Creating a higer wattage resistor

The short answer is: Yes, and none of the things that you mentioned are likely to happen because of it. To be a "really good boy or girl " you should de-rate the resistors by about

50 percent. The bundle that you make is not as efficient at heat dissipation. We made dummy loads like this all the time. No they are not pretty but they function.

Tom

It works for me.

de W8CCW John

John Ferrell W8CCW

Yup. No problem at all. Neither illegal, immoral or unethical.

Most series/parallel setups like this place the resistors in close proximity. That hotboxes the ones in the middle, which can lead to failure.

Be sure to leave enough free air between for normal power dissipation.

Cheers (and happy Friday) Chris

--
Sure.

Once upon a time when surface mount parts were new, I had a design where I needed a 2.5 ohm SMD resistor but the company didn't have any in stock so I used four 10 ohm resistors in its place. No problem.

There is another benefit of using multiple parts to create a composite value and that is tolerance. Using two 100 ohm, 5% resistors to make a 50 ohm load will reduce the tolerance to 3.5%. This reduction in tolerance occurs regardless of whether the resistors are used in parallel or series. The reduction in tolerance can be calculated by dividing the tolerance of the individual part by the square root of the number of parts used. (They must all be the same value and tolerance for this formula to work.) Eg. 5% / sqrt(2) = 3.5%.

"phaeton" schreef in bericht news: snipped-for-privacy@p79g2000cwp.googlegroups.com...

Ever made a double euro card filled with resistors as an extra load to make sure the power would not switch off due to too light a load. It dissipated

100W at 5V if memory serves. (I know, 20A looks like a lot these days but a 19" rack filled with old TTL could easily consume some hundreds.)

petrus bitbyter

I have done exactly this.

Graham

--
That\'s not true.

suppose you have two 100 ohm resistors that are both 5% high.  That
means that their actual resistance will be 105 ohms each.

Now put them in parallel, and their total resistance will be 52.5
ohms instead of 50 ohms.  That 2.5 ohms is 5% of 50 ohms, so the
tolerance has stayed the same.

But its ok to have a higher wattage resistor. Just use 50 1/4 watt resistors and you'll be ok johnny.

Sure. This is a direct consequence of Kirchoffs current law and voltage laws. By running the resistors in parallel you are splitting the applied current to each. So in effect each resistor see's less current and hence less power.

P = I^2*R

By decreasing the current you decrease the power dissipated if in parallel. If they are in series then you have to decrease the resistance but by the square to get the same effect. So its much better solution to do what you are saying in parallel than in series except possibly if that is the only choice.

You can take this a step farther to determine how n resistors in parallel work:

the total power dissipated is Pt = n*(I/n)^2*R = I^2*R/n while the power dissipated in a single resistor is Pr = I^2*R/n^2

So, Pr = Pt/n. Hence putting n resistors in parallel each with resistance R will divide the power through each by n. So you just have to choose n and R such that you get your n*R ~= total resistance and Pr*n > total dissipated power

i.e., say you have a bunch of R W resistors. If you take n of them in parallel then its equivilent to R/n nW resistor.

i.e., suppose R = 1k and W = 1/4, then 10 in parallel is equivilent to a

100ohm 4w resistor. 20 in parallel is equivilent to 50ohm 8w resistor.

Also, if you have two types of resistors then you can combine them into other combinations by just taking the equivilent "bulk" resistor in parallel(or possibily series):

If, say, you have a bunch of R1 W1 and R2 W2 resistors then R1/n1*R2/n2/(R1/n1 + R2/n2) = R1*R2/(n2*R1 + n1*R2) = RT and n1*W1 + n2*W2 = PT.

Since this is 2 equations in 2 unknowns you could solve for n1 and n2(or atleast approximately).

There might be derating needed and such so it might be more complicated than this.

Anyways... Jon

If this were true then all resistors would have 0% tolerance. Why? Because resistance is "serial" in that you can think of a resistor as a composite of several smaller resistors in series. It would be quite easy to get a resistor with a tolerance that is arbitrarily small.

--
I see you missed the joke.  Oh well... ;)

--
Not true.

If you need a 100 ohm resistance capable of dissipating 100 watts
you can get there with:

1. A single 100 ohm 100 watt resistor.

2. Two 200 ohm 50 watt resistors in parallel:

3. Two 50 ohm 50 watt resistors in series.


In the case of the single resistor the current through it will be 

               P          100W
     I = sqrt --- = sqrt ------ = sqrt (1) = 1 ampere
               R          100R

and the voltage across it will be:


     E = sqrt(PR) = sqrt (100W * 100R) = sqrt (1E4) = 100V


In the case of the two 200 ohm resistors in parallel, the voltage
across each of them will still be 100 volts, but the current will
split and each one of them will have to carry 0.5 ampere.

