Diodes - General Question

Thread Starter

NewToElectricity

Joined Jul 6, 2023
7
Hello,

I am hoping to understand a concept that I cannot seem to wrap my head around…

I understand that a diode conducts one way (anode to cathode assuming convention current flow), and requires a certain forward voltage potential across anode and cathode to conduct.

Where I am lost is in the attached image.

When a diode is used in a simple DC circuit consisting of the diode in series with a DC relay; and the positive potential side wire is disconnect, why do I read that the anode of the diode is at 0V potential? I can understand the cathode being at 0V potential since it is directly connected to the relay which has both terminals at ground potential, but I find myself confused that the anode is at the same potential because I’m thinking it should be blocked somehow. And if I flip the diode around to have the anode connected directly to the relay, then anode is at 0V potential but the cathode is not At 0V potential.

how I am taking this potential reading: positive meter lead is on the 24V power supply as a reference and the negative meter lead is on either anode or cathode.


appreciate all responses :)
 

Attachments

crutschow

Joined Mar 14, 2008
34,037
If there is no current other than leakage going through a diode, then the anode is at the same potential as the cathode or more negative than the cathode (reverse biased).
Does that make sense?
 

Thread Starter

NewToElectricity

Joined Jul 6, 2023
7
If there is no current other than leakage going through a diode, then the anode is at the same potential as the cathode or more negative than the cathode (reverse biased).
Does that make sense?
Thank you crutschow! Kind of makes sense but not really. Maybe I would better understand it if you could explain what would happen if the diode is flipped in the attached image of the original post?

Also why would the anode be more negative than the cathode?
 

MrChips

Joined Oct 2, 2009
30,466
You have to take into account that your testmeter has internal resistance, albeit very high resistance, usually 1MΩ or 10MΩ.
So even though the switch is open, the meter is creating a path for current to flow.

That is an unusual method of checking voltages.
Normally, one would connect the BLACK (-) lead of the meter to the negative side of the power supply. This is the reference node.
Then one would use the RED (+) lead of the meter for probing.
 

vu2nan

Joined Sep 11, 2014
341
It is standard practice to make voltage measurements with '0V' as reference.

Case 1: When the switch is open

1.png

The voltage is 24 V before the switch and 0 V after it.

Case 2: When the switch is closed

2.png

The voltage is 24 V before and immediately after the switch but 23.3 V after the diode,
which drops 0.7 V.
 
Last edited:

MrAl

Joined Jun 17, 2014
11,263
Hello,

I am hoping to understand a concept that I cannot seem to wrap my head around…

I understand that a diode conducts one way (anode to cathode assuming convention current flow), and requires a certain forward voltage potential across anode and cathode to conduct.

Where I am lost is in the attached image.

When a diode is used in a simple DC circuit consisting of the diode in series with a DC relay; and the positive potential side wire is disconnect, why do I read that the anode of the diode is at 0V potential? I can understand the cathode being at 0V potential since it is directly connected to the relay which has both terminals at ground potential, but I find myself confused that the anode is at the same potential because I’m thinking it should be blocked somehow. And if I flip the diode around to have the anode connected directly to the relay, then anode is at 0V potential but the cathode is not At 0V potential.

how I am taking this potential reading: positive meter lead is on the 24V power supply as a reference and the negative meter lead is on either anode or cathode.


appreciate all responses :)
Hi,

That is kind of like asking, "Why do I measure 0v when I did not connect the meter to anything yet".

As others have said, the diode is not a true open circuit when it is not forward biased. There is an equivalent parallel resistance. For many rectifier diodes it could look as high as 500 Megohms, but if it is a Schottky diode it could measure around 20k or something. There is always some leakage through the diode.
In fact, diodes with a clear glass package will even PRODUCE a voltage all on their own if there is any significant light hitting the die inside. It acts a lot like a solar cell then.

Perhaps you can report what the diode part number is that you are using, and if it has a clear glass package or black epoxy.

Also, as someone else said, connect the negative lead of the meter to the negative terminal of the power supply when you take readings, and then repeat the experiment and report the results back here so we can take a look.

It's a good question anyway.
 

