There are two ways to make an OR gate:

OR gate using diodes and: OR gate using transistors

But with diodes it looks easier to make than with transistors, isn't it? What are the advantages of the transistors version and drawbacks of the diodes version?

  • 5
    \$\begingroup\$ What will happen when you chain a few diode OR gates to other diode OR gates? What will happen to the output voltage? (although the transistor OR-gate you've shown probably isn't much better in terms of output voltage drop) \$\endgroup\$
    – tangrs
    Commented Feb 19, 2017 at 13:39
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    \$\begingroup\$ Those are two very different configurations. It's like asking if a truck is useful when you can use a sedan instead. Just look at the transistors diagram. You can see that the output is 6V, but the input to the transistor gates does not have to be 6V. That's just one example of how they are different. \$\endgroup\$
    – Bort
    Commented Feb 19, 2017 at 13:40
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    \$\begingroup\$ @Bort the transistors are used as emitter-followers, hence the innput must be 0.6V higher than the output, just like for the diodes circuit. (And the base resistors can be omitted). \$\endgroup\$ Commented Feb 19, 2017 at 13:48
  • 2
    \$\begingroup\$ It may also be noteworthy, that these are wired-or gates. For most modern applications (and in most modern chips), one would instead use a NOR gate of 4 transistors, actively driving the high and the low level, and then invert that output with another two transistors. See for example en.wikibooks.org/wiki/Practical_Electronics/IC/4071#/media/… \$\endgroup\$
    – mox
    Commented Feb 19, 2017 at 21:56

2 Answers 2


"There are two ways to make an OR gate" and they are both crap unless combined with other (voltage amplifying) stages.

In both circuits the output voltage will be 0.6V lower than the input voltage, hence the need for apmplification stages.

The difference is where the current comes from: for the diodes version, all output current must be provided by the input(s). For the transistor version the inputs deliver only the base current, which is (very) small compared to the output current. And BTW the base resistors are not needed.

  • \$\begingroup\$ Why are the base resistors not needed? I work with breadboards and an Arduino (if that's relevant) and the tutorial I followed said to include a base resistor. \$\endgroup\$
    – Arc676
    Commented Feb 20, 2017 at 10:32
  • \$\begingroup\$ For the gate as it's shown the base resistors are not needed because the emitters are serially connected to another resistor. But are they really not needed if Out circuit has small input resistance? Wouldn't then base current be basically unlimited? Or is the question just senseless since connecting such a circuit is abuse of the gate? \$\endgroup\$
    – Ruslan
    Commented Feb 20, 2017 at 10:58
  • \$\begingroup\$ The essense of this way to use a transistor is that the base current is the emitter current / Beta, hence any sensible emitter current yields a sensible base current. \$\endgroup\$ Commented Feb 20, 2017 at 16:00
  • \$\begingroup\$ @arc676 if your tutorial shows base resisters for a common collector circuit (== emitter follower) they are simply wrong. Which is, sadly, far from unlikely. \$\endgroup\$ Commented Feb 20, 2017 at 16:01

The transistor OR gate, using emitter followers, progressively degrades the noise margin. If 0.7 volts per OR, after 3 ORs the Vout is 5v-3*0.7 = 2.9 volts, which is dangerously close to the Vdd/2 value.

This popular OR ( actually 3-input-NOR, converted to OR by the final Inverter) preserves the logic levels, with ~ 0.0 volts and ~ +5 volts, at all times.


simulate this circuit – Schematic created using CircuitLab


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