Understanding this oscillating relay with a capacitor.

[Spork Schivago]

Hello,

I'm trying to learn how to make circuits and I'm using this book, Make: Electronics. I'm having trouble understanding this circuit I built. I did it using a breadboard, however, the book firsts tells you how to do it using alligator clips and wires. I've attached two files, both PDFs. One is a drawing of how it's all hooked up, the other is the schematic. I believe I have a good understanding of how the relay works.

We have two LEDs (I used a red and a green one) and I have a relay and a momentary switch. When I push the momentary switch, the relay clicks really really fast and the green LED lights up. When I add the capacitor between the coil pins, it slows everything down and the green LED will light up when the RED is off, then turn off and the red LED will light up when the green is off and then turn off. It just repeats. Essentially, when one is on, the other is off and they just blink.

I just don't understand why when I add a 1000 uF capacitor in between the coils it slows the coil down. I know capacitors are like batteries and they block DC voltage from going through the circuit. That's about it though. I also don't understand why the electricity isn't still going through the relay with the capacitor there. The cap is just between the two pins on the relay. However, the wires are still connected to those two pins. Could someone please help me understand this a bit better? Thank you.

[stj]

Re: Understanding this oscillating relay with a capacitor.

jokers.

i'm too drunk to think about the capacitor right now.
but i will tell you 2 things.

1: the old trick of using a relay to disconect itself creating a buzzer is a very old way of generating a shitload of RF to jam tv's and radio's!!! - dont do it.

2: the inductive reverse spikes from the coil will eventually destroy the capacitor.
a relay coil should always have a reverse-connected diode in parallel to supress this current.

[paulstef]

Re: Understanding this oscillating relay with a capacitor.

1. Press the button, left LED is lit, voltage on coil rises (slope is dictated by RC time constant of the capacitor and the 12V internal resistance)
2. Relay turns on when ON voltage is reached, contacts turn over, left LED OFF, right LED ON
3. 12V input voltage is now disconnected from relay
4. energy contained in capacitor discharges through coil and maintains it energized for some time
5. capacitor voltages drops below holding voltage for relay, contacts turn over again

cycle repeats.

[Spork Schivago]

Re: Understanding this oscillating relay with a capacitor.

Thank you Stj. The book does say that if you hold the button down (before the capacitor is added) and it buzzes a lot, to only hold it down for a very short period of time because it will ruin the capacitor.

So when it's setup in oscillating mode (the buzzer), it generates radio frequencies and will jam a radio / tv? Now that everything is digital on the TV signals, will it still jam them? I don't plan on using it to jam anything. I'm just trying to learn by building the circuits they have me build. Just curious.

When you say the inductive reverse spikes from the coil will eventually destroy the capacitor, does that mean when I have a circuit with current flowing through it and the circuit is broken, the inductor will produce a high frequency voltage spike?

I don't think I know enough about circuits yet to hookup a reverse-connected diode in parallel to suppress this current. I haven't learned about diodes in this book yet. I do know what a diode is though. We haven't really gotten into parallel circuits yet either. I read about them on other websites though. Would the diode just go in the same place where the capacitor is? Thanks.

[paulstef]

Re: Understanding this oscillating relay with a capacitor.

When the coil is de-energized the collapsing magnetic field will/would create a reversed voltage on the coil terminals. The capacitor is polarized, it is not supposed to support inverse voltage, the coil will try to drive negative current through it, not good.
A simple diode with at least the same current carrying capability of the the relay coil from minus to plus of the coil will solve this problem.

Look here, same principle: http://en.wikipedia.org/wiki/Flyback_diode

[Spork Schivago]

Re: Understanding this oscillating relay with a capacitor.

1. Press the button, left LED is lit, voltage on coil rises (slope is dictated by RC time constant of the capacitor and the 12V internal resistance)
2. Relay turns on when ON voltage is reached, contacts turn over, left LED OFF, right LED ON
3. 12V input voltage is now disconnected from relay
4. energy contained in capacitor discharges through coil and maintains it energized for some time
5. capacitor voltages drops below holding voltage for relay, contacts turn over again

cycle repeats.

Okay! I finally think I got it! I ripped apart the other relay I had with the kit like the book suggested just to get a better understanding of how it all works. A few more questions if you don't mind though. RC time constant, that's just the amount of time it takes the capacitor to charge up to roughly 63.2 percent?

With capacitors, what determines how much time it takes them to fully charge? Am I right in thinking the more current, the quicker they charge? Also, once they're fully charged, they'll only discharge if I stop supplying current / voltage to them, is that correct?

How do I determine how quickly a capacitor will discharge? Do they all discharge at the same rate? I'm guessing the larger the farad, the longer it will take to discharge because there's more energy inside them.

When the capacitor is supplying voltage to the circuit, will it be the same voltage that I charged it with?

Thanks for helping and explaining everything to me. The book I'm following is okay. I wish they'd go a bit more in detail about how it all works though.

[budm]

Re: Understanding this oscillating relay with a capacitor.

Look up R-C time constant.

[stj]

Re: Understanding this oscillating relay with a capacitor.

digital tv/radio AND PHONES wifi etc are more easy to jam than analog signals.

