Schottky Diode Question

Hello everyone. This is my first time here. I wanted to ask a quick question.

I am rather new to electronics and am almost done with my first solo project which is a solar battery charger.

My circuit uses a schottky diode to keep the battery from discharging via the panels when it is dark.

For the diode I used Vishay SB120-E3/54. The only issue is when I tested the diode, I got the same voltage both forward and reverse bias. I have several of the same diodes and they all test the same.

To test I placed the diode on the positive side of 1.2v AA battery and metered both forward and reverse. I get the same voltage both ways.

It is my understanding that reversed shouldn't allow anything to pass. Is my understanding wrong here, did I use the wrong part or am I looking at this wrong?

Thanks for your assistance!

 

You were measuring the leakage current of the diode with no load, which is apparently more than enough to register on your meter. Add a load resistor to dump the leakage current and you will get a valid reading.

 

Thanks for the reply.

Here is my test:



D1 is the Schottky Diode I mentioned before.

When I meter the circuit I am getting 1.24v. When I reverse the diode I get the same 1.24v.

When I swap out the Vishay SB120-E3/54 for a Fairchild 1N4001 I get 1.02v and 0v when reversed.

 

Apparently your diode is broken.

 

Is it possible that the entire lot that I bought could be shot? I tested all of them and they all react the same way.

 

You have heard from me. The only thing to do is wait and see if somebody else knows why your test is invalid or I missed something.

 

Where are you measuring the voltage?

 

I'd say he's measuring across the voltage source.

 

Between the diode and negative terminal on the battery also between the resistor and diode.

 

You should use the Ohm's setting(resistance) on your meter to check the diode and it should NOT be powered or in circuit when you check it.

The reading on a good diode will be a low resistance and when the meter leads are reversed you should find a high resistance.

You are currently measuring the voltage across your voltage source and this will only tell you the level of voltage present in the circuit. With the resistance shown(1k) the voltage reading will be almost identical whether the diode is conducting or not.

 

There are two ways (maybe more) that I can think of to test diodes:

1:
Many meters have diode testers you can use them to test diodes out of circuit. A good schottky should read ~0.3V-0.5V forward bias and OL reversed biased (when the test leads are in the other direction).

2.
If you are measuring the voltage drop across the schottky in circuit (as you show in your picture), then you could reverse the direction of the schottky to see if it's broken. If you reverse the direction of the schottky then the voltage across it should be the same voltage as your source (or close to it).

 

What Kermit2 said.

Or you can modify your circuit to measure the voltage drop across the diode.
Here is a diagram:

 

Thanks everyone.

I did what WhatsThatSmell said. This time I used 3v. Both forward and reverse I got 3v.

When I run the diode in diode mode on my meter I get .200 forward and OL reverse.

Using the ohm setting I get 1kΩ forward and 850kΩ reversed.

Running the diode in circuit with an LED the LED lights when the schottky is forward and goes dark when it is in reverse.

I think what is confusing me is the voltage present when running in the setup WhatsThatSmell showed and the diode is reversed. I Thought that the diode just blocked any voltage like a switch.

 

You need to measure the voltage directly across the diode with the diode connected in both directions, in series with the resistor. In the forward direction it should measure about 0.5V. In the reverse direction it should read the battery voltage.

 

I agree with crutschow: You need to measure the voltage just across the diode. Also, don't swap the leads of the probe to measure forward and reverse. Instead swap the diode around, and measure across the diode again.