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Resisitors, question on wattage.

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MikeT1982

Mon Apr 16 2007, 08:40AM

Registered Member #621 Joined: Sun Apr 01 2007, 12:37AM
Location:
Posts: 119

If I get 30 KOhm resistors to place across my caps, and instead of going with 5 watts like was mentioned I go with much bigger like 30 watt wire wound ones, does this mean they will draw more power or be of any downfall? I'm guessing it'd be OK so long as the ohms is still 30K but want to be sure. I just have an opportunity to get some but they are 30 watts.

Sulaiman

Mon Apr 16 2007, 09:13AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Assuming that the size or weight is not a problem it's ALWAYS better to use a higher wattage resistor than needed.

I try to never use resistors at their rated wattage because they will get very hot.

The circuit/electrical performance will be the same.

(Except if you use a very large wirewound resistor at very high frequencies where the inductance may be undesirable)

Electroholic

Mon Apr 16 2007, 09:31AM

Registered Member #191 Joined: Fri Feb 17 2006, 02:01AM
Location: Esbjerg Denmark
Posts: 720

as long as you have a good way to mount them, yea.

Dr. Slack

Mon Apr 16 2007, 12:49PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

As long as their voltage rating is OK as well.

Wirewound on a ceramic core will almost certainly be OK, as the insulation on the wire has only to cope with the turn to turn voltage. If the rated voltage seems low, it may be that the manufacturer couldn't be bothered to test to a higher voltage, and you might get away with much more stress. OTOH, there may be a real reason like metallic ink on the body, conductive glue or frit under the wire, or voids in the encapsulant.

Wirewound in a metal case subjects the insulation at each end to the rated voltage, so with these I would respect the rating.

Simon

Tue Apr 17 2007, 12:54AM

Registered Member #32 Joined: Sat Feb 04 2006, 08:58AM
Location: Australia
Posts: 549

To make it absolutely clear, any 30k resistor will draw the same current as any other in an application like this. A 30W rated one will run cooler than a 5W one.

Steve Ward

Tue Apr 17 2007, 03:48AM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

To make it absolutely clear, any 30k resistor will draw the same current as any other in an application like this. A 30W rated one will run cooler than a 5W one.

how about to make it even clearer, we suggest he learns some basic electronics: OHMS LAW.

If you have a voltage source (your capacitor) and a resistor across it, you can find the current through the resistor with the relationship:

V = I*R

Now that we know the current, there are several power equations for the power dissipated in the resistor:

P = I*I*R = I*V = (V*V) /R

so for example, at 400V your 30k resistor dissipates 5.33W (you can get this directly from the last eq).

The wattage rating of a resistor is strictly the amount of power it has the *ability* to dissipate, not how much it *will* dissipate.

GreySoul

Tue Apr 17 2007, 04:12AM

Registered Member #546 Joined: Fri Feb 23 2007, 11:43PM
Location: Albuquerque, NM
Posts: 239

You could also use 2 60k 5w resistors in parallel for 10w @ 30k, if you have extras.

-Doug

MikeT1982

Tue Apr 17 2007, 05:53AM

Registered Member #621 Joined: Sun Apr 01 2007, 12:37AM
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Posts: 119

Ok cool, now I understand why you guys said go for one OVER 5 watts, because at 30kohm and 400 volts I get just over 5 watts. They don't have a voltage rating on them but are 25kohm 30 watt Ohmite wirewound and look to be ceramic with a wirewound core. I thought they were 30kohm but are 25kohm. So using those Ohms law equation I get 6.4 watts. So at 30 watts I'll be fine as heck. I apologize being behind with my mathmatical skills, my first year of college was for Electrical Engineering but I changed to Medical Imaging (i'm an x-ray tech) because I dreaded math and equations, but I love to tinker. With this laser project I'm learning I have no choice but to do calculations if I want the components to work together. I am glad I'm taking on this project (complicated for me) because it further serves to shoot down any regrets I had that i didn't complete my EE degree...I wouldn't be able to stand doing it all the time! I have ALOT of respect for Engineers of any type though as I myself couldn't handle it as a job!!

... I just had to edit this post and add this....I just took those equations and played around with them. if I have a 25,000 Ohm Resisitor and it is running at it's max power of 30 watts, using those Ohm's law equations, I get a voltage across it of 866.025volts! Does that mean that each resistor would have that as its actual voltage rating since with a fixed resisitance of 25,000 ohms, there is no way to get to the 30 watt limit by any other way as increasing the voltage....right? I mean you can't increase the current draw because the resistance is a fixed resistance. So I'm deducing from this that that is why there is no voltage limit on the resistor's side....because 866.025 volts gets you to its max 30 watt's and it HAS to hold that voltage in order to have that watt rating....

Sulaiman

Tue Apr 17 2007, 08:52AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

For hobby use you may assume that the dc voltage rating is at least V = SQRT(power x resistance)
this is what you determined yourself.
This rating could be continuous voltage (dc)
OR
rms for a sinewave, which gives a PEAK voltage of SQRT(2xpowerxresistance)

However;
Manufacturers usually give voltage ratings for a particular range/type of resistor
so you have to ensure that both power and voltage ratings are not exceeded.
For each type of resistor there is also a peak voltage rating
which is applicable in circuits that produce pulses of voltage.
There is also a much lower voltage rating for resistors which is the "safety critical" rating
this is for situations where the failure of a component could harm civilians.

Be aware that resistors also have an insulation voltage rating
this is the voltage from any part of the resistor to nearby/touching components.
Ceramic wirewound resistors have a very poor insulation rating, especially when hot, due to the glaze used.

So even though resistors are very simple devices, there are quite a few interesting details.

Steve Conner

Tue Apr 17 2007, 09:43AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

wrote ...
So I'm deducing from this that that is why there is no voltage limit on the resistor's side....because 866.025 volts gets you to its max 30 watt's and it HAS to hold that voltage in order to have that watt rating....

Yes, you are right. Resistors have typically two voltage ratings. One is determined by the rated power dissipation, like you figured out. The other is the maximum voltage the resistor can stand before arcing over internally. The value of the resistor determines which limit you hit first.

"350V or the square root of (rated power*resistance), whichever is less"

Resistor power ratings can be exceeded a lot for a short time, but only so much. I have had those gold coloured metal cased resistors blow their ends out and spew smoke, while developing a soft-start circuit for a high-powered audio amp.

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