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I breadboarded some AND, OR, NAND and NOR logic gates (schematics below) using:
- 2N2222 transistors
- 10K resistors
- 4.7K resistors

Can anyone explain (or provide a link that explains) why 10k and 4.7K resistors are used?

I'm sure this topic has been covered before, but I just cannot find any relevant information, most likely due to Googling the wrong keywords...

AND OR

NAND NOR

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  • \$\begingroup\$ Have you tried "Vbe" and "Vce"? \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 9 '13 at 23:44
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Why 4k7 and 10k?

enter image description here

When you are designing non critical circuits, such as these Resistor-transistor logic gates, you select values for resistors that you know are common types rather than calculate a specific value. The selection comes from experience - you know this will work as you point out other values would work too.

Firstly your dealing with a small signal general purpose transistor so there will be a range of current that you can safely pass through it, (Ic). Suppose we decide to limit this current range to between 1mA and 50mA . This means that for a 6V supply the resistor can range in value between 6k (for 1mA) and 120R (for 50mA).

A 4k7 resistor limits the current to about 1.3mA - a value near the lower end of our current range (low power requirement/dissipation) and so it would be deemed as a suitable value. 5K6 may be a better choice.

Having set this value of resistor we then look at a suitable value for the input resistor. The current gain of a general purpose signal transistor is going to be in the range 50 - 300. Lets assume the worst case and take 50. Then for a current of 1.3 mA we will need at least 26 microamps (.026mA)

The worst case for input is this circuit.

enter image description here

Input A has to be close to the 6V line to get any current into the base of the top transistor. The 10k value will produce a voltage drop of 0.26V @ 0.026mA. A lower value would produce a lower drop but would also mean higher base current in the other configurations. Once again it becomes a compromise between too much or too little. The 10k value sits in the 'goldilocks zone' and is just right.

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  • \$\begingroup\$ Your answer makes it look really simple! But I would never have figured out this myself... Thanks for your time, really helped me! \$\endgroup\$ – Anne Jun 10 '13 at 14:34
  • \$\begingroup\$ @Anne - no problem, here to help \$\endgroup\$ – JIm Dearden Jun 10 '13 at 14:50
  • \$\begingroup\$ For the last circuit - the and gate - couldn't you just leave out the 10k resistors? You wouldn't need them for current limiting, the 4k7 would do that already. Or is it to prevent ringing like with mosfets or something else entirely? \$\endgroup\$ – Nebula Jul 31 '15 at 5:02
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The resistors are required to prevent an infinite current from flowing. What does this circuit do?

schematic

simulate this circuit – Schematic created using CircuitLab

(hint: it's not much different from this circuit)

schematic

simulate this circuit

How much current flows? A lot. This is an NPN transistor, so called because it is constructed from a sandwich of N, P, and N doped semiconductor. That is, it's two diodes back to back sharing a cathode (a diode being just a PN junction). The base-emitter junction is exactly a diode, and that's why it has an arrow. The base-collector junction is also a diode, and the only difference is that the asymmetrical construction of the BJT means you won't get a very good transistor if you try to use it backwards like this.

Like any diode, the current that flows in the base-emitter diode, once forward biased (about 0.65V for any silicon junction) is limited only be the internal resistance of the diode1, and the resistance of the wires. This is the same reason you need a resistor or other current limiting device when driving an LED from a voltage source.

Likewise, consider what happens here:

schematic

simulate this circuit

Hint: it's not much different from this:

schematic

simulate this circuit

(\$0.2V\$ being the collector-emitter voltage of a typical BJT in saturation, and the base current being small enough to be negligible).

Again, you get smoke. The resistors are there to limit the current through the transistor.

1: this is a first-order approximation. Really the voltage-current relationship for an ideal diode, with no internal resistance, is an exponential function, but at some point, the current shoots up so rapidly for any increase in voltage that the current might as well be infinite.

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  • 2
    \$\begingroup\$ An excellent and very clear answer as to why the resistors are required (so +1 from me) but I think the question is asking why those particular values (10K and 4k7) \$\endgroup\$ – JIm Dearden Jun 10 '13 at 8:26
  • \$\begingroup\$ +1 I greatly appreciate your extensive explanation and truly wish I could accept both answers! The art of explaining something step by step! Suddenly everything makes sense! \$\endgroup\$ – Anne Jun 10 '13 at 14:36

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