Why didn't I get the voltage drop expected on diode rectifier?

[red913]

I got some 1N4001 diodes and putting 3 in series gave me a voltage drop of about only .4V. I don't understand why I didn't get more? I need about 3V drop. I'm not quite sure what to look for on the data sheet either. If someone could direct me in what I did wrong or direct me to the proper diode(I think the one I ordered may be incorrect) I would be very grateful.

 

[Harald Kapp]

What current was flowing?
A diode has not a sharp corner in the characteristic but a more or less strong knee.
If you measured the output voltage with a high impedance multimete, current will have been so small that the voltage drop is almost negligible.
Add aload (e.g. 1kOhm resistor) to the output and measure again.

 

[red913]

What current was flowing?
A diode has not a sharp corner in the characteristic but a more or less strong knee.
If you measured the output voltage with a high impedance multimete, current will have been so small that the voltage drop is almost negligible.
Add aload (e.g. 1kOhm resistor) to the output and measure again.

When adding a resistor it won't function at all. I tried a 1k and 470 ohm one. I Don't have anything to measure current at the moment.

 

[davenn]

3 diodes should give you ~ 2V drop with about 0.6 - 0.7 V per diode


Dave

 

[BobK]

Actually, if you look at the data sheet, the drop is 1.1V at 1A.

Bob

 

[Harald Kapp]

When adding a resistor it won't function at all. I tried a 1k and 470 ohm one. I Don't have anything to measure current at the moment.

How did you connect the resistor? It sounds like you connected the transistor in parallel to the diodes. That, of course, won't help. Connect the resistor from the output to ground, so it acts as a load to the diodes.

I Don't have anything to measure current at the moment.

How then do you measure the voltage drop? Any decent multimeter should have a current range. If not, measure the voltage across the load resistor. From Ohm's law you get I=V/R.