Wendy

Joined Mar 24, 2008
23,396
The way I remember it is the arrow always points to the negative. This puts the device in a conductive state, it also works for transistors.
 

Thread Starter

NewToElectricity

Joined Jul 6, 2023
7
Hi,

That is kind of like asking, "Why do I measure 0v when I did not connect the meter to anything yet".

As others have said, the diode is not a true open circuit when it is not forward biased. There is an equivalent parallel resistance. For many rectifier diodes it could look as high as 500 Megohms, but if it is a Schottky diode it could measure around 20k or something. There is always some leakage through the diode.
In fact, diodes with a clear glass package will even PRODUCE a voltage all on their own if there is any significant light hitting the die inside. It acts a lot like a solar cell then.

Perhaps you can report what the diode part number is that you are using, and if it has a clear glass package or black epoxy.

Also, as someone else said, connect the negative lead of the meter to the negative terminal of the power supply when you take readings, and then repeat the experiment and report the results back here so we can take a look.

It's a good question anyway.
Thank you all...very interesting to learn more. Here is my thought process as to why I am not referencing the negative or "0V" terminal of the power supply: in order not to have a situation where i read 0V even though the potential of a point is not at zero volts (i only measured this way for this particular circuit, circuit #5). Please correct me if I am mistaken. I will explain my thinking below.

The diode I am using in the below examples is an IN5406.

REFER TO ATTACHED IMAGE


Circuit #1 makes sense to me.
Circuit #2 also makes sense to me.

Circuit #3 does not make sense to me. The readings tell me the the potential at point b and point a are very close to the potential at N (with respect to point P). Essentially this scenario is what we have in Circuit #5, which is where I am discovering all this and trying to understand why/the simple things I am clearly missing. The input i1 is active when keyswitch K1 is closed because we are grounding the active low i1 input (makes perfect sense). The input i2 is active when the relay contacts close because we are grounding the active low i2 input (also makes perfect sense). But why is i2 active when K1 is closed??? This is the reason why i am choosing my reference point as P which is 24 VDC, and I am probing with the negative lead. This way I can tell if i2 is truly at ground potential or not and why the input is activated.

Circuit#4, if I reference the usual way with negative lead on the negative potential, then I will read very close potential at point a and at point b, and it will almost be like they are both at 0 volts potential with respect to point P. But if I reference the positive point P, then i know P is at 24 V potential with respect to a so a must be at "N" potential. But point b shows 0.05VDC telling me that it is not at ground potential, and it is likely not at the P potential either (situation where you measure 0V on meter when leads aren't connected to anything).

Sorry my explanation is a bit all over the place. My intention is to explain my thought process in hopes that you can better see the flaws in my thinking so I can better understand what I am missing.

I guess i'll try again with maybe a resistor.

If i had a resistor in circuit 1, i'm expecting to see 24VDC on either side.
If i had a resistor in circuit 2, again I am expecting to see 24VDC on either side. It makes sense why the diode wouldn't have positive/24V potential instead on the side "b" side since it is blocking the 24VDC.

If i had a resistor in circuit 3, I know that the resistor is at N potential on both sides with respect to P. But the bloody diode example instead, the side connected to the N potential is obviously at N potential but how does the b side get the N potential also? I am basically expecting the results in circuit 4 to be for circuit 3. Clearly from the readings point b is at a very slightly higher potential than point a with respect to the negative in circuit 3, but how is it at a very slightly higher potential if it is not connected to anything and it is on the blocked side?! This is how my i2 input goes active when K1 is closed without the relay needing to close.

If i had a resistor in circuit 4, I know that the resistor is at N potential on both sides with respect to P.

There is clearly a parallel between circuit 2 and circuit 4. I just need to see that.

Thanks very much for your patience and help all!!!

View attachment 297901
 

MrAl

Joined Jun 17, 2014
11,263
Thank you all...very interesting to learn more. Here is my thought process as to why I am not referencing the negative or "0V" terminal of the power supply: in order not to have a situation where i read 0V even though the potential of a point is not at zero volts (i only measured this way for this particular circuit, circuit #5). Please correct me if I am mistaken. I will explain my thinking below.

The diode I am using in the below examples is an IN5406.

REFER TO ATTACHED IMAGE


Circuit #1 makes sense to me.
Circuit #2 also makes sense to me.

