Draw 2-Input AND Gate Circuit Diagram | Transistors & Resistors

[TsAmE]

Homework Statement

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

Component values are not required.

Homework Equations

None.

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?

 

[berkeman]

Homework Statement

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

Component values are not required.

Homework Equations

None.

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?

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?

 

[TsAmE]

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?

Sorry but I don't understand. How can its base be pulled high or low?

 

[berkeman]

Sorry but I don't understand. How can its base be pulled high or low?

This page on Resistor-Transistor Logic (RTL) should help you:

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?

 

[TsAmE]

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?

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.

 

[berkeman]

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.

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.

 

[vk6kro]

In the following diagram:

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

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.

 

[TsAmE]

A: I understand but don't 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 isn't 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.

 

[vk6kro]

A: I understand but don't 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 isn't 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.

 

[TsAmE]

fNot 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.

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.

 

[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

 

[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?

 

[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?

 

[TsAmE]

Now, how did you go with the AND gate?

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.

 

[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.

 

[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?

 

[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.

 

[TsAmE]

Oh ok. I read somewhere that for a darlington (made up of 2 transistors), when it is fully on, there isn't 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 doesn't 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 don't get why there isn't a voltage between x and y.

 

[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.

 

[TsAmE]

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.

Sorry but I still don't 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.

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

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

 

[vk6kro]

Sorry but I still don't 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).

The 2N5770 has a saturation voltage of 0.190 volts when used on its own. Yes, that is a lot smaller than the base emitter voltage of 0.7 volts. This is quite normal.

If the second transistor in the Darlington pair did have only 0.190 volts on its collector, don't forget that the first transistor would also have this voltage (because they are joined together.) But it has 0.7 volts on its emitter so it would cease to function because it would have the wrong polarity across it.

This would mean the second transistor stopped getting base current and hence was not driven into full saturation. The current it draws stops at the point where its base current is starting to be cut off if it drops any further.

This is a disadvantage of Darlington pairs as the higher minimum voltage results in less available voltage swing.

 

[TsAmE]

Oh I see. And for the direction of the current flow in the darlington would it be like this (referring to my last attachment):

*At base of first darlington if Vin > 1.2, IB flows through first diode, into second diode, to earth.
*IC starts flowing, at X it splits into 2 currents: one goes down to earth, the other one goes through X then to earth.

 

[vk6kro]

That is fairly correct.

The initial base current must flow through both bases, but it is the amplified version of this from the first transistor which gives the main drive to the second transistor.

In practice, it all happens smoothly and you don't see sudden jumps in current.