24VDC to 9VDC relay


  #1  
Old 05-28-04, 06:33 AM
ChrisHelvey
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24VDC to 9VDC relay

Hi all, I have a small project I am building that will turn on my sprinkler pump whenever a timer turns on a zone. (3 of them.) Essentially, I am taking 24VAC from the timer, rectifying it to 24VDC, filtering with a cap, and want to hook it up to a 9VDC relay (because I already have them.) The relay calls for 18mA nominal coil current. Someone at Radio Shack calculated that I would need a 150 ohm resistor between the + side of the 24VDC and the 9VDC relay. Can someone confirm for me that this is correct and will be OK over the long term?
It works OK right now. The relay gets warm but not hot, even after 30 minutes. (It will not be on more than 20 minutes at a time - while that zone is sprinkling.)
Can someone explain how it is OK to run a 9VDC device with 24VDC? Does it have to do with current draw overall?

Thanks
 
  #2  
Old 05-28-04, 07:10 AM
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If the relay pulls 18ma at 9VDC, and you are feeding it with exactly 24VDC, then I calculate that you need an 820 ohm resistor. I calculated by

24VDC-9VDC/18ma=833 ohms. The closest standard size is 820 ohms.


With a 150 ohm resistor, you would have approximately 100ma of current going though the relay. It might hold up. but I doubt it.

This all assumes that you have exactly 24VDC created by your "power supply" measure the exact voltage and then plug that value in for the 24VDC above to get a better picture of where your resistance needs to be.
 
  #3  
Old 05-28-04, 07:58 AM
J
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This may take some testing. Measure the voltage across the relay. If it is more than 9 volts then you need a bigger resistor. Scott's calc are correct. However the relay may draw a little more or less current so some adjustment may be required. The relay should not get hot or even warm.
 
  #4  
Old 05-28-04, 08:20 AM
ChrisHelvey
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Ah, I measured an actual 21.5V. So, using the formula Scott provided, I come up with 694ohms. (Or, the closest thing to it.) Thanks for the formula.
So, am I correct in saying that, although the resistor does not drop the VOLTAGE in the circuit, it does drop the current in enough of a way that the relay doesn't care about the voltage so much?
I read the same voltage regardless of the size of resistor unless it is a really large one like 100K, then it drops to about 15V. I don't understand it but I would like to. Any explanation is appreciated.
 
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Old 05-28-04, 08:51 AM
K
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Rather than just using resistors for voltage dividing calculations. I highly recommend using a shunt regulator (zener diode, resistor). It will be WAY more reliable and won't be nearly as influenced by temperature swings. You can buy diodes that have specific voltage ratings. I assume you could find one for 9 volts. here is a picture I found from googling "zener shunt regulator" Shunt regulator
 
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Old 05-28-04, 09:44 AM
K
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I was mistaken in my previous comment about the shunt regulator, if your relay needs a specific current, not necessarily 9Volts input. So ignore the shunt regulator, unless it does need 9V input for some reason.

Originally Posted by ChrisHelvey
So, am I correct in saying that, although the resistor does not drop the VOLTAGE in the circuit, it does drop the current in enough of a way that the relay doesn't care about the voltage so much?
I read the same voltage regardless of the size of resistor unless it is a really large one like 100K, then it drops to about 15V. I don't understand it but I would like to. Any explanation is appreciated.
Actually the resistor does drop the voltage to the relay. There is ALWAYS a voltage accross a resistor. You can think about a circuit as a circle. Your supply is 24VDC, and you can add a resistor in the circle. Say the voltage accross that resistor is 12 volts, now you have 12 volts at the relay. If you put in a resistor that drops the voltage 6 volts you have 18 volts at the relay. But you cant just buy "6 volt resistors", you have to calculate the voltage drop based on the current that is used. Thats where the formula that Scott gave you comes in: V=IR. similarly, I=V/R and R=V/I

So anyway, to solve your problem, you would want to figure out how to input 0.018 Amps into your relay. To do this you are going to need to figure out what the resistance of the relay will be: So unless you were given an effective resistance in your manual with the relays, you have to do some calculations. You can calculate the resistance of the relay by doing some measurements. You can use a known resistor, "R1". Place R1 before the relay and measure the resistance accross that resistor.

