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Electronic Theory and Principle Discussion of electrical theory, law, etc.

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  #11  
Old 10-23-2006
Rubycon Rubycon is offline
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Lightbulbs make terrible load banks for precision testing. They are much more difficult to start. A cold filament has a very low resistance. This is why if you have a tungsten driver that ramps up the power (as opposed to direct across the line starting) the lamps will last MUCH longer. Inrush current on a tungsten lamp is very high. This makes them candidates for tweeter diaphragm protectors too.
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  #12  
Old 10-25-2006
talc talc is offline
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The wattage is the max the resistor will handle. Beyond that it breaks down and becomes for all intents a fuse. Since the resistor is a fixed value (and we are going to deal with purely resistive loads here and not get involved with impedance etc.) the wattage detemines the maximum voltage that it can handle before the current passing through rips it apart. Your example of the 120 W and 120 ohm resistor is a special case. We all know ohms laws etc. but you can see how special this case is if you look at it as P=I^2 R. Thus 120=I^2 120 or 1=I^2 Thus Max current is 1 amp. 120 V is the point at it will start to turn into a fuse.

If you are thinking about a tester then you will be dealing with a constant Voltage (IE 12V). With this you want to use the formula of W=(V^2)/R. Thus W=144/R. Thus if you want 100 W @ 12 V you plug it in and get 100=144/R or R=144/100 or 1.44 Ohms. Since resistors only come in standard sizes you never get an exact match. Most test equipment is probably set up as an active circuit which is sinking or sourcing a constant current. In this case VI is the easy solution. If you want to put a 100 watt load on a 12V source you need to have a circuit that will sink 100/12 =8.3 Amps.

Using an active circuit with a transistor to shunt the current makes more sense and is probably an easier and cheaper alternative to using a bank of resistive loads. This is because it can be set up with a variable resistor and probably an op amp so that you can dial in what you want. With resistor banks you run into the problem of being limited to just the designed load for each bank. If you want each load to be 100 watts you need to create a bank that matches 1.44 Ohms and make sure that the current through each resistor in the bank does not exceed it' s individual wattage. To get near a 100 watt load at 12 volts using standard sizes you need to do the following. 1/R = 1/R1 + 1/R2 + 1/R3 ... a quick try is to use the standard values of 3 and 2.7 Ohms. 1/3 + 1/2.7 = .703 and 1/.703 gives 1.42. Current through the bank is 12V/1.42ohm = 8.45 A. 8.45 x 12 = 101.4 watts. Close enough to be within the allowed tolerance levels of components. Current through the 3 ohm resistor is 12/3 = 4 and its watt rating should be no lower than 4x4x3 or 48 watts. That through the 2.7 Ohm resistor is 12/2.7 or 4.444 A and its wattage should be no lower than 53.333 watts. (Always choose the next higher watt value to give headroom) Obviously we are starting to involve precision resistors and high wattage values. It only becomes more difficult when you consider that if you want to be able to test to 1000 Watts (which is what some PS are coming out at today) you need an effective bank resistance of .144 ohms that can handle 1000 Watts and is handling over 83 A.
I am sure that the above is known in general if not specifics for many of those frequenting this site.

Last edited by talc; 10-25-2006 at 11:47 AM. Reason: Further clarification
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  #13  
Old 10-27-2007
cypherpunks cypherpunks is offline
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Default Wattage ratings

Actually, this is a bit more complex than you think.

The wattage rating of a resistor is the maximum heating power that you're allowed to pump into it assuming "normal" heat sinking. Given a voltage V across it and a current I through it, which are related by V = I*R, the heat dissipated in the resistor is P = V*I. This must be less than the resistor's power rating or the resistor may be damaged. Since V = I*R, you can shorten the computation using the equivalent formulas P = V*I = V*V/R = I*I*R. You'll get the same answer all three ways.

A brief digression about Ohm's law. Ohm's law is, as its name says, a law of nature, and applies at all times and all places. Human laws say what you may do, what is permitted. A law of nature says what you can do, what is possible. You can't (it is not possible to) break it any more than you can break the law of gravity. At high frequencies, you need to add complex reactance to the mix, which turns your simple resistance into a more complicated impedance, but with that detail, it's still true.

