voltage drop question??
voltage drop question??
I have a simple series circuit that consists of, a spring loaded push button, a 150ohm resistor and a relay coil. and the connecting wires. ( the relay is rated at 48 volts). The supply voltage is 54volts DC from a transformer and rectifier. The relay does not pull-in. The voltage measured at the input of the resistor is 51 volts. ( when pushing the button) and At the output of the resistor measured voltage is .235volts. I did not measure resistance of the relay and do not have a known good one to compare with. Maybe the relay is shorted and acts as a straight connection to the ground side of the circuit and all the voltage is being dropped by only the resistor?? My question is, what is the general rule for figuring out individual voltage drops of components in a series dc circuit?
- Chris Smith
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- Location: Bieber Ca.
Its current, not voltage that is MORE relevant.
If your voltage is up to specs, what is your current ability?
IF the current isn’t there, nothing happens.
Yes a dead short will bring down everything, even a low current proper voltage.
You can supply a 50 volt system of 50 milli amps with 100 volts, rated at 25 ma and get the same current.
The only real disadvantage here is the higher voltage MAY jump its protection level of insulation.
However, a 25 volt rating at any amperage above what is required may not flow due to the internal resistance, thus not deliver the correct amount of current to operate the switch/ relay.
Bottom line is the voltage is relevant as long as the current is there so that the voltage can over come the natural resistance of the object, and the amperage can deliver the amount of work necessary to that unit as a matched pair.
If your voltage is up to specs, what is your current ability?
IF the current isn’t there, nothing happens.
Yes a dead short will bring down everything, even a low current proper voltage.
You can supply a 50 volt system of 50 milli amps with 100 volts, rated at 25 ma and get the same current.
The only real disadvantage here is the higher voltage MAY jump its protection level of insulation.
However, a 25 volt rating at any amperage above what is required may not flow due to the internal resistance, thus not deliver the correct amount of current to operate the switch/ relay.
Bottom line is the voltage is relevant as long as the current is there so that the voltage can over come the natural resistance of the object, and the amperage can deliver the amount of work necessary to that unit as a matched pair.
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The voltage drops are directly proportional to the resistors in the series circuit. In other words - two 10 ohm resistors in series with 10 volts across the string will divide the voltage in half for each resistor ( 5 volts across each one) because each resistor has equal value and the consequent voltage drops are of equal value, In your relay circuit the voltage ratios are 218:1 and the resistances will also have this ratio - so 150 divided by 218 equals 0.69 ohms for the relay coil resistance. This is way to low for a relay of even the higher power handling capabilities. If this were a relay of average size ( about 2 cubic inches) I would think the relay coil would be in the area of 1000 ohms or so. You either have something wired wrong or the coil has a near dead short across it. Just be sure this relay was not intended for in line current pull-in only, in which case the coils are very low resistance and operate on high line currents as they are operated in series with the load. BTW, Ohms law is very simple and will answer many of your questions regarding this.l
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Ohm's Law memory aid:
E is Electromotive force in Volts
I is current in Amps
R is resistance in Ohms
So:
E = I times R (E = I * R)
I = E over R (I = E / R)
R = E over I (R = E / I)
How did you determine that the resistor is 150 Ohms? If the resistor is really 150KOhms (150000) you will see a fraction of a volt across the relay coil.
Cheers,
E is Electromotive force in Volts
I is current in Amps
R is resistance in Ohms
So:
E = I times R (E = I * R)
I = E over R (I = E / R)
R = E over I (R = E / I)
How did you determine that the resistor is 150 Ohms? If the resistor is really 150KOhms (150000) you will see a fraction of a volt across the relay coil.
Cheers,
Dale Y
- Chris Smith
- Posts: 4325
- Joined: Tue Dec 04, 2001 1:01 am
- Location: Bieber Ca.
Hi.
Connect the power supply direct to the relay coil terminals to confirm the health of the relay and the validity of identifying its coil terminals. The extra ~15% voltage will not harm it.
When proven to work, make sure as Robert says that resistor is the value you said, and insert it in series with the circuit.
Seems you have a multimeter as measured the voltages; measure the resistance of the coil and come back.
Miguel
Connect the power supply direct to the relay coil terminals to confirm the health of the relay and the validity of identifying its coil terminals. The extra ~15% voltage will not harm it.
When proven to work, make sure as Robert says that resistor is the value you said, and insert it in series with the circuit.
Seems you have a multimeter as measured the voltages; measure the resistance of the coil and come back.
Miguel
- Abolish the deciBel ! -
With the need for transient surpression being so pervasive, due to just about everything being digital nowadays, I've been coming across a LOT of solenoids and relays that have snubber diodes already connected, deep inside the coil. Something to keep in mind.
However, your voltage drop (.2V ?) is a bit low for that. Should be .5 top .7 if that's the case, unless the diode itself is semi-blown. I've come across that a few times in my Motorola Tech days-- diode tests "soemwhat OK", but goes wide open with high powered RF in the vicinity.
However, your voltage drop (.2V ?) is a bit low for that. Should be .5 top .7 if that's the case, unless the diode itself is semi-blown. I've come across that a few times in my Motorola Tech days-- diode tests "soemwhat OK", but goes wide open with high powered RF in the vicinity.
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