My question is this -

Can a diode still work if it has electricity coming in from both ends? For example if I have a diode, and connect a DC generators output to one end of the diode, and do the same for the other end as well what will happen?

 

A diode conducts in one direction, acts as an open circuit in the
other direction (a little leakage). This is polarity sensitive, and V
dependent. Here is a "typical" diode V vs I curve of operation -

Regards, Dana.

 

It will only conduct when forward biased.!
Entender?
Max.

 

The graph shown in post #2 is misleading. You observe a kink in the curve at the origin.
The reason this is drawn like that is because the vertical scale for positive current (perhaps mA) is different from the scale for negative current (maybe microamp).

If the scales were the same for both forward and reverse current, the graph would look as follows:

In reality, the curve is continuous at the origin. If you were to zoom in at the origin, the graph would look as follows:

 

My question is this -

Can a diode still work if it has electricity coming in from both ends? For example if I have a diode, and connect a DC generators output to one end of the diode, and do the same for the other end as well what will happen?

If you are putting the output of the SAME generator at both ends, then neither the diode nor any other device will conduct any current because the voltage across the device is zero.

 

Good points about the graph. But then the new graph, visually, implies Irev = 0
which of course it is not. Maybe best if graph done log log......

I guess we have to choose which misleading graph we want to perpetuate .


Regards, Dana.

 

Thank you for all the responses, but to clarify - what if generator "A", producing less voltage is connected to the Anode and generator "B" producing higher voltage is connected to the cathode?

In case that was confusing here is the same question worded slightly different:
What if the generator producing less voltage is connected to the Anode and the generator producing higher voltage connected to the cathode?

 

More voltage or less voltage doesn't say anything about polarity and that is what is important for a diode.
If the cathode of the diode is positive of the anode then it is not conducting.

 

Thank you for all the responses, but to clarify - what if generator "A", producing less voltage is connected to the Anode and generator "B" producing higher voltage is connected to the cathode?

In case that was confusing here is the same question worded slightly different:
What if the generator producing less voltage is connected to the Anode and the generator producing higher voltage connected to the cathode?

How are the generators connected?

If I take a 120 V battery and a 12 V battery and connect the two positive terminals together and nothing else, I will have two batteries sitting there looking stupid with a voltage of 108 V appearing between their negative terminals.

Try sketching a circuit what exactly what you have in mind.

 

How are the generators connected?

If I take a 120 V battery and a 12 V battery and connect the two positive terminals together and nothing else, I will have two batteries sitting there looking stupid with a voltage of 108 V appearing between their negative terminals.

Try sketching a circuit what exactly what you have in mind.

Let me know if that helps! I did two slightly different versions of what I'm thinking of, but I'm not sure if the difference matters. The right side battery flips polarities between the top and bottom drawings.

 

Think of a diode as a one way valve in a water line. Then your voltage sources as water tanks. The water (current) will flow from the high pressure to the low pressure, BUT only if the one way valve if the correct way around.

 

This discussion continues here: https://forum.allaboutcircuits.com/...icity-to-where-i-want-it.156537/#post-1352413

 

Let me know if that helps! I did two slightly different versions of what I'm thinking of, but I'm not sure if the difference matters. The right side battery flips polarities between the top and bottom drawings.

What's the circle with the X in it?

 

What's the circle with the X in it?

That is a symbol for lamp.

 

That is a symbol for lamp.

Yep. It's been so long since I've seen that symbol that I didn't recognize it.

 

Let me know if that helps! I did two slightly different versions of what I'm thinking of, but I'm not sure if the difference matters. The right side battery flips polarities between the top and bottom drawings.

In both circuits the diode is forward biased. In the top circuit you are attempting to put 11 V across it and in the bottom you are attempting to put 13 V across it. In neither case it is likely to do what you want it to do, but we can't be sure because you haven't told us what it is you want to do.

 

In both circuits the diode is forward biased. In the top circuit you are attempting to put 11 V across it and in the bottom you are attempting to put 13 V across it. In neither case it is likely to do what you want it to do, but we can't be sure because you haven't told us what it is you want to do.

I want to power the light bulb with 13 Volts, with the batteries in that specific arrangement. I put the Diode that way so that the power from the 12V battery didn't go into the 1V battery.

 

I want to power the light bulb with 13 Volts, with the batteries in that specific arrangement. I put the Diode that way so that the power from the 12V battery didn't go into the 1V battery.

It's getting really hard to keep track of all your circuits, what with all the threads and in some it's 12 V and 2 v, in another it's 24 V and 2 V, and now here it's 12 V and 1 V.

If you want to apply 13 V across the lamp from a 12 V and a 1 V battery (and, again, what is this battery that is only outputting 1 V?), then the two batteries have to be in series. The same current flowing in one battery WILL flow through the other. There is no need for a diode at all. In fact, it will only eat up a good portion of that additional 1 V you are trying to get.

Have you actually powered the light with 13 V instead of 12 V and found it to be sufficiently brighter to be worth the hassle?

Is there a reason why you can't just use a bulb that is rated for 12 V with something like 10% to 20% greater wattage?

 

It's getting really hard to keep track of all your circuits, what with all the threads and in some it's 12 V and 2 v, in another it's 24 V and 2 V, and now here it's 12 V and 1 V.

If you want to apply 13 V across the lamp from a 12 V and a 1 V battery (and, again, what is this battery that is only outputting 1 V?), then the two batteries have to be in series. The same current flowing in one battery WILL flow through the other. There is no need for a diode at all. In fact, it will only eat up a good portion of that additional 1 V you are trying to get.

Have you actually powered the light with 13 V instead of 12 V and found it to be sufficiently brighter to be worth the hassle?

Is there a reason why you can't just use a bulb that is rated for 12 V with something like 10% to 20% greater wattage?

Thank you for the help, and guidance as far as proper etiquette goes on this forum! I will ensure to apply better more thought through questions from now on.

 

Thank you for the help, and guidance as far as proper etiquette goes on this forum! I will ensure to apply better more thought through questions from now on.

Sounds good -- live and learn.

But what I'm really interested in is the answer to the last two questions:

Q1: Have you actually powered the light with 13 V instead of 12 V and found it to be sufficiently brighter to be worth the hassle?

Q2: Is there a reason why you can't just use a bulb that is rated for 12 V with something like 10% to 20% greater wattage?