Circuit #3 does not make sense to me. The readings tell me the the potential at point b and point a are very close to the potential at N (with respect to point P). Essentially this scenario is what we have in Circuit #5, which is where I am discovering all this and trying to understand why/the simple things I am clearly missing. The input i1 is active when keyswitch K1 is closed because we are grounding the active low i1 input (makes perfect sense). The input i2 is active when the relay contacts close because we are grounding the active low i2 input (also makes perfect sense). But why is i2 active when K1 is closed??? This is the reason why i am choosing my reference point as P which is 24 VDC, and I am probing with the negative lead. This way I can tell if i2 is truly at ground potential or not and why the input is activated.

Circuit#4, if I reference the usual way with negative lead on the negative potential, then I will read very close potential at point a and at point b, and it will almost be like they are both at 0 volts potential with respect to point P. But if I reference the positive point P, then i know P is at 24 V potential with respect to a so a must be at "N" potential. But point b shows 0.05VDC telling me that it is not at ground potential, and it is likely not at the P potential either (situation where you measure 0V on meter when leads aren't connected to anything).

Sorry my explanation is a bit all over the place. My intention is to explain my thought process in hopes that you can better see the flaws in my thinking so I can better understand what I am missing.

I guess i'll try again with maybe a resistor.

If i had a resistor in circuit 1, i'm expecting to see 24VDC on either side.
If i had a resistor in circuit 2, again I am expecting to see 24VDC on either side. It makes sense why the diode wouldn't have positive/24V potential instead on the side "b" side since it is blocking the 24VDC.

If i had a resistor in circuit 3, I know that the resistor is at N potential on both sides with respect to P. But the bloody diode example instead, the side connected to the N potential is obviously at N potential but how does the b side get the N potential also? I am basically expecting the results in circuit 4 to be for circuit 3. Clearly from the readings point b is at a very slightly higher potential than point a with respect to the negative in circuit 3, but how is it at a very slightly higher potential if it is not connected to anything and it is on the blocked side?! This is how my i2 input goes active when K1 is closed without the relay needing to close.

If i had a resistor in circuit 4, I know that the resistor is at N potential on both sides with respect to P.

There is clearly a parallel between circuit 2 and circuit 4. I just need to see that.

Thanks very much for your patience and help all!!!

View attachment 297901
Hi,

Sorry your attachment is not working. You have to click "attach files" to upload a picture.
 

MrAl

Joined Jun 17, 2014
11,263
Sorry about that!
Hi,

That's better now.

For #2 and #4 it looks like your meter could be 1Megohm for it's input impedance.
With that presumption, if you replace the diode with a 479Megohm resistor and use a 24v DC source, you get 0.05 volts DC for the meter reading. That would not be unusual i don't think. 479Megohms is roughly 500Megohms. So you are seeing the voltage divider effect of a 500Meg resistor in series with a 1Meg resistor and reading the voltage across the 1Meg resistor with an ideal voltmeter that has infinite input impedance. That's the equivalent. The diode leakage current, at the voltage that is across it at the time, makes it look like a very high resistance resistor because it is reverse biased. The diode leakage current would be about 50 nanoamps. You could check the data sheet for that diode see if it matches up somewhat.
#1 and #3 i think you understand already.

For #5 you don't seem to give any readings.

Here is the leakage current plot. The red dot is at 24v and there the leakage current is around 50na.

DiodeLeakage_20230708_151444.png
 
Last edited:

BobTPH

Joined Jun 5, 2013
8,661
Measuring the voltage is equivalent to placing a 1M resistor where the probes are connected.

Your readings are what I would expect. If the diode us forward biased, it draws the current if a 1M resistor and sees a small voltage drop.

If it is reverse biased, it sees the smaller leakage current.
 

