Schottky Diode Passing Voltage When Acting as Blocking Diode

1. Try using an analog voltmeter, instead of the digital one, and see if the voltage is still there.
2. Try wiring a 27kΩ resistor across the digital voltmeter terminals and see if the voltage is still there.

Was there any special reason to go for Schottky rectifier? Yes, they have greater leakage current and much higher current with increasing temperature. You already have these points discussed now.

What is your maximum current requirement?

What is the switching time you need?

Main reason to select the Schottky Rectifier is its fast response time. They also switch relatively higher current and have lower forward drop voltage.

Right thing to do is to select a part for the desired specifications. As you have not specified your goal, nothing can be suggested to improve upon the current design.

I do give you credit here. You show the fundamentals of becoming an engineer here by applying some rudimentary testing and measuring to solve a personal mystery. I recommend that you include some more thought and detail description when you find a problem that initially does not make sense to you. For instance, you do not specify the low value resistance you used to us. In this problem it was not critical for us to help you but in another case it might have helped for us to know this. When you added the low resistor (100 ohm?) and then decided to try some other resistor values (100K, 10K, etc.). A little calculation using the voltages that appear across these known resistor values and you would've found that a relatively constant but tiny current was going backwards through the Schottky diode. Bingo reverse leakage current.

My point here, you were so close to solving this problem on your own. All you had to do was apply some other resistor values and applied a little analysis of these results. Have faith in your curiosity and ability to understand circuitry.

Oh, and welcome to the madhouse.

Various diode types are made for various purposes, and widely different characteristics. All of those are tradeoffs. Schottky in your case happened to be selected for low voltage drop in forward direction. A price for it is significant leakage in reverse direction. That goes up sharply with temperature on the top of it.

Unless something (a resistor maybe) provides a low impedance path to ground, that leakage shows up as (full) voltage. A short circuit -with a current meter in series - to ground proves to you the (low) magnitude of the leakage.

On the other hand, if there is miswiring, or a defective diode, the effects can get interesting.

Hi Nigel. Welcome to CR4.

Two schottky diodes or'd together

Do you mean in series? Same polarity? opposite polarity? If you were to scan the schematic and show it to us with resistor values shown and voltages measured, then we might be able to help you.

Hi,

The Two diodes have the output of the individual batteries to the Anode of each diode. The Cathodes are tied together which acts as the output for this circuit. The diodes are connected this way to stop one battery charging up the other battery.

There are no resistor values in this circuit, the actual load is a camera and accessories which could up to 10 amps in total. The need for Two batteries is to stop the camera powering down due to a flat battery, and the need not to have to keep rebooting.

Some of the replys seemed to have missed the point I was getting at.

Namely, why is there full battery volts across the blocking diode of the lower voltage battery.

Ok, I believe we have a failure to communicate here. The only way I can see you should have (close) to a full battery voltage across a reverse biased blocking diode (voltage across anode to cathode) is if one battery has been replaced by a short circuit. Obviously this battery should either be dead or replaced by the short circuit. Now if you said that you had two times the battery voltage across one diode I'd tell you one battery was installed backwards. The voltage drop across these "OR"ing diodes should only be the drop from current flowing through the junction.

Now you might not have the circuit topology you expect. The connection between battery and diode anode may not really be there. Thus measuring the voltage to ground at the anode of that diode you get close to zero volts but the voltage to ground at the battery is still the battery voltage. This can be due to a bad crimp, broken wire, bad solder or a lot of other similar conditions.

10amps, you will smoke this. This is good for 1 amp, and again look at the spec page 3 figure 5. You can put pulse current through it, but that is in MICRO seconds.

Quote "Namely, why is there full battery volts across the blocking diode of the lower voltage battery."

What are you measuring? Right across the diode? The common cathode to the ground of the minus battery terminal? Your not making it clear just what you are measuring. My assumption was from my first reply, that the open unconnected anode (where you had not connected the backup battery) was measured to ground, and because of diode leakage current and a ohm meter input resistance, that is an order of magnitude (by 1000 or more) greater then this leakage current resistance, you were measuring the voltage at the cathode.

You need to clarify this.