I have been reading the book Practical Electronics for Inventors, and one of the examples pictured below has confused me.
I understand that the inputs, when set high, will mean there is no voltage potential across the diode, and so current will not flow to ground through RD. And that when all inputs are high, current will only flow through the output.
But what I don't understand is how the LED will actually light? the resistor RU (10k) will reduce the current to well below what the LED requires?
tunacheese:
But what I don't understand is how the LED will actually light? the resistor RU (10k) will reduce the current to well below what the LED requires?
Well, many years ago that would have been correct, but LEDs now are extremely efficient and will in fact, light quite well with a third of a milliamp. They are very linear in terms of light output as a function of current and eyes are quite sensitive.
What a poor circuit. The Ru value should be 330, Rled is superfluous, as are the Rd's. Any input pulled low
pulls the output low too and turns the LED off.
MorganS:
10.33k looks like a very high resistance to power a LED. Is there some kind of amplifier or transistor inside that circular symbol labelled "output"?
Hmm, possibly I suppose - still doesn't explain the spurious input pull-down resistors...
