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I know that this is a very basic question but I really do not know the answer.

I was making simple logic gates using transistors and I used PNP transistor as NOT gate.

I made the following circuit: (A) AND (B) AND (NOT C)

This circuit does not work without the resistors R1,R2 and R3. Why are these resistors important ?

schematic

simulate this circuit – Schematic created using CircuitLab

The Following NOT gate does not need a base resistor. I connect Vcc or ground directly to base and it works.

How can it work without resistors although it is very similar to the previous circuit ?

schematic

simulate this circuit

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  • \$\begingroup\$ Most people don't like it when their magic smoke escapes... \$\endgroup\$ – PlasmaHH Sep 24 '15 at 20:16
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    \$\begingroup\$ If you drive "D" with something that can sink significant current, and use a battery with low internal resistance, then the second circuit is not likely to work for very long. \$\endgroup\$ – The Photon Sep 24 '15 at 20:35
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    \$\begingroup\$ Resistors in the bases might just save the LED from failing on over-current. Have you heard of BJT current gain? \$\endgroup\$ – Andy aka Sep 24 '15 at 20:38
  • \$\begingroup\$ @Andyaka yes, I've heard of BJT current gain. "Resistors in the bases might just save the LED from failing on over-current". The LED did not light at all, It is not burnt. So, I wonder why the led is not "on" at the first circuit when I removed base resistors, it should receive big current ? \$\endgroup\$ – Michael George Sep 24 '15 at 20:46
  • \$\begingroup\$ Can you show us what is driving the bases of the transistors? As the others have hinted at, you need to limit the current through the LED. If it's a red, green or yellow LED it will pass a very large current if you power it from 6V as shown. A resistor of about 470 ohms would limit the current to a safe value - about 20 mA. \$\endgroup\$ – Transistor Sep 24 '15 at 21:59
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Without resistors, your circuit looks like

schematic

simulate this circuit – Schematic created using CircuitLab

The problem is that there is no voltage across the emitter-base of Q1, so it cannot turn on.

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The second circuit works because when you connect the base to ground you have shorted out the base-collector junction and, effectively, turned your transistor into a diode (between the emitter and the base. Since that "diode" is forward biased current will flow.

As per my comment above, we need a little more info on what was driving the transistor bases to understand what was happening with the first circuit.

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  • \$\begingroup\$ Wires were driving the transistor bases. I connect them manually to vcc or ground. The circuit does not work without resistors. It should draw large current without resistors, shouldn't it ? \$\endgroup\$ – Michael George Sep 24 '15 at 23:05

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