Question about open circuit/current/diode/capacitor

Thread Starter

rfpd

Joined Jul 6, 2016
101
Hey there, just to make sure I have all my concepts right. Voltage is the potential energy between two points, or the 'work needed to push an eletron', and current is the flow of eletrons, bigger the current, bigger the number of eletrons flowing. Current needs to flow from a higher charge to a lower charge, in some way to balance things, if there's not a second charge there will be no current, therefore eletrons will not move. So my question is, imagining we have a diode connect to an ac source, how are the eletrons 'moving past' the diode (since the eletrons are not moving) ? In a situation where there's not an open circuit, and with an increasing positive charge, the depletion region will shrink, due to the accumulation of positive charges. So it would be necessary 0.7 volts to shrink the depletion zone to make the diode a short circuit, coming x from one side and leaving x-0.7 from the other. I don't get how an open circuit bypasses this, take the capacitor as an example, it blocks all DC current even in an open circuit, what's the difference.

BTW: What does it mean, in AC current, for it to have many directions and DC just one?

When I say 'higher charge to a lower charge', I don't know if my terms are correct, I meant like from DC power to ground.


These question have been bugging my head all day, if you could shed some light, I would appreciate it.
 

recklessrog

Joined May 23, 2013
985
Hi, if you take a look at the heading "education" at the top of the page, there are tutorials that will explain in clear terms all you need to know :)
 

Thread Starter

rfpd

Joined Jul 6, 2016
101
Hi, if you take a look at the heading "education" at the top of the page, there are tutorials that will explain in clear terms all you need to know :)
I've seen them all, just wanting to know if I got my facts straight. I've searched in them, and it only says that an open circuit means that there is no current, therefore dc power voltage is equal in all the circuit. I just don't get how it 'goes through' the diode, or bypasses it...

In a circuit like this:


AC source----diode---(measure here)

I mean, no current, the eletrons are not moving, right?
 

MaxHeadRoom

Joined Jul 18, 2013
23,606
They will if your meter draws any kind of current from the circuit.
There has to be sufficient current to forward bias the rectifier.
This is why with modern meters you cannot read a diode with the normal resistance meter, there is insufficient current to forward bias the diode/rectifier.
Max,
 

Thread Starter

rfpd

Joined Jul 6, 2016
101
They will if your meter draws any kind of current from the circuit.
There has to be sufficient current to forward bias the rectifier.
This is why with modern meters you cannot read a diode with the normal resistance meter, there is insufficient current to forward bias the diode/rectifier.
Max,
But if we use a meter to measure the voltage after the diode, we get the same voltage as in the power source. How is it 'bypassing' the diode (all of this still considering the open circuit scenario), and why doesn't that happen after an open circuit capacitor?

Thanks for the answers so far.

EDIT: By theory I get it, since no voltage is dropping, but I don't get how it is pratical, since the eletrons are not moving.
 
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MaxHeadRoom

Joined Jul 18, 2013
23,606
It is not 'bypassing' the diode, once you introduce any device/load on the diode you are forward biasing it and measuring the potential after the diode.
What is meant by open circuit capacitor?
In the absence of load, a capacitor once charged will result in zero current flow.
Max.
 

Thread Starter

rfpd

Joined Jul 6, 2016
101
It is not 'bypassing' the diode, once you introduce any device/load on the diode you are forward biasing it and measuring the potential after the diode.
What is meant by open circuit capacitor?
In the absence of load, a capacitor once charged will result in zero current flow.
Max.
What I mean by bypassing it is 'ignoring the effect of it'. Imagine an AC signal that oscilates between -2 and 2, if the the circuit is open, the signal would be unchanged after the diode, that's what I meant. If I apply a DC signal of 0.1V (for example), it will remain untouched after the diode, but if I do it to a capacitor, it will result in zero current flow like you said. Again, I'm assuming a scenario with an open circuit.