That means that they must each dissipate:

    P = IE = 0.5A * 100V = 50 watts.


In the case of the two 50 ohm resistors in series, each one must
carry the total current, but the voltage will split and each one
will have 50V across it.

That means that they must each dissipate:

    P = IE = 1A * 50V = 50 watts.  

So, there is no difference in the power dissipated by the
resistances, whether they\'re connected in series or in parallel.

No, you are not understanding what I'm saying. Decreasing n by the square. notice how 200 = 2^2*50. If you have 3 resistors in parallel or series then there will be a factor of 3^2. i.e. something like 3 90 ohm resistors in parallel will dissipate the same power as 9 10 ohm resistors in series... i.e.

Since you are looking at the equation P = I^2*R you have two variables that can change. Since I've already shown that P = (I/n)^2*R for n resistors in parallel, you have P = I^2*R/n^2. So by cutting the current into n branches is equivilent to cutting the resistances by a factor n^2. Its much easier to cut things into n than n^2. This has nothing to do with what you are talking about.

This is the same idea about power transmission and why its more important to cut the current than to cut the resistance but you could do both... just a bit impractical to cut the resistance.

I suppose. Didn't seem like a joke to me but an oversight ;/ Oh well... people tell me I have no sense of humor ;/

On Sat, 26 Aug 2006 19:16:05 -0500, in message , "Abstract Dissonance" scribed:

I don't know if it was so much of a joke, as an interesting way to present a design decision.

On Sat, 26 Aug 2006 19:15:30 -0500, in message , "Abstract Dissonance" scribed:

That makes two of us.

What is 'n'? What equation does that factor of 3^2 apply to?

I take it you are assuming a constant voltage, let's call it Vs. So, in your example, Pcircuit = Vs * (Vs/90) = Vs * (Vs/30). That's not right - the second circuit dissipates 3x the power. Maybe you mean, as opposed to the entire circuit power, that each resistor will dissipate the same power? So Pr1 = Vs * (Vs/9/90) = Vs * (Vs/90). That's not right either, because each resistor in the first circuit draws 1/9 the power of each resistor in the second circuit. The only firm deduction is that *one* of the parallel resistors will dissipate the same power as the total power dissipated by the series combination, and that's as simple as PIE. So what are you trying to say?

Could you show the work behind P = (I/n)^2 * R = I^2 * R/n^2? Yes, it works, but let's show the rest of the class. Then, and this is the part where I am confused, of what use is the identity? What does 'n' signify in each case?

Here's what your original post in the thread stated:

Which, of course, applies to a series circuit as well.

Why? Why can't you decrease current to decrease power dissipation? What is the basic premise behind your equation juggling?

I'm sorry, I simply can't parse that.

-----

Are you talking about some specific situation? Why do you want to cut current on a power transmission line? Why is it important? Are you talking about design or operation?

n is the number of resistors that you plan on using.... should be clear from the context of my original post.

Nope... look you can have 2 possible configurations: All n of the identical resistors in parallel or all in series.

Obviously your eating the wrong pie because your forgetting that you can have constant current.

I didn't know I was going to have to demonstrate it in such simple terms. I thought most people here would be educated enough to see the what is going on and I wouldn't have to take baby steps.

Sheesh... I think you are being a little dense.

I will, for the sake of your enlightenment be more precise and clear. I figured most here would have basic knowledge of elementary circuit analysis but i guess I was wrong ;/

Suppose we have n1 resistors each with resistance Ra and wattage Wa in parallel. I'm not going to try and draw it because its obvious and if you can't picture it in your brain then maybe you don't need to be in electronics.

Now. The total current into the node of the resistors is I. Since each branch is identical we know the current splits identically. This means that the current going through each resistor is I/n1. This is basic kirchoff's current law. What is the power dissipated by each resistor? Simply Pa = (I/n1)^2*Ra. (i.e. current squared * resistance). But now power adds so the TOTAL POWER is TPa = n1*Pa = I^2*Ra/n1.

---------- (Second part)

Suppose we have n2 resistors each with resistance Rb and wattage Wb in series.

