How to "create" high load?

I can't translate so I could be wrong, but that looks like a switch-mode power supply with variable settings (can you confirm?). In which case if you think you can build one cheaper then start Googling terms like 'switch mode power supply'. There are plenty of circuits out there. For measuring current you need to either use a current sensor or a shunt with a voltmeter across it to measure the voltage drop. However I don't like your chances of building one cheaper/as nice.

Ken

Ken Taylor's log on stardate 03 lis 2005

Yeah, those Czechs... ;) Just kidding, there was another German link, but since I'm from Croat, I "understand" Czech a bit more than German. Anyway, sorry for inconvenience.

Actually, no, i need _load_. So, instead i put 30W stepper motor (naturally, I am just being descriptive) that runs on 10V (thus needs 3A), i'd put my mean machine that would produce 30A load on 10V.

Huh, i hope i managed to put it right this time. I guess that, consequently, that could be some sort of supply, however, when connected to another supply, it would be treated as a load. On the other hand, i might just be talking nonsense. :) Thanks for your time!

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Okay, I originally thought you were talking about a load then saw the site and guessed it was actually a supply. No problemo! Regarding a resistive load as suggested by Bob you could use stove or heater elements and tap off for variability. Again, a shunt in line with the load will permit a voltage drop to get a current reading. It still won't be "all-singing and all dancing" like that in the link but also significantly lower cost. I had a dummy load of about 33kW made up to test load our site generator once, fitted into a large-ish box with some fans fitted (most important!).

Cheers.

Ken

Bubba, the simplest way to make load is to just hang resistors of the right value on whatever source you're trying to load down. 30A at 10V is simply

10V/30A = 0.333 ohms,

so what you need is just a resistor of about 1/3 ohm. The tricky part whenever you're making loads like this is the power - 30A at 10V is also 300W, so whatever you create has to be able to dissipate that much power without (a) melting down and (b) creating a hazard. This is what makes loads somewhat more expensive than you might first imagine. However, they don't have to be TOO costly. For years, Heathkit sold a "dummy load" kit to ham radio operators that was nothing more than a large 50 ohm resistor (already made for high power dissipation in the first place) suspended in a paint can filled with mineral oil (which acted as additional heat sinking and further increased the resistor's ability to dissipate power. It was good for a full kilowatt for at least several minutes, and as I recall could handle at least a couple of hundred watts essentially forever.

You could do something similar; the specifics of the heat sinking will depend on just what sort of resistors you use (some are designed with metal cases and are to be mounted directly on a larger piece of metal for a heat sink), and you're likely going to be using some series/parallel combination of resistors to get both the value you want and sufficient power-handling.

Bob M.

Try the circuit at

and see if it fits with your idea. if you need it to dissipate more power, lower the 0.33 ohm resistor's value and increase its wattage, and add more MOSFETs. Of course, a big heat sink for the MOSFETs and the sense resistor is mandatory. Use a fan to move cool air across the heat sink for cooling. Remember, this is a load, and will be dissipating lots of raw power, and wasting power creates heat, which has to be removed in some way. Cheers!!!

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Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in 

DaveM's log on stardate 04 lis 2005

This is _exactly_ what i had in mind.

Right. There is one thing, though. I can't seem to find so small 50 W shunt in my local stores. The closest thing to 0.33 ohms is only 5W. So, what is the minimum power that i should put instead of that shunt (i plan to increase power with more MOSFET-s)?

Can i do it with a parallel resistors? In that case, does resistor powers stack (eg. if I have 4 1.5 ohm 20W resistors, would it be 60W?).

Naturally, proper cooling will be installed. Many thanks for your assistance!

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Kupio sam pistolj od svercera na crno,
Na tebe cu rado da potrosim zrno...

I don't read that language but the machine looks like a laboratory (or bench) power supply.

much

for one with that capacity 75V and 50A that's not an excessive price.

there have been kitset versions of that sort of thing. and low-powered versions (say 20V 2A) can probably be done for around $100.

there are a number of ways to convert a current into a voltage... (like use a low resistance resistor or a hall-effect current sensor) then you can use a milivoltmeter unit to read the volts, and from that determine the amps, (if you use a 0.1 ohm resistor 1mv = 10 mA...)

