Diode ORing voltage measurement

[vinod]

For the below circuit, I have placed multimeter +ve probe at 'B' wrt GND and measures 4V against the expected value of 0V.

Hence I have terminated the 'B' end with 10K to GND temporarily before taking the measurement and reads 0V.

Why this happens?

What happens to the anode of D2 while I connect a multimeter to 'B' end wrt GND ?

I have replaced the 10K temporary resistor with 1M, then again 4V measured !

[Robert Reed]

All modern day DMMs have at least 10 meg input inpedance and some have much higher. It is not uncommon to measure a voltage of some sort in an apparent "open circuit", if the scenario is right. That being that even a small amount of leakage current in those diodes will develop some voltage across the DMM. However if you are still seeing that with a 1 meg load, something else is going on. For starters check your diodes for excessive leakage.

[MrAl]

Hi,

As Robert says, it sounds like too much leakage in the diode(s). It could be just that one diode.

Some switches and relays show higher than expected values of leakage too, so you could check the relay also. Try removing the 28 volt source just to see if it has any effect with the 1M resistor. If not, then it's the diode and the other diode might have high leakage too.

Keep in mind that the diode(s) would have to have about 100 times the normal leakage for that type of diode, so check the part number too. You might expect this with a Schottky diode but with a normal 1N4148 you shouldnt see that much leakage.

Oh yeah, this all assumes you are working near 25 degrees C. At 150 degrees C you'd see 1000 times the normal leakage which would easily explain what you are seeing here. 10ua to 1M means 10 volts across the resistor, and that would be seen at elevated temperatures while 30na is more likely around 25 degrees C.

[haklesup]

Diodes have current leakage, not voltage leakage, there has to be load to drop that current across to measure a voltage. With no current flowing through, you will measure the same voltage on both sides of a diode, Reverse leakage current can cause at most mV drops.

you can expect near zero current at point B but not zero voltage. When you pass current through diode D1 from 30V source to point A (assuming branch A has current flowing out of it) you will see the forward voltage of Diode D1 at point A. You have a lot of current flowing through D1 to get 4V but it is possible. Since D2 has no current flowing there is no voltage drop across it and you get the voltage at point A. Since no current is flowing, it won't matter what resistor you put in the path, you are measuring V with I=0

Try measuring the differential voltage across the diodes instead of WRT ground as a comparison.

Now if branch A is an open circuit then you should have no current flowing through either diode unless your meter is broken. Even with leakage current, the voltage measured should be very small, much smaller than 4V in any case. I think there has to be current flowing down branch A

I think the purpose of the two diodes it to prevent accidental shorting of the + and - supplies through the relay should it want to arc while closing one way or the other.

[Lenp]

Vinod
Question? Is this a design of yours or from someplace else?
Unless I am missing some design requirement it seems that you are using a DPDT relay when a SPDT would work.
If the +30V went to the N/C contact (01) and the +24 went to normally open contact (S1) the outout would be from the common contact (P1).
With only a 2V difference between the two supply voltages, any arcing would be most unlikely, so even the diodes could be eliminated.

YES/NO...COMMENT!
Len