As just an example R1 = 1000 Ohms. Voltage Source = 24 VDC

Relay + R1: Voltage accross R1 is 20 volts,
V/R=I, 20volts/1000Ohms=.020 Amps
If the Voltage accross the resistor is 20 volts, then the relay has 4 volts (24-20=4).


Since the current is constant through the circuit, the relay's resistance is:
V/I=R or 4/0.020=200 Ohms.

Now you would decide how large you want the resister to be in order to be in order to get 0.018 Amps into the relay. V/I=R = 24 V/0.018 A= 1333 Ohms. Since the Resistance of the Relay is 200 Ohms, then R1 should be 1333-200 = 1133 Ohms or as close as you can get to it.
 

Last edited by kuhurdler; 05-28-04 at 09:58 AM.
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Old 05-28-04, 10:26 AM
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scott e's calculations are flawless, assuming the relay conducts 18ma at 9V, and that is a fairly good assumption. You need an 820 ohm resistor, but also pay attention to the wattage. I believe you will need a 1/2 watt or better.

BTW, I'm a bit confused by your second post. For all intents an purposes, a 100K resistor should drop all the voltage in the circuit, allowing less that .24ma.

Fun ain't it?
 
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Old 05-28-04, 11:04 AM
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Originally Posted by ChrisHelvey
Ah, I measured an actual 21.5V. So, using the formula Scott provided, I come up with 694ohms. (Or, the closest thing to it.) Thanks for the formula.
So, am I correct in saying that, although the resistor does not drop the VOLTAGE in the circuit, it does drop the current in enough of a way that the relay doesn't care about the voltage so much?
I read the same voltage regardless of the size of resistor unless it is a really large one like 100K, then it drops to about 15V. I don't understand it but I would like to. Any explanation is appreciated.

It depends on what voltage you are measuring. I assume that you are measuring the voltage produced by your "power supply". Actually the voltage, resistance, and current are all related by V=I*R. The resistor drops voltage in proportion to the current flowing through it.
 
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Old 05-28-04, 11:06 AM
ChrisHelvey
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Excellent responses. Thank you. Now, pardon my ignorance as I ask a silly question: Am I correct in saying that when this circuit is all put together and the "circle" is completed, I should be able to put the leads of my meter on the input of the relay (the coil input) (the other side of the resistor from the 24VDC power supply and read about 9VDC?
 
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Old 05-28-04, 11:13 AM
ChrisHelvey
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I see. I posted my question just as Scott answered it. That makes some sense to me - that the voltage drop is proportional to the current flowing through it. So, direct voltage readings from the meter are less dramatic than I first thought?
 
  #11  
Old 05-28-04, 11:20 AM
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Possibly you could connect three 9-volt relay-coils in series across 24 volts which would result in 8 volts across each coil.I'll guess that the relays will operate satifactorly on 8/9 = 90% of the "rated' voltage.

Good Luck & Enjoy the Experience!!!!!!!!!!!
 
  #12  
Old 05-28-04, 11:28 AM
ChrisHelvey
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That is an excellent idea! I'll try that.
 
  #13  
Old 05-28-04, 12:38 PM
ChrisHelvey
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Wait a minute. That won't work - all the relays must operate separately or all the valves will be open at the same time. Each valve has it's own 24VDC supply.
 
  #14  
Old 06-05-04, 07:56 AM
ChrisHelvey
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Hi all, I wanted to let everyone know that my little project works flawlessly. Thanks to everyone who had some input on this circuit. In the end, I did this all with parts I could get at radio shack (because it was convenient and the cost difference was not great enought to price shop
Each 24VAC valve controller wire goes to a small rectifier where it is converted to DC, it passes through 820 ohms of resistance where it connects to a small 9VCD relay. I have a 24VAC wall wart transformer that has one side connecting to the "common" side of all the relays along with one of the pump relay's wires and the other pump relay wire connecting to the other side of the 24VAC.
So, when any one zone valve is turned on from the timer, it's corresponding 9VDC relay completes the circuit from the 24VAC wall transformer and my pump relay gets that power...pump is on any time a zone valve is open. Yeah!
It all fits in a small little project box that mounts on the wall under the timer and looks very neat and professional.
I know it's just a simple thing, but it saved me 130.00 from having to buy a "pump capable" sprinkler timer, I learned a thing or two in the process, and I feel great about creating something that works!
Thanks again for all your input.
Chris
 
 

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