Signal --- Capacitor -- (measure here)
 

Tonyr1084

Joined Sep 24, 2015
6,249
HMMMmmmm.

Diodes and open circuits - I think that may be where you're missing something important. First, an open circuit is a circuit that is not complete. For a moment consider a battery, a switch and a light bulb. When the switch is open (off) no current flows. It's an open circuit. No current would flow in any direction. Now, consider the same circuit but with an AC source. With the switch off - no current flows. In any direction. Close the switch and current flows in both directions and the light comes on. Now lets throw a diode in there just for fun. With the switch closed and using an AC power supply, current flows through the diode in one direction but when the current reverses it blocks the current. Provided you remain within the specs of the diode you won't see any current flow against the diode. However, when the current is flowing WITH the diode then you will see a (typical) 0.7 volt drop. Meaning if the supply and bulb are 12 volts, with the diode in place you will see 11.3 volts across the light bulb.

Capacitors are different beasts. True, no current flows through a capacitor. However, current will flow into and out of a capacitor. But never "Through" it. (unless shorted internally). Imagine that circuit again, using an AC supply and a diode; the capacitor is across the light bulb. When you turn the switch on immediately the diode is blocking the reverse current, giving only the forward current to the light and cap. With the cap starting off with zero charge its internal resistance is very very low. Current will run at its highest when the cap is completely discharged. But as the cap starts charging then the resistance starts going up. As the capacitor reaches full charge (full from what the supply is giving it) then the resistance goes up to near infinity, meaning the capacitor is no longer accepting charge current.

Funny thing with AC: It's measured at its RMS value (Root Means Square). So a 12 volt transformer with an RMS of 12 volts, when pushing through a diode and charging a capacitor (ignore the light bulb for a second) the voltage across the capacitor will charge to 1.414 times the value of the transformer. AH, BUT WAIT A SECOND: There's a diode in between. So 12 x 1.414 = 16.97 volts. Subtract that forward voltage from the diode and you get 16.97 - 0.7 = 16.27 volts of total charge across the capacitor (ignoring any internal volt leakage).

Remember, a diode isn't an "Open" circuit during half of the sine wave, it's an extremely high resistance to the flow of current in the opposite direction. And a capacitor never conducts current through it. It merely takes a charge or gives it back. Just like in the last circuit with the AC supply, the switch, diode, capacitor and light bulb, when you turn the switch off the light will remain lit until the charge in the capacitor is bled off.
 

Thread Starter

rfpd

Joined Jul 6, 2016
101
HMMMmmmm.

Diodes and open circuits - I think that may be where you're missing something important. First, an open circuit is a circuit that is not complete. For a moment consider a battery, a switch and a light bulb. When the switch is open (off) no current flows. It's an open circuit. No current would flow in any direction. Now, consider the same circuit but with an AC source. With the switch off - no current flows. In any direction. Close the switch and current flows in both directions and the light comes on. Now lets throw a diode in there just for fun. With the switch closed and using an AC power supply, current flows through the diode in one direction but when the current reverses it blocks the current. Provided you remain within the specs of the diode you won't see any current flow against the diode. However, when the current is flowing WITH the diode then you will see a (typical) 0.7 volt drop. Meaning if the supply and bulb are 12 volts, with the diode in place you will see 11.3 volts across the light bulb.

Capacitors are different beasts. True, no current flows through a capacitor. However, current will flow into and out of a capacitor. But never "Through" it. (unless shorted internally). Imagine that circuit again, using an AC supply and a diode; the capacitor is across the light bulb. When you turn the switch on immediately the diode is blocking the reverse current, giving only the forward current to the light and cap. With the cap starting off with zero charge its internal resistance is very very low. Current will run at its highest when the cap is completely discharged. But as the cap starts charging then the resistance starts going up. As the capacitor reaches full charge (full from what the supply is giving it) then the resistance goes up to near infinity, meaning the capacitor is no longer accepting charge current.