Now. The total current through each resistor is I. Hence The power dissipated by each resistor is Pb = I^2*Rb. The total power TPb = n2*I^2*Rb

----------

So, if you get the above then if we want TPb = TPa which means we want the series version to dissipate the same power as the parallel version then we have

n2*I^2*Rb = I^2*Ra/n1

or

n2*Rb = Ra/n1

This can be seen from another much simpler method.

In parallel with n1 resistors of resistance Ra the total resistance is Ra/n1 and in n2 resistors of resistance Rb the total resistance is n2*Rb.

If we want equivlent components then we must have n2*Rb = Ra/n1.

Its quite simple and any 3rd grader should have no problem with the math and electronics to this point.

What does this mean though? It means that if we are trying to convert from one parallel to series then the resistance must change by the square(or sqrt) to still make the two equivilent.

i.e., If I have n resistors with resistance R in parallel then how many resistors will I need and of what resistance to get the equivilent power dissipation?

well,

R/n = n2*Rb. So Rb = R/n/n2.

I choose n2 = n the original problem to make it easier to work with. (else you are dealing with two possible choices to make the circuit equivilent(I did point it out though).

For some real world examples to make all the egg heads understand? Ok...

Given 10 resistors of 100 ohms in parallel how can we get an equivilent circuit(same resistance and wattage dissipation) in series?

100/10/n2 = Rb ==> Rb = 10/n2.

n2 = 1 then Rb = 10, n2 = 2 then Rb = 5, n2 = 5 then Rb = 2, n2 = 10, then Rb = 1.

i.e. 10 resistors of 100 ohms in series will dissipate the same power as 10 resistors of 1 ohm in series. This does not take into account the wattage of the but I explained that in the original post.

Now, to the problem at hand... why is it better to do it in parallel than in series?

Well, Ra = n1*n2*Rb.

If we are working with the same number in each case, i.e., n1 = n2 = n (its still called the square in ither case though)

then

Ra = n^2*Rb.

Rb represents the resistance in series and Ra in parallel.

So, what happens as we use 10 resistors in parallel of resistance R? then our resistors in series need to be R/100 ohms to get the same equivilent circuit.

I guess it doesn't matter though as large resistances are easy to come by too. Ultimately the equation fails at both ends though but if your trying to do something like 100 resistors in parallel then you have a factor of 10000 if you want to convert that circuit to one in series. (you could use that to your advantage if you happen to be working with extremly large or small resistances to convert to the other topology and get resonable results)

If, say, R is 100ohms and you have 100 in parallel but decide you need to do it in series then you need 100 1/100 ohm resistors. (since 100/100 = 1 =

100*1/100)...

But on the other hand if you have 100 100 ohm resistors in series and want to convert that to a parallel(for whatever reason) then you need 100 100000 ohm resistors for parallel. (sinec 100*100 = 1000 = 100000/100).

So ineffect going from parallel to series will decrease the resistance you need while going from series to parallel will increase it(obviously).

If we look at the above resistors and look at the power they are dissipating then they have to be the same because resistance adds... doesn't matter the applied voltage or current because they are equivilent topologies. (they are effectively the same bulk resistance) It doesn't matter how many resistors you use or how they are configured because if they are equivilent to another configuration then they will dissipate the same total power. BUT they will not necessarily dissipate the same power PER resistor.

Thats the mistake you seem to be making. Its easy to show a counter example:

------R------ | |

--- R ---- --------- | | ------R------

The first resistor does not dissipate the same power as the other two(individually).

The first is 2 the power dissipated in either of the others. Why is this important? It should be obvious. (This goes to the part when I talk about combining the two configurations)

Anyways, What I said is this:

If you read carefully what I say you will realize it makes perfect sense(although I'm not saying its completely clear). You have to understand that I didn't say that the power changed when the topology changed.... The first sentence is an obvious statement. The second is is obvious too. I'm saying that if you want the same equivilent circuit(same bulk resistance) then you have to reduce the "new" resistors by a squared factor. What this means is that you might end up having to use extremly low resistor values if you plan on using a series circuit. To do the job. The logical converse is obviously true in that you have to use "extremly" large resistor values in parallel as compared to the equivilent series topology. You can see this in the two examples I gave. In the parallel topology they the resistors are way larger than in the series. (Ofcourse you might be able to get away with choosing different numbers in each topology to make it easier to do)

--
Not true.

three 90 ohm resistors in parallel will yield a total resistance of
30 ohms, while nine 10 ohm resistors in series will yield a total of
90 ohms.

Connected to the same voltage source, the parallel string will
dissipate 3 times more power than the series string.