Calculate how to size a resistor for following circuit?
Have a DC supply that varies from 48-54VDC that is supplying a relay that works between 12-48VDC. We have a 475 ohm resistor in series to drop the voltage but we continually keep blowing up the electronic relay.
We are now looking at connecting a zener Diode (39V 1W) and keep the series resistor. Will we need to change the size of the resistor?
Could someone show me how to calculate this?
- jrrymillerLv 510 years agoFavorite Answer
Ohm's Law is a powerful tool that will solve this easily. First a comment - for a 475 ohm resistor to reduce 54 volts to 48 volts the relay would have to draw 12 milliamps minimum. Your relay may not have been that great of a load.
Now for your proposed solution (which is a good one) the resistor will drop 15 volts max and you want the zener to not exceed 1 watt say 1/2 watt so the zener current will be limited to 33 milliamps.Now for Ohm's law R=E/I or R = 15/.033 R = 450 ohms. You could probably still use the 475 Ohm resistor as long as it can safely handle the power of about 1/2 watt. The 39 volt zener diode will keep the voltage on the relay within spec.
- RogerLv 710 years ago
Calculate the maximum voltage dropped in the series resistor 54 -39 = 15 volts
Calculate the maximum current in the 1 watt 39 volt zener, 1/39 = 25 milliamps
Assuming the electronic relay only needs 10 milliamps maximum, if it is more than this please let me know
Set the zener current to slightly less than 1/2 the maximum 10 milliamps
The series resistor will need to handle 20 mA
The resistor must be 15/0.020 = 750 ohms minimum, this is a standard 5% value
use a half watt minimum 750 ohm 5% resistor
At 48 volts, only 13 volts are dropped across the resistor, the current through the series resistor is 13/750 = 17.3 ma, 10 ma will flow through the relay and 7 ma will flow through the zener
At 54 volts, 20 ma flow through the 750 ohm resistor, 10 ma through the relay and 10 ma through the zener.Source(s): If you know the relay's current draw I can give you a more exact answer.
- veeyesveeLv 710 years ago
Come again: 12 to 48V relay? If I take a 24 V relay one can apply perhaps 5 to 6V more and have it operted at 29V. It is unlikley that it can be used at less than 18V. ANyway, it looks your relay requires a sort of constnat current. If you provide four times rated current the relay can burn off. Is that what is happening? You should probably find out what is the votage across relay when it is just able to operate with minimum voltage, and find the resistance of the relay. That will enable you to calculate the current required for its operation. SInce your voltage is 48V to 54V, ensure that the resistor is chosen to provide this voltage across relay at 48V. If that voltage is say 16V, and resistance is say 400ohms, current is 40mA, and you can afford to drop 48-16 = 32V across resistor. That means a resistor of 800 ohms, 2watts min.
Do not use the 39V zener that will mean that the relay may not operate at 48V as the voltage across relay will be 48-39 = 9V. The resistor should be fine if chosen as has been suggested here.
- 異域秦後人Lv 710 years ago
Put 750 ohm rate 1W resistor in series with the +54V. Other end of resistor connects to 39V zener cathode. Zener anode connects to negative power source. 39V source is taken across the zener diode.This circuit only able to support 20mA to drive the electronic relay. Check the relay data sheet to make sure 20mA is sufficient,otherwise it never work. To increase the current rate,let the zener diode drive a Darlington power transistor BASE. Output obtains from its emitter can supply up to 2 amperes.
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- Anonymous5 years ago
The LEDs are arranged in twelve series strings of three. By your data, they will drop 3 * 3 = 9 volts and require 12 * 24 = 288mA. With 288mA flowing through a 330 ohm resistor, there would have to be 95 volts across it! And it would be dissipating 27 watts! I think the resistor value is wrong. It should be nearer 3.3 ohms (orange, orange, gold). That should give you 288mA, if the PSU is actually putting out 10V, and will dissipate 0.27 watts. (Are you sure you didn't mistake gold for brown?) What voltages are you seeing (a) across the resistor and (b) between the common + terminals and the common - terminals that connect to the resistor ?
- ChuckLv 610 years ago
The relay operates by current, so you need to know what the resistance of the coil is and what current it requires to operate it. If this resistance is R (you can measure it with an ohmmeter) and you choose a voltage in the middle of the 12-48 volt operating range - lets say 30 volts. And pick a supply voltage in the middle of its range: 52 volts. Then the relay will draw 30/R Amperes and the external resistor you need to drop 52 volts to 30 volts will be (52-30)*R/30. Be careful to scale the current correctly: if R is in K-Ohms then the current is in milli-Amperes.Source(s): engineer