Funny thing with AC: It's measured at its RMS value (Root Means Square). So a 12 volt transformer with an RMS of 12 volts, when pushing through a diode and charging a capacitor (ignore the light bulb for a second) the voltage across the capacitor will charge to 1.414 times the value of the transformer. AH, BUT WAIT A SECOND: There's a diode in between. So 12 x 1.414 = 16.97 volts. Subtract that forward voltage from the diode and you get 16.97 - 0.7 = 16.27 volts of total charge across the capacitor (ignoring any internal volt leakage).

Remember, a diode isn't an "Open" circuit during half of the sine wave, it's an extremely high resistance to the flow of current in the opposite direction. And a capacitor never conducts current through it. It merely takes a charge or gives it back. Just like in the last circuit with the AC supply, the switch, diode, capacitor and light bulb, when you turn the switch off the light will remain lit until the charge in the capacitor is bled off.
Thank you for this amazing answer. Like you said, no current flows in an open circuit, so imaging an open circuit with a DC source going through a resistor (or a diode), and you get the voltage after that point, imagine like a transistor switch or something like this:

DC source --- Resistor
And you take the voltage after the resistor.

My question is, if there is no flow, how is there charge after the resistor? I know that through the equation R = V / I, it makes sense, I just want to understand it, sorry if it's kind of a 'stupid' question.
 

Tonyr1084

Joined Sep 24, 2015
6,249
Two things are different: Potential and current flow. The potential between the battery negative terminal and the open end of the resistor that is connected to the battery. There is "Potential" between the resistor and the battery negative (assuming the resistor is connected to the positive end of the battery. When you connect your test meter the circuit is no longer open You are reading the potential between the resistor and the battery neg terminal. For all intents and purposes, there is no current flowing (though there actually IS some current flowing through your meter. Otherwise your meter would read nothing.

Imagine a balloon with air in it. That's the same as voltage; a potential of energy. Imagine now a hose connected to the balloon, but the end of the hose is pinched off so that no air escapes. The potential energy is the same at the balloon and at the end of the hose. NOW imagine a valve (potentially restricting air flow) half closed. Since no air is escaping, the potential at the end of the hose is still the same.

Now imagine two balloons both blown up to the same amount of pressure. You release one end and the air rushes out. The other balloon is connected to the hose and valve. You release the plugged end of the hose and the air comes out but at a slower rate because the valve is resisting some of the air flow (current).

So when you see a "Potential" at the other end of the resistor you have to remember that the resistor "Resists" current. Not voltage. If you were to connect the other end of the resistor back to the negative end of the battery then you would have current flowing through the resistor. If you measured across the battery you'd read the full potential of the battery. Same would be true if you measured the voltage across the resistor, because one end is connected to the positive terminal and the other end to the negative end.

NOW: Imagine you have TWO resistors in series, and not connected back to the battery. You'll STILL have the same "Potential", but no current flowing. Now connect the second resistor back to the battery and now you have current flowing. Measure the potential across one of the resistors and it will NOT be equal to the full potential of the battery because of the second resistor. Assuming both resistors are the same exact value, the potential across just one of the resistors will be exactly half of the supply voltage. But if you disconnect the negative end of the battery then you will again have the full potential across all the resistors. Weird, I know. But true.
 

Tonyr1084

Joined Sep 24, 2015
6,249
I was being curious, wondering what the voltage reading would be through a diode. I have several Amazon LED's. By "Amazon LED" I mean I have no idea what their specs are. A few months ago I decided to measure their forward voltage at 10 mA. Some cases they were 3.2 Vf (forward Voltage). But with your question I began wondering, so I put an LED on my power supply and dialed it up to 50 VDC and checked the voltage after the diode. I saw between 2.5 and 1.7 Vf using a 5 mm white, green, red and yellow. I was surprised to see the yellow come in at 1.7 volts (same as the red). I had much higher readings when there was a current through them.