[Spork Schivago]

Re: Understanding this oscillating relay with a capacitor.

Look up R-C time constant.

Okay, I'm reading up on it again, this time from a source other than wikipedia. From what I've gathered, time constant is the amount of time that it takes for a capacitor to charge or discharge. A capacitor will only charge up to the voltage that's supplied to it. So if I have a 12vdc source, I can use a 25v capacitor and that capacitor will only charge up to 12vdc not 25vdc.

Time constant is just the amount of time it takes the capacitor to charge up to 63.2% or to discharge to 36.8%. It's abbreviated as τ. If you addf a resistor to the capacitor in series, you form an R-C circuit and the capacitor will charge up more slowly through the resistor. The time to fully charge the capacitor is equivalent to about 5 time constants (5τ) and is called the transient response time.

The website I'm looking at, http://www.electronics-tutorials.ws/rc/rc_1.html , shows
τ = R x C, in seconds, for transient response time but then says it can be looked at as 5 * RC, where R is the value of the resistor, in ohms, and C is the capacitance in farads.

I have a 663 ohm resistor and a 1010 uF capacitor. Does that mean for it to be charged 63.2% of the supply voltage, it would take 663 * 0.00101 seconds? 0.66963 seconds? And for it to be fully charged, it should take 0.66963 seconds * 5 or 3.34815 seconds? I'm pretty sure I still don't fully understand this, even after reading up on R-C time constant. The red (left) LED blinks real quick. I would say roughly every half second, maybe less. Definitely not 3 seconds though.

[budm]

Re: Understanding this oscillating relay with a capacitor.

"Does that mean for it to be charged 63.2% of the supply voltage, it would take 663 * 0.00101 seconds? 0.66963 seconds?" correct, that is 1 time constant. The cap may not have a chance to charge to the 12V since the relay pull-in voltage is about 75% of the coil Voltage rating, and the release coil Voltage is about 25% of the coil Voltage rating.
The internal of the power supply is the equivalent to the series resistance for charging the capacitor, but you do not know what the impedance of the power supply is. the capacitor is then discharged through the relay coil when the switch goes open, that will be the time constant of the discharging of the cap plus the back EMF of the relay coil also comes into affect.

[Spork Schivago]

Re: Understanding this oscillating relay with a capacitor.

"Does that mean for it to be charged 63.2% of the supply voltage, it would take 663 * 0.00101 seconds? 0.66963 seconds?" correct, that is 1 time constant. The cap may not have a chance to charge to the 12V since the relay pull-in voltage is about 75% of the coil Voltage rating, and the release coil Voltage is about 25% of the coil Voltage rating.
The internal of the power supply is the equivalent to the series resistance for charging the capacitor, but you do not know what the impedance of the power supply is. the capacitor is then discharged through the relay coil when the switch goes open, that will be the time constant of the discharging of the cap plus the back EMF of the relay coil also comes into affect.

That's great! I was reading up on the RC Time Constant and thought I was getting a good understanding until I calculated the numbers and thought I was way off. I didn't take into account the relay switching on and off. When they say τ = RC, for R, they're talking the value of the resistor, not the resistance of the whole circuit, right? And just so I'm 100% sure on this, if I didn't have the relay there, just the cap and resistor, it would roughly 3.34815 seconds for the capacitor to fully charge to the supply load, right? And then another 3.34815 seconds to fully discharge?

[budm]

Re: Understanding this oscillating relay with a capacitor.

The circuit you show has power supply, switch, and cap in parallel with the relay coil, there is really no resistor in series with the cap except the internal resistance of the power supply, the switch contact resistance, the wiring resistance so the resistance is very low.
if you want to experiment, why don't you use 10V power supply, 1K resistor, and the 1000uF then calculate the Tc. and use your meter to monitor the DCV across the cap to see how long the time and the the Voltage are in 1 Tc, you can measure the capacitance so you will more accurate reading.

[Spork Schivago]

Re: Understanding this oscillating relay with a capacitor.

Okay, that's a great idea. So remove the relay, just have a 10V power supply and hook a 1k resistor in series to a 1,000uF cap? Gonna sit down to watch a movie with the wife. I'll start the experiments either after the movie or tomorrow when I wake up.

[LLLlllou]

Re: Understanding this oscillating relay with a capacitor.

1: the old trick of using a relay to disconect itself creating a buzzer is a very old way of generating a shitload of RF to jam tv's and radio's!!! - dont do it.

It was also used to run step up transformers in early automotive radios that used tubes.

[goontron]

Re: Understanding this oscillating relay with a capacitor.

It was also used to run step up transformers in early automotive radios that used tubes.

+1
http://en.wikipedia.org/wiki/Vibrator_%28electronic%29
https://www.youtube.com/watch?v=Fp6PkRTmb8U
Learn your old Technology there mates (not you LLLlllou)

[rievax_60]

Re: Understanding this oscillating relay with a capacitor.

There will be no high voltage back EMF spike or reverse voltage to worry about while there is a large capacitor across the coil.
Relay contacts don't take kindly to being made to directly charge or discharge capacitors. The very large inrush current can quickly wear out the contacts or cause the contacts to weld.
If you have trouble, just put a small value resistor in series, one to ten ohms will do.