[plast744]

hi all

1 have 2 resistor 2.5ohms 75watts hookup in series, would it be 5ohms 150watts or 5ohms 75watts

[jerber]

That would be 5 ohm 75 watt.

[gar]

050709-0634 EST USA

plast744:

It is very simple. If a resistor is rated at 75 watts, then it can operate at this power dissipation. However, I would generally prefer to run below its rating. Its rating is based on some assumed ambient temperature, freedom of air flow, and orientation.

Since you have two equal resistors in series the voltage drop will be the same for both resistors. Thus, each resistor will have the same power dissipation. Power P = V squared / R.

Your two 2.5 ohm 75 watt resistors in series result in a 5.0 ohm 150 watt resistor.

Were the two resistors unequal, but of equal power rating, then you would calculate the maximum current for the given power rating for the larger resistor. Imax = sq-root of ( P / R ) where I is in amperes, P in watts, and R in ohms.

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[plast744]

thanx you gar
so the final anwser is 5ohms 150 watts.

i have a Superior Electric Slo-Syn stepper motor 6 wire 4.6amp 2.5 volt
my power supply 24v 12amps hook up to the unipolar board that require resistor. i use the stepper resistor calculate fr http://www.eftechusa.com/downloads.htm
came up with optimal resistance 4.67ohms max resistance 6.08 ohms
minimum power 129 watts. it's ok to use two 2.5ohms 75watts hook up in series = 5ohms 150watts

[gar]

050709-0920 EST USA

plast744:

Basically yes on two 2.5 ohm 75 w resistors in series..

Mount the resistors so their axis is horizontal. Avoid mounting the resistors one above the other, because the lowest will run coolest and the top hotest. Provide good vertical convection air flow. You could mount in other ways and use a fan, but fans fail. You could also go to resistors with a higher power rating and this would reduce surface and internal temperature rise.

The rough calculations on your required resistance are:

Assume source voltage is Vs = 24 v, it may be higher depending upon load and line voltage. Motor resistance cold is Rm = 2.5 v / 4.6 amps = 0.543 ohms, from the nameplate. Assume 1 v drop in your switch ( transistor or whatever ), then ( Vs - 1 ) / 4.6 = 5.0 ohms. Now subtract Rm and your current limit resistor is 5.0 - 0.543 = 4.457 ohms. Unless you have a Dividohm resistor you need to stay with standard values, and these in power resistors probably have a 5% tolerance.

You do not need to be too precise because you are not dealing with absolute threshold limits. But generally you want to work below maximum ratings.

Using the nominal 5 ohm current limit resistor above will give you 23 / 5.543 = 4.15 amps motor current. Thus torque will be 4.15/4.6 of torque at 4.6 amps, or 10% lower than max rating. At this operating point the power dissipation in the current limit resistor is 4.15 squared times 5 = 86.1 watts. This is equally split between the two 2.5 ohm resistors.

If you used the exact resistor to get 4.6 amps, which is ( 23 / 4.6 ) - 0.543 = 4.457 ohms, then I squared R is 4.6 squared times 4.457 = 94.3 watts. If you used a 5 ohm resistor at 4.6 amps then the power is 105 watts.

I do not what is at the site you referenced, but my guess is that the 129 watt figure you mentioned includes some derating factor.

(edit 050709-1004)
The only disadvantage of a higher resistance for the current limit resistor for the same source voltage is lower torque. There really is no maximum resistance limit until you get much higher.

Why not try a chopper driver?

Study and experiment with the equations V = I * R, P = V squared / R, and
P = I squared * R, P = V * I . For example you can derive V squared / R from P = V * I and V = I * R.

The algerbraic sum of the voltages in a closed loop = 0. The algebraic sum of the currents at a node (junction) = 0.
(end edit)

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[smarbaga]

_____ jerber is correct _____
2.5 +2.5 = 5 ohms
and the watts is still 75 watts
in parallel it would be
2.5 / 2 = 1.3 ohms
and 75+75 = 150 watts
- twice the current paths (2 resisrors in parallel)
- twice the watts

same as if you add 10 feet of wire to a 10 foot wire ( series)
it can still only take X amount of amps
but if you parellel them, your wire is thicker ( more amps)

[gar]

050710-0521 EST USA

I believe the question by plast744 is ---

How much power can be dissipated by two resistors of equal resistance and power rating when connected in series in relation to the power rating of one of the resistors?

The question is not how much total power is dissipated when the same voltage is applied to either the series or parallel combination. In this case the answer is a 4 to 1 ratio.

It makes no difference whether the two equal valued resistors are in series or parallel the combined power dissipating capability is twice that of a single resistor.

Power dissipation capability is a function of thermal resistance from a device to some reference point. As the power rating of a resistor goes up the surface area also goes up to lower the thermal resistance so that some design criteria for maximum internal hot spot does not exceed a design temperature maximum.

Note all 1/2 watt carbon composition resistors are of the same physical size independent of resistance. Also note that a power wire wound resistor will generally be in a smaller package then a carbon comp of the same rating. In particular a 2 watt carbon comp is larger than a 2 watt wire wound.

Continuing on the series parallel resistor combination. If I have N equal valued resistors each of power rating P, then whether in series or parallel the combined power rating is N * P. This of course requires that the resistors be physically separated so that the ambient temperature of any one resistor is not affected by the power dissipated by the other resistors.

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[jerber]

Thanks Smartbaga but gar is right. I must read better (or don't building my machine while being on cnczone at the same time, but this place is so inspiring ).

[gar]

050710-1419 EST USA

Warning:

To make things absolutely clear.

The power dissipated in a linear resistor is P = V * I.

Where P is in watts, V is in volts across the resistor, and I is in amperes thru the resistor. This is basic high school physics, colloge physics, and beginning electrical engineering.

This applies for instantaneous values, or steady state RMS values. A steady DC value is an RMS value.

The power dissipated in a resistor has nothing to do with the power rating of the resistor. The power rating is a vlaue established by the manufacture that implies a moderately safe maximum operating point under normal applications conditions. If you want to operate the resistor in an ambient temperature of 100 deg C., then you must greatly derate an ordinary resistor. A normal rating will be based on an ambient somewhat above room temperature, maybe 50 to 60 deg C. In general for a reliable system I would not design to operate a resistor at its power rating, but maybe at 50 % of that rating.

Important if you have a parallel or series resistor network made up of unequal resistors, but equal power ratings, then you must design for the resistor that has the maximum power dissipation. In a parallel network this is the one with the lowest resistance, and in a series network it is the one with the highest resistance.

Unrelated warning: Never put electolytic capacitors in series to try to achieve a higher voltgae rating. I will not explain the reason here.

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