So in answer to my puzzlement was would there be a voltage drop at the cathode end of a diode. It would appear so. Given that my test did not incorporate a complete circuit. I probably should repeat the test with a set voltage and set resistance and then measure the voltage across each type of LED and make notes.

[edit] I used 50 volts to see how the numbers compared to using 5 volts. The results were the same. The potential dropped 2.5 volts (white), 2.2 (green), 1.7 (red and yellow).
 

Tonyr1084

Joined Sep 24, 2015
6,249
Using 5 V @ 10µA I got
2.8 Vf across the white and green LED's,
2.2 Vf across the Yellow and
1.9 Vf across the red LED.

The other way - I got
2.5 Vf at the end of the white LED
2.2 Vf at the end of the green
1.7 Vf at the ends of red and yellow.

Don't ask me why, I don't know why. But like Crutchow said, my meter completed the circuit, so it wasn't a true open circuit. It probably has something to do with the resistance of the meter completing the circuit.
 
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Tonyr1084

Joined Sep 24, 2015
6,249
I know why I'm getting different readings: My "Amazon" LED's each (of the same color) have different Vf's. Just tested four yellow LED's. Vf read 1.98, 2.07, 2.08 & 1.99 Vf. So I'm going to chart 10 LED's of each color on Excel and come up with a random average and mark the containers I keep them in accordingly. Buggers when you measure something then come back to it and can't repeat the same results. Not unless you use the same exact LED every time. Hence, the need for extensive testing. And maybe 10 is not a large enough sample, but each container holds close to 100 LED's.
 

Thread Starter

rfpd

Joined Jul 6, 2016
101
I know why I'm getting different readings: My "Amazon" LED's each (of the same color) have different Vf's. Just tested four yellow LED's. Vf read 1.98, 2.07, 2.08 & 1.99 Vf. So I'm going to chart 10 LED's of each color on Excel and come up with a random average and mark the containers I keep them in accordingly. Buggers when you measure something then come back to it and can't repeat the same results. Not unless you use the same exact LED every time. Hence, the need for extensive testing. And maybe 10 is not a large enough sample, but each container holds close to 100 LED's.
http://oksolar.com/led/led_color_chart.htm

https://forum.allaboutcircuits.com/threads/led-forward-voltage-drop.22329/
 

ebeowulf17

Joined Aug 12, 2014
3,282
My question is, if there is no flow, how is there charge after the resistor? I know that through the equation R = V / I, it makes sense, I just want to understand it, sorry if it's kind of a 'stupid' question.
I would love to hear some more expert response to this - I'm absolutely not an expert, but here are my thoughts.

I believe that charge exists at equal levels on both ends of that hypothetical open circuit resistor. In other words, if it's a 50VDC source relative to ground that's charging one side of the resistor, then both ends of that resistor are at 50VDC relative to ground. There wouldn't be any voltage drop across the resistor while there's no current flowing through it, so the voltage should be the same on both ends.

Perhaps I was misunderstanding the phrasing, but it seems like it was implied above somewhere that the voltage wouldn't appear on the open side of the resistor until there was a complete circuit allowing current to flow. I'm pretty sure this is untrue, and that the voltage is there even before you complete the circuit with a meter.

Think about any spark generating scenario - lightning, static electricity, spark plugs in your car engine. If the voltage didn't climb to extremely high values, the spark would never form. Clearly, even before there's a complete circuit, voltage has reached the extreme end of wherever the spark will jump from. So it seems voltage can "move" without a complete circuit.

My attempt at understanding how this works is that there is a very small amount of current that flows whenever the source voltage changes, due to stray capacitance. The wires, resistors, spark plug bodies, and every other part of the circuit all have there own small amount of capacitance. When you change the source voltage, tiny amounts of current flow to charge those minute capacitances up until they match the source voltage.

I wouldn't be too surprised to hear that I'm way off on this. If I am, I'd appreciate any attempt to help me understand where I've gone wrong.
 
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