Without knowing much more about the problem (the voltage of the motor drive? AC or DC? Is switching time important? Hostile environment?) it isn't possible to offer an exact solution.
However, here are some guidelines:
0.5 mA isn't very much. Very often switching circuits are set up so that the switching element sinks current rather than sourcing it; I suspect you may be looking at the wrong spec on your data sheet. Just reverse the inputs and you will reverse the sense of the output so you will maintain the same switching logic. -10 mA would be a pretty reasonable current sinking capability.
If not, you will probably need to add a buffer amplifier to boost up the current capability a bit.
Look into solid state relays; check the Digikey on-line catalog (Omron has a large line and there are many others). They typically require 2~5 mA to operate. For high voltage or high power equipment, if safety is any concern at all, solid state relays should always be in series with a mechanical relay or lockout switch because they can fail closed.
For a motor that large you will certainly need to pay attention to a snubber circuit, or else you will likely blow the relay the first time you try to turn it off.
For a DC motor, a diode connected across the motor so that it is reverse-biased when the motor runs is the best snubber. Be sure the reverse voltage rating is higher than the power supply voltage by a healthy margin and be sure the surge rating of the diode can handle the motor current.
For an AC motor of that current, you should probably use a zero-switching type relay. If the voltage is low you might get away with back-to-back Zener diodes for the snubber. It is pretty high current for a resistive snubber.
I know the Omron website has some application notes on snubbers for solid state relay circuits.
2007-01-04 16:46:25
·
answer #1
·
answered by AnswerMan 4
·
0⤊
0⤋
As some have answered before, a high power solid state or mechanical relay will work. You will probably need to place a line driver or open collector TTL gate between the comparator output and relay in order to get your relay drive current up to around 7-8 mA.. The type of relay will vary depending on whether your motor is single or three phase (assuming it is AC). Use a single relay that will handle the current load required, with some margin. I would not recommend using two smaller relays in parallel. If one should fail, this could be a fire hazard. If you are messing with 40A of current, I recommend keeping your experimentation to the low power stuff and feed your control line into a Variable Frequency Drive(VFD).These devices gracefully power up and brake motors, and gets you out of the loop of trying to do design on the dangerous high power business end of the circuit. VFDs are available from companies like Hitachi and GE, and will prolong the life of the motor. But for a 40A application, they won't be cheap.
2007-01-04 17:30:59
·
answer #2
·
answered by Tech_Toot 2
·
0⤊
0⤋
Sounds like an interesting project. I've looked through my sources and can find 40A relays. But none that will drive with 500uA. Email me with your comparator output voltage and your supply voltage. I will design for you a simple buffer stage to boost the comparator output to drive relays like these.
They can switch AC or DC loads.
AZ also makes solid state relays for AC only.
2007-01-04 17:36:22
·
answer #3
·
answered by charley128 5
·
0⤊
0⤋
You can use either a SCR (silicon controlled rectifier) or a relay.
For very small control currents or voltages, you can use a prestage using a darlington transistor which triggers the SCR.
Have used that circuit design to control an air horn in a car.
If using a relay and you can't find any that have enough current carrying capacity, you can use two relays in parallel.
2007-01-04 16:30:47
·
answer #4
·
answered by Anonymous
·
0⤊
0⤋
If the motor is unidirectional DC, a mosfet transistor can handle it. If it's an AC motor, easiest solution would be a SSR (solid-state relay), tho you might need to amplify that meager 1/2mil current to drive one. Are you sure your comparator is that limited? Most I'm familiar with can drive at least 10 times that amount. Google!
Feel free to email if further info needed
2007-01-04 16:37:05
·
answer #5
·
answered by Gary H 6
·
0⤊
0⤋
You could use your trigger to send voltage to an led light. The light could trigger a photoelectic eye. The photoelectric eye could close the circuit on a relay. This may sound complicated, but the parts are cheaper than a microchip made to do the job. All can be found at Radio Shack.
2007-01-04 16:35:21
·
answer #6
·
answered by Mythos 2
·
0⤊
1⤋
You will need to construct a transistor amplifier front end for a 40Amp Relay to provide the nessesary drive current. This should not be hard. Especially if you use a FET as the current required is 0mAmp. What voltage do you have for the driving voltage?
2007-01-04 16:32:46
·
answer #7
·
answered by rscanner 6
·
0⤊
0⤋
You omit the factor. extreme voltage is used, in many situations, to cut back the present requirement for potential transmission. you desire a definite volume of potential. that individual same potential could be introduced using low voltage and extreme cutting-edge, or extreme voltage and espresso cutting-edge. you like the low cutting-edge selection each time obtainable, with the aid of fact cutting-edge produces IR losses. Do the math. in case you have a similar volume of resistance (R), do you lose greater potential with a greater robust or decrease cutting-edge? i.e., for a 1000W requirement (or load), you will have: 100v x 10A = 1000W or 1000V x 1A = 1000W. yet what have you ever gained? assume the transmission line has a resistance of .a million ohm. 10A x 10A x .a million ohm = 10W loss (I^2 x R = W) or 1A x 1A x .a million ohm = .1W loss. the final wager is the better voltage for a greater effective flow of potential to the burden. The call for from the potential furnish is set through the burden standards. you could not plug a 110v gadget right into a 1000v outlet. i'm no longer asserting that. we are speaking approximately moving potential from the potential station to the top consumer. There are different concerns, including secure practices, yet this is the fee you pay to flow the potential. You alter for that. i desire this helps.
2016-11-26 20:28:00
·
answer #8
·
answered by ? 4
·
0⤊
0⤋
Your question is the definition of a relay switch.
2007-01-04 16:29:35
·
answer #9
·
answered by Jerry P 6
·
0⤊
0⤋