Thread Starter

NewToElectricity

Joined Jul 6, 2023
7
For #2 and #4 it looks like your meter could be 1Megohm for it's input impedance.
With that presumption, if you replace the diode with a 479Megohm resistor and use a 24v DC source, you get 0.05 volts DC for the meter reading. That would not be unusual i don't think. 479Megohms is roughly 500Megohms. So you are seeing the voltage divider effect of a 500Meg resistor in series with a 1Meg resistor and reading the voltage across the 1Meg resistor with an ideal voltmeter that has infinite input impedance. That's the equivalent. The diode leakage current, at the voltage that is across it at the time, makes it look like a very high resistance resistor because it is reverse biased. The diode leakage current would be about 50 nanoamps. You could check the data sheet for that diode see if it matches up somewhat.
Measuring the voltage is equivalent to placing a 1M resistor where the probes are connected.

Your readings are what I would expect. If the diode us forward biased, it draws the current if a 1M resistor and sees a small voltage drop.

If it is reverse biased, it sees the smaller leakage current.
Thank you both very much for a fantastic explanation, I never would’ve thought about leakage current in a reverse biased diode at such small voltage.

I think I am still a bit lost regarding circuit #3 in particular (#5 is same as #3 sorry)

The cathode of the diode is directly connected to the 0V potential point. Am I safe to say that the anode is pulled to the 0V potential because the diode is forward biased? This is the concept that I am confused about; I would’ve thought that the anode is blocked by from the ground potential.
 

Wendy

Joined Mar 24, 2008
23,396
It's all about direction of current flow. If the anode is more positive then the cathode then there will be flow. a side thought most meters come with a diode scale, this usually will tell you the full forward voltage drop of the diode when it is conducting. Silicon devices typically drop 0.6 volts. Shut the diodes typically drop 0.2 volts. Germanium diodes typically drop 0.3 volts.
Then there are LEDs whose Ford voltage drop is all over the place. And cannot take much back by voltage without damage.

The main reason I posted this is that what the diodes are made of and their geometry matters.
 

MrAl

Joined Jun 17, 2014
11,263
Thank you both very much for a fantastic explanation, I never would’ve thought about leakage current in a reverse biased diode at such small voltage.

I think I am still a bit lost regarding circuit #3 in particular (#5 is same as #3 sorry)

The cathode of the diode is directly connected to the 0V potential point. Am I safe to say that the anode is pulled to the 0V potential because the diode is forward biased? This is the concept that I am confused about; I would’ve thought that the anode is blocked by from the ground potential.
Hello again,

The anode is only theoretically at 0v because there is nothing connected to it and the cathode is at ground. In reality, the slightest current will make the anode slightly positive with respect to ground (the negative terminal of the battery) and see figure 3A. That means that when you go to actually measure it in real life you will see something other than a purely theoretical result (as in 3B).

If you connect almost any meter from +24v to the anode, there will be a small forward bias current and that will raise the anode voltage slightly. It can be almost anything between just above 0.0 volts to around 0.7 volts. In figure 3B we see the meter itself biases the diode slightly and so we will see a small voltage drop. In your case you seem to have measured 23.7 volts, but you can expect just about anything up to about 0.7 volts or so. It usually won't go any higher than that unless the current is fairly large, like 100ma or maybe 500ma or 1 amp or higher. Since 24/1000000=24ua you will just see a small voltage drop. [Edit: corrected typo 24/100000 to 24/1000000]

DiodeBiasing-1.jpg
 
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BobTPH

Joined Jun 5, 2013
8,661
In your figures 2 and 4, the ones where the diode is reverse biased, you are reading 0.05V.

A typical multimeter has a resistance if 1MΩ.

So, what you are seeing is 50nA of current flowing in the circuit. Or, another way of looking at it, the diode is acting like a 19MΩ resistor.

Edited to add: The max leakage current of a 1N4000 diode is 50uA at 50V, 1000 times higher than what you are measuring.
 
Last edited:

MrAl

Joined Jun 17, 2014
11,263
In your figures 2 and 4, the ones where the diode is reverse biased, you are reading 0.05V.

A typical multimeter has a resistance if 1MΩ.

So, what you are seeing is 50nA of current flowing in the circuit. Or, another way of looking at it, the diode is acting like a 19MΩ resistor.

Edited to add: The max leakage current of a 1N4000 diode is 50uA at 50V, 1000 times higher than what you are measuring.
Hi,

Read some of the other replies he is not using a 1N4000 series diode.
The leakage current being seen here is reasonable for his actual diode, but these specs are always just rough estimates anyway.
 
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