I have a circuit with an inductor, a diode and a neon light, a switch and 10V. I need to explain it.

Discussion in 'Homework Help' started by Antonella Puricelli, Jul 28, 2016.

1. Antonella Puricelli Thread Starter New Member

Apr 24, 2016
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Hello, good morning/afternoon/night/everythingelese,
I have the following circuit and some questions to answer about its functioning but I'm not exactly sure I'm right about what I thought.

What we had to do, is connect the circuit accordingly and see what happens.
We've seen that the neon light lit up when we opened the switch (but didn't when we closed it). But I don'n know is why this happens.

My theory (after reserching) is that when we close the switch, the inductor generates a tension that goes against the diode and that's why it doesn't light up. Then, when we open the switch, the direction of the tension goes the other way allowing the neon to light up.

However, then, we had to reverse the direction of the diode (to reverse-biased), we could observe that the neon worked the same way, only that its light was brighter. I have no clue as to why this happens. My guess is that it's related to the functioning of the neon light (the thing is that I haven't completely understood how it works).

I know this is a lot, but any kind of help or explaining could help, even providing some links for me to research would be great.
I want to clarify my native language is Spanish so I apologice in advanced for any mistakes and I'm new to the page so I hope I put this in the right place.

Thank you all.

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2. wayneh Expert

Sep 9, 2010
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With the switch closed, the neon bulb can never see a voltage greater than the battery voltage less the diode drop. That's not enough to light it.

After current has been flowing in the inductor, it contains the stored energy of the magnetic field. Opening the switch causes nearly immediate collapse of the field. Voltage across the inductor can reach very high levels, plenty to light the bulb.

Do you know the breakdown voltage of the diode?

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3. Antonella Puricelli Thread Starter New Member

Apr 24, 2016
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It's a 1N4007, the datasheet (Vishay) says its breakdown voltage is 1000V.

4. Antonella Puricelli Thread Starter New Member

Apr 24, 2016
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It's a 1N4007, the datasheet (Vishay) says its breakdown voltage is 1000V.

5. wayneh Expert

Sep 9, 2010
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I think the diode is in the circuit to demonstrate the polarity of the spike coming from the inductor. The spike can light the bulb with the diode in either orientation because the spike is over 1000V, but the light is brighter with the diode in the proper orientation. There's only ~0.7V drop in voltage in that direction, and a drop of about 1000V in the "wrong" direction.

6. Papabravo Expert

Feb 24, 2006
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What is a vuelta? Is it a unit of inductance in some other language?

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7. RichardO Well-Known Member

May 4, 2013
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L1 = 1200 turns

8. Antonella Puricelli Thread Starter New Member

Apr 24, 2016
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I get what you're saying. However, the following step in the project was to see what happened when we inverted the polarity of the diode. In this case, what happened was that the light was actually much brighter.

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9. wayneh Expert

Sep 9, 2010
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As predictable. Think about the polarity of the spike. Note the evidence in the photo as well. The bulb has two electrodes, and only one is lit.

I think this is called a relaxation oscillator?

Last edited: Jul 28, 2016
10. Antonella Puricelli Thread Starter New Member

Apr 24, 2016
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Right, okey, I think I got it.
Thank you so much

11. Papabravo Expert

Feb 24, 2006
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Thanks. Good to know.

12. Antonella Puricelli Thread Starter New Member

Apr 24, 2016
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I don't know about the name, but I think that's because we're using DC instead of AC; I've read something that when using DC only the negative terminal is lit up as opposed to AC in which both are lit up.

13. crutschow Expert

Mar 14, 2008
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But it's not oscillating.

14. crutschow Expert

Mar 14, 2008
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When the switch is closed, current starts to flow through the inductor which stores energy in its magnetic field equal to ½LI².
When the switch is opened this stored energy in the field tries to keep the current flowing in the same direction until the energy is dissipated in the external load (the inductor now acting as a current source).
The only path for this current is the neon bulb and diode so a negative voltage (note the current direction to see why it's negative) will build up until the current starts flowing through the bulb and diode.
The current only flows for a short time until all the inductive energy is dissipated in the bulb and diode voltage drop, so just a short flash of light is generated.
If the diode is reverse biased than much more energy is dissipated in the diode compared to the bulb.
If the diode is forward biased than much more energy is dissipated in the bulb.

15. wayneh Expert

Sep 9, 2010
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I wasn't sure. If the inductor were a capacitor, it would. So says the wiki about neon bulbs.

16. Bordodynov Active Member

May 20, 2015
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See

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17. Tonyr1084 Active Member

Sep 24, 2015
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Inductors and capacitors are similar in nature in that they store electrical energy then give it back. The difference with an inductor is that it can store 10 volts and give it back as a much higher voltage whereas a capacitor can never give more than it has stored.

The capacitor will charge to 10 volts. When the switch is opened the capacitor will supply its voltage (10 volts) until it discharges completely. Depending on the resistance across the capacitor the rate of decay will vary.

The inductor will generate a magnetic field. Its rise time will depend on how much current is supplied. It may take several milliseconds to charge (depending on the inductor). But when the switch is opened the magnetic field is at liberty to collapse completely in just a few microseconds. The changing magnetic field is what generates a voltage. The faster that changes the higher the voltage it generates. So when you open the switch the magnetic field collapses and generates a much higher voltage - one capable of lighting the Neon lamp.

Neon typically requires 70 volts to ignite. (unless things have changed since I messed with them) 10 volts into a coil (inductor) can easily light the neon when its current is interrupted. And as electrons flow from one electrode to the other one will glow. With AC, current is flowing one way then the other. So it appears as if the neon bulb is lit on both electrodes at the same time. We can't see events happening that fast, so our brain assumes both electrodes are glowing at the same time. By reversing the diode you change the direction of flow. One electrode will glow in one orientation, the other electrode will glow when you reverse the system. And the voltage drop across the diode is insignificant to the voltages your coil is generating.

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18. crutschow Expert

Mar 14, 2008
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In this circuit, reversing the diode does not change the direction of current flow.
That's determined by the inductor current direction which does not change.
If the diode is connected with anode to (-) source, the inductor current will flow in the forward direction through the diode when the switch is opened, with low voltage drop.
If the diode is connected with the cathode to (- ) source, the inductor current will flow through the diode in the reverse direction when the switch is opened, causing reverse avalanche breakdown of the diode at high voltage.

19. crutschow Expert

Mar 14, 2008
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Yes.
Replacing the inductor with a capacitor and adding a high value resistor in series with a high voltage source will create a relaxation oscillator.