, a larger base resistor is used, so that the transistor is not driven to saturation but its collector sits at about half the supply voltage.

Now, how did you go with the AND gate?

14. Jun 14, 2010

### TsAmE

If you mean how I think the current flows then this is what I say:

*In the beginning if Vin > 0.7 for the first inverter, then the IB will flow through the resistor, to earth, setting up the IC from positive rail to earth. Vc (output) = 0 (as all the supply-voltage dropped across the resistor).

*If for the second inverter Vin < 0.7, then IB = 0, thus IC = 0 therefore Vc = 1.

*For the NOR-gate, its first input will = 0 and its second one = 1.
*By the time these to inputs reach the common point at the base, the '1' will overide the '0', causing the final Vc of the AND-gate = 0 (as the transistor at the end switches on).

I think this is right, but please tell me if I am wrong somewhere.

15. Jun 14, 2010

### vk6kro

That could work.

Having either input low would make the same transistor output high which should be enough to turn on the final transistor and give a low output.
Only if both inputs are high would the final transistor turn off, giving a high output.

16. Jun 15, 2010

### TsAmE

Oh I see. Another thing I am curious about is why is it that although the output at the collector is a '0' it is actually 0.7V?

17. Jun 15, 2010

### vk6kro

That is because we are using bopolar transistors.
It isn't really a problem and you could drive a following stage with such an output, if you were careful.
You just have to make sure that the 0.6 volts is not enough to be counted as a "1" by the following stage.

CMOS transistors are much better at this and their saturation voltage is much lower. Their output goes much closer to the + and - supply rails than the bipolar devices.

18. Jun 15, 2010

### TsAmE

Oh ok. I read somewhere that for a darlington (made up of 2 transistors), when it is fully on, there isnt a voltage between x and y, but there is always 0.7V between y and z, and thus between x and z (refer to new attachment for diagram).

This doesnt make sense to me, cause since a darlington contains 2 diodes, it should have 0.7 x 2 = 1.4V between x and z? I dont get why there isnt a voltage between x and y.

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19. Jun 15, 2010

### vk6kro

Using the values of resistors in the 3rd diagram above (10 K bias resistor and 1 K load and a 12 V supply), here are some voltages for a Darlington made of individual 2N5772 transistors. These have a current gain of about 72.

X 785 mV

input 1.271 V

y 685 mV

So, there is a voltage across the 1st transistor, as you would expect, but it is only 100 mV.

Notice that the "saturation" voltage for the Darlington is 0.785 volts which is much higher than for a single transistor.

20. Jun 16, 2010

### TsAmE

Sorry but I still dont get why the first transistor has almost no voltage drop, while the second one still does. To me it looks like both the transistors (that make up the darlington) are the same.

The saturation for a single transistor is 0.7V, right? (cause of the forward-biased diode at the base-emitter).