Need some ohms law help
metron9
Posts: 1,100
I am going to be hooking up an optical sensor. It uses 24 volts and has a contact rating of 150 milliamps
I want to drive a Macromatic time delay relay (TR-56128-03) Series B Delay Interval relay with it.
The manufacture says the current draw is about 100 milliamps but how do I verify that
When I measure the coil I set my digital multimeter to 20M I get 4.88 on the meter, (4,880,000 ohms?)
when I set my meter at 2M I get .983 (983,000,000) OHMS
What do I need to do to find out how much the coil draws? Figure out the voltage drop? or something?
I want to drive a Macromatic time delay relay (TR-56128-03) Series B Delay Interval relay with it.
The manufacture says the current draw is about 100 milliamps but how do I verify that
When I measure the coil I set my digital multimeter to 20M I get 4.88 on the meter, (4,880,000 ohms?)
when I set my meter at 2M I get .983 (983,000,000) OHMS
What do I need to do to find out how much the coil draws? Figure out the voltage drop? or something?
Comments
the "coil resistance" with a meter because the input may actually be a transistor, or some other type of circuit,
not a coil of wire.
I suggest that you simply apply 24VDC to the relay and measure the current draw. If you don't have a
current meter then put a 10ohm resistor in series with the relay and measure the voltage drop across it.
83ma should give you an 830mv reading.
Dave G
I am interested in how the math works using the 10ohm resistor to measure the voltage drop.
1. I am not sure what 2VA means on the spec sheet.
2. How do you get the 83ma from that number
2. How do you know to use a 10ohm resistor
3. When I measure the "voltage drop"
I put a resistor in series with the relay, There are three points to take a reading
+
/\/\/\/\/\/\
}{
|||:
24V Resistor Relay Ground
Point A Point B Point C
I measure the voltage drop between points B and C
and subtract it with the voltage I get at points A and C ? and that should be around 83ma?
(I have not tried it yet so I may answer some of this through experimenting but I want to understand it down to the electron level and what LAWS are applied here)
Yep, If I dont understand how you get the answer to a problem I won't be able to solve the next problem.
This is the kind of help that really gives me a boost in my understanding of circuits. I have two daughters in college, one at the UofM and one at Minnesota State. I never had the chance to go to college but the ability to access all the information on the net and get past sticking points with the good people that take time from their lives to help others makes me feel a bit like I am in college right now. Keep the information flowing and I will absorb as much as I can.
To calculate the current: Relay Volts x Relay Amps = 2VA. Or Relay Amps = 2VA/Relay Volts
Relay Amps = 2VA/24 volts = .083 amps = 83 milliamps
If your digital multimeter has a current measuring capability you can measure the current directly,
without using the 10 ohm resistor. Simply set your multimeter to measure DC amps and connect it
IN PLACE OF the 10 ohm resistor. This will allow the current to flow through your meter AND the relay.
The meter should read approx. .083 amps.
If you choose to use the 10 ohm resistor, you should measure the voltage "drop" between Points A and B.
The formula is I = E/R. Current = Measured voltage / 10 ohms.
Expected values are: I (amps) = .83 volts / 10 ohms = .083 amps = 83 milliamps
As you know, all measuring devices affect the circuit they are measuring, to varying degrees. The trick is to reduce
the effect to an acceptable level. I chose a 10 ohm resistor because it will only take away (drop) .83 v from the relay.
There will still be 23.17 V across the relay. You could use a 1 ohm resistor which would be even better, but 10 ohm
resistors are easier to find and produce less lead wire error.
Dave G
On to the next project. I plan to put up a little web site showing my progress on this project soon but I will just tell you where it's at now and ask one question.
I bought a stepper controller from stepper3.com and a motor.
Manual for "stepper3 plus 1" http://www.stepper3.com/manuals/stepper3Plus1.pdf
Motor spec: 5.1V 1.0A 5.0ohms Stepper 200S/R
I installed windows2000 on an old 750 mhz machine and put on Mach2 software.
Today I finished the wiring and plugged in the board and hooked up the parallel line to the computer.
The motor runs perfect with the Mach2 software. Waiting now for my "Driven linear unit" to finish the Y table for a machine I am building.
My question has to do with page 14 of the manual (linked above)
I am using an old PC power supply and box to hold the stepper controller
The power supply is rated at 5V 24A
When I have the system ON, I realize power is always being used by the motor wether its moving or not just to hold it in place.
It gets quite warm just sitting there and my concern is on page 14 having to do with current.
I don't have any idea wether the power resistors it talks about are necessary (let me say I don't think i need one) as I am using a 5 volt supply. It's example starts with "You decide you want 3.5 Amperes", I don't know why one would decide on 3.5 amperes and the text is not saying what the motor voltage rating is only that the motor in the example is rated at .8 ohms.
I assume if I am using 5 volts supply (with my motor at 5.1 ohms that the motor will only consume 1.0 amps maximum as 5.1 volts / 5.1 ohms is 1A. (my power supply indeed shows 5.1 volts)
Just rambeling on here....
Is it saying IF I were to use a 6 volt supply the current would be 6volts / 5 ohms or 1.2 amps and that would overheat the motor?
I would then have to put a resistor on the center tap of the motor to reduce that current to 1.0A Maximum?
I would then need a total resistance of 6 ohms with a 6 volt supply and would require a 1ohm resistor?
What would happen if I used a 12 volt supply, would a 7 ohm resistor work understanding that a lot of heat would be given off by the resistor and it would not be very efficient but would indeed run the motor without burning it up?
If I don't need a full 78 ounces of tork from the motor and I wanted to reduce the current on the motor so it runs cooler to say .769A using a 1.5 ohm resistor?
That's more than one question but I think I am set for a while now, as a matter of fact I think I need a break as the motor is turning, the software works, and I feel like my brain needs som R&R.
Thank's a bunch, you know what happens here is when I get an answer to a question, I know more, I apply that to something else and whoops I end up with more questions. It just kinda gets out of hand.