I need a diode that has a constant voltage drop at either 0.9 or 1.0 v. What diode should i use?

 

I don't think any have CONSTANT voltage drop, they vary with temperature and current.

What are you trying to do?, and why do you need a 1V drop?.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

It is for the circuit you helped me with, the current limiting circuit. The voltagedrop over the diode/diodes are critical. I cant use 2 1N4148 (0.7v each). Then the output current gets to low under 12V input.

But if i have a lower voltage drop, say 1V and a lower resistance value, then 12V works fine.
So it don't have to be constant, but it have to be rather close to 1V to work fine.

Someone said to me that a 1N4001 had a 1V drop. Would that diode work?

 

How many LED's are you putting in series?, you should have something like 11V available with a 12V input. The voltage drop acros the diodes isn't critical at all, but if you're trying to run too many LED's in series you will run out of voltage.

If you're running out of voltage, changing the diodes (assuming you can find something at 1V?) will only increase the voltage headroom by 0.4V - which isn't very much.

You could always split your LED's into two chains, and have two transistors and two 22 ohm resistors - you can connect the two bases together, and use the same pair of diodes and feed resistor to set both base voltages.

It's a standard silicon rectifer, it will have about 0.7V drop, like any standard diode junction. Under high current demand it will increase somewhat, as all junctions do - but it's not running with high current.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Well the led's are in a module(6 leds), with only 2 pins out, Anode and Cathode. So splitting up isn't an option.

 

In that case add a potentiometer across the two diodes, and connect the slider to the base of the transistor - then you can adjust the base voltage to 1V, it will still be stabilised as it's fed from the 1.4V reference across the diodes. This pot will then act as a current control as well! - so be careful, you may be able to adjust it high enough to blow the LED's.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Hmm... constant 1v?
Well, you can't use a zener because the lowest voltage used with Zeners is 1.8v.

You could put a normal rectifier diode in series with a Schottkey. Normal rectifier is 0.7v and Schottkey is 0.3v. The voltage will rise a bit as current rises though, it is not extremely tightly controlled. Temp changes it a small amount too.

But you mention LEDs? So I assume you're doing that common emitter circuit. For this, so you see the tricky problem in these. You need either a really specific driving voltage or a really specific low ohm resistor.

 

I was thinking Vbe multiplier circuit. 1 tran & 2 resistors.

 

Try

 

That still drops 0.7V across the emitter resistor, just like the two diode version.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

You can bias a single NPN transistor to act like a 1V diode.
Connect 1K B to E, and connect 390ohms C to B.
Use C and E as connections as a diode.