(The entire complexity about "power factor" and "VA" vs. "Watts" is because Ohm's law holds true at each moment, but if current and voltage are changing, then the average current times the average voltage is not the same thing as the average (current times voltage).)

If you ever do a computation with Ohm's law that gives you a ridiculous or impossible answer, then it's telling you that building the thing you're trying to compute the behavior of is ridiculous or impossible. You've made a mistake and overlooked something. If you could build it, Ohm's law tells you what would happen, and if that can't happen, that means that it's not buildable.

(This happens a lot if you assume resistances that are zero or truly infinite. Wires have a small, but non-zero resistance, and air and insulation has a large, but non-infinite resistance.)

Some people seem to think that voltage and current are independently controllable parameters of a power supply or other circuit. That you can somehow change one without changing the other. They're independently controllable if and only if you're allowed to change the resistance. If the resistance is fixed, then voltage and current are in lockstep.

Anyway, a "12 V, 18 A" power supply rail will deliver 12 V at up to 18 A, or 18 A at up to 12 V, but you will get both simultaneously if and only if you use a 12/18 = 2/3 Ohm resistor. If the resistor is higher, you'll get 12V at 12/R amps. If the resistor is any lower, you'll get 18*R V at 18 A. The power supply can't force you to connect a particular load to it, so it can't force the output to 12 V and 18 A simultaneously, despite what some people seem to think.

Anyway, back to resistors... the real limit on a resistor is how hot it can get. If you can keep it cool enough, there is no limit on the amount of power you can pump through it. On the other side, if you put something in a perfect thermos flask, even a little bit of heat will build up and up until you'll eventually melt something.

The power rating of a resistor is assuming some sort of "normal" heat sinking. If you wrap it in insulation, it will stop working at a lower power. If you bolt it to a giant block of finned aluminum in a wind tunnel, it will take a staggering amount of power.

For a classical axial-lead resistor (those little things with the color codes and a wire sticking out each end), most of the heat leaves through the lead wires. the power rating assumes they're connected to some other wires or circuitry that isn't too hot. Making sure those wires don't actually get too hot is the problem of you, the user. Go see the Yaego CFR series datasheet for an example: Notice the spec on "ambient temperature" and "resistor rise above ambient". If you let the resistor heat the ambient above 70 C, you're no longer allowed to put full load through the resistor.

Higher-power resistors assume they're bolted to a heat sink or have a certain amount of air flow to cool them. Typically "in still air" is used as a test standard, so air flow is generated only be convection from the resistor's heat. Any moving air (from a fan or whatever) will help, but additional heat-generating components nearby will make things worse. It often comes out in the wash.

So you can dump 1/4 watt into a 1/4 watt resistor indefinitely, as long as air can flow to it, and it's got some clear space around it. If you put 100 1/4 watt resistors shoulder-to-shoulder and try to pump 1/4 watt into each of them, without serious cooling effort, your little experiment will glow like a 25 watt light bulb and if you complain to the resistor manufacturer, he'll laugh in your face.

Then there are things like surge ratings: you can put much more power into a resistor if you stop quickly enough; this depends on the thermal mass of the resistor. This is generally higher for lower-precision resistors. 0.1% metal film have low thermal mass, while good old fashioned 20% tolerance carbon composition can tolerate staggering surges.
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  #14  
Old 10-27-2007
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mpilchfamily mpilchfamily is offline
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I just wanted to put my $.02 in on this since I've been trying to build my own load tester.

Lets say i want to put a 1A load on the 12v rail. According to Ohms law i need to have a 12 Ohm resistor. Now since P=I*E we know at 12V there will be about 12W on the resistor so the resistor would have to be rated to handle 12W or better. In the case of one of my load designs I'm using a 12 Ohm 12.5W resistor. So when given proper cooling and a HS the resistor will easily handle the load. I have a 6 Ohm 25W resistor to give me a 2A load. 12v/6 Ohms gives me 2A. 12v*2A=24W so a resistor rated at 25W will be sufficient for my needs. All the resistors i'm using as +/-1% tolerance but there will still bee a margin of error. So even if i have both the 12 Ohm and 6 Ohm resistors on my load may be under or over 3A. Course the higher the load the more heat is involved inside the load tester and the further off the load can be from its expected value so measurements must be taken to maintain accuracy. My new idea adds a way to tweak the load a bit to get a more accurate load.