Bye. Jasen

hooking two supplies together is unlikely to give the desired result. you're likely to damage one or both or cause one or both to shut down and not produce the load you want...

for a 10V 3A load you could try using a pair of 12V 21W automotive indicator lamps, that combo should come fairly close and only cost a few dollars.

if want a load, that you can turn up and down a big wire-wound , or carbon-pile rheostat might be the answer.

Bye. Jasen

Cheap man's load is either:

1. A bunch of power resistors in series, parallel, or series-parallel as necessary to take the load power.

1. A bunch of light bulbs in series, parallel, or series-parallel as necessary to take the load power.

A traditional way is to do (2) and just parallel more and more light bulbs until you reach the desired load current.

(1) has the advantage that the resistance is ohmic, almost independent of voltage applied.

(2) has the advantage that the resistance changes with applied voltage to tend to keep current very roughly constant. And you get visual feedback, too :-).

If you really need to keep current constant the olf-fashioned way was ballast tubes, but you probably cannot find enough of them anymore...

Tim.

You can stack resistors and add the power (leave enough room for air to circulate)

You can make your own resistor by winding wire on a form. In this case your mosfets do the regulating so the temperature rise in your resistor wire (and subsequent resistance change) would be compensated for. -Providing the resistance of the wire didn't go too high, or mosfet power dissipation wasn't exceeded, and your winding technique didn't leave enough inductance to cause the circuit to oscillate.

I find it more convenient to wind current shunts than search for just the right resistor. Copper wire tables give the resistance per linear measure of wire of a particular gauge. I measure out some heavy gauge wire #16 - #12, Double it over so I don't wind an inductor and wind it on some PVC pipe.

Place I worked at - we would get wire for the purpose of winding shunts. We used a phosphor bronze alloy that had a negligible resistance change with temperature. Used them to set the current limits on the power supplies we made.

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Good idea, especially for his application Light bulbs are inexpensive and 50 watts would take up very little space.

I'm usually concerned with temperature/resistance change so didn't think of using light bulbs - but with a current regulator it isn't a concern.

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For the resistor specified in the circuit, you can use any combination of resistor values that will give you 0.33 ohms at 50 watts or greater. For instance, three 1-ohm/20-W resistors in parallel will give you the right combination, as will six 2-ohm/10-W resistors, etc, etc.

Power dissipation capabilities of resistor combinations such as these are additive, in other words, three 20 watt resistors will give you a 60 watt rating, as will six 10 watt resistors. And the power rating doesn't change whether they are in series or parallel.

A fan will be mandatory to cool these components, as getting rid of 100 watts of heat isn't trivial in a small space.

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Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in 

DaveM's log on stardate 04 lis 2005

/snip

I see. That's what I wanted to know.

Two high RPM Sunons are already prepared. ;)

There is one more thing, though. Could I, instead of relay, put an LM317 or such to control voltage "externally". In the figure 1, it is clearly showed that you can only control voltage _or_ current, however, i would like to be able to control both of them simultaneously. As I've understood, Iload is changed by changing reference voltage on the op-amp, and instead of putting a relay I'd simply remove the voltage divider and bring external voltage from another source. Is that possible?

Thanks for your help once more.

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Kupio sam pistolj od svercera na crno,
Na tebe cu rado da potrosim zrno...

Jasen Betts's log on stardate 04 lis 2005

That's a nice one. However, is there some sort of integrated circuit ampermeter, that would exclude need for (additional) digital IC-s?

OTOH, i could always buy el cheapo multimeter and measure current with it...

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Kupio sam pistolj od svercera na crno,
Na tebe cu rado da potrosim zrno...

snipped-for-privacy@trailing-edge.com's log on stardate 04 lis 2005

That sort of producing load would be indeed visually attractive, however, i really need something "a bit" more precise than 50W +-... But thanks for the idea. :)

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Kupio sam pistolj od svercera na crno,
Na tebe cu rado da potrosim zrno...

You just substitute the load resistor (that