If i'm way off base here please let me know.
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  #15  
Old 10-27-2007
cypherpunks cypherpunks is offline
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Default Tolerances

It's all a matter of what level of precision you're trying to achieve. 1% is generally pretty good. For example, if your power supply voltage varies by 1% (and it's allowed to vary by 5%, so it will change 1% as the load changes), then the current varies by the same 1%, and the power in the load tester changes by 2%.

So if you only have a 0.5% voltmeter, that's 1% of power measuring error right there, and there's no point making everything else better.

How do you intend to measure the current to better than 1%? I noticed you took the shunts out of the design, and even if you still had them, high-power shunts better than 1% are expen$ive. You can measure them with a good meter when not in use, but the resistance might change as they heat up in "real use". The manufacturer will assign part of that 1% error tolerance to initial manufacturing, and part to change with heating. So it'll stay within 1% over temperature, but it might be -1% when cold and +1% when hot.

If you have a spec, it's possible to make something to meet it, but as you get below 1%, it may involve more manual steps. (NIST can measure resistance down to 0.1 part per million, but see NIST TN1458 for how much work it is.)

For example, if you want to calibrate your 1% resistors yourself to the 0.1% level, then you need a 0.1% ohmmeter and should arrange to use double-throw switches to switch them into the circuit. The common should go to the resistor, and the other side should switch between the power supply and your ohmmeter. Then connect the resistor to a power supply, get it nice and warm, then switch it over to the ohmmeter, and quickly take a measurement before it cools down.

Repeat this several times to convince yourself that the resistor's value, when warm, is stable at the 0.1% level. For a wire-wound resistor, it should be.

The other question is how settable you want your load. Is it okay if you can only set it within 5% of the desired load, but you know what the value of that load is to within 0.1%? Or do you need (this will be notably more expensive!) to be able to set it to any 0.1% value?

Do you need continuous adjustment over a range, or is it okay if you have to flip three switches to go between 8 A and 11A, so you'll never be able to make a glitch-free transition?
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  #16  
Old 10-27-2007
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It will never be a smooth transition since the Load will be created by closing 2 or more switches to get the desired load as i have mapped out in my other thread.
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  #17  
Old 12-06-2007
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The voltage stability of an resistor is not related to the power rating. Most low wattage resistors can not sustain a higher voltage then 60v to 100v. So they are not rated for AC lines or similar high voltage. But most power resistors are in fact wire wound resistors, which can handle high voltage without problem.

For an accurate and adjustable load, I would simply use a high power MOSFET bank.
Then put an 1k to 10k pull down resistor, a potentiometer (for accuracy i would suggest some cermet types) and a voltage source (5v for logic level MOSFET, 10v for standard N channel types) at the gate pin.
With this setup you would have an adjustable power resistor. The higher the voltage, the lower the resistor of the transistor until a short (the lowest RDSon the MOSFET could have).
A small power resistor in series with this load, would be not a bad idea to add a little short protection.
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  #18  
Old 02-28-2014
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Quote:
Originally Posted by jonnyGURU View Post
Ok... I know Ohms law, but it doesn't help me when shopping for resistors.

If a resistor is labeled in Watts, does that mean the load is whatever wattage regardless of voltage?
You are close but the wattage is the maximum it can dissipate before burning up. So the total wattage becomes a function of the current and voltage the resistor sees in the circuit. For example some circuit designers will overspec a part, i.e use a 1 watt instead of a 1/2 watt to provide a margin for the component. Since Power dissipated can be expressed as P = I x I x R then its possible to calculate the power being dissipated. Say you have a 100 ohm resistor in a circuit with 120 volts across it. The current is 12 amps and the power dissipation is 144 watts. So the resistor in this case will need to dissipate the amount of power and should be selected based on the expected power.
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  #19  
Old 02-28-2014
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You realize this is a seven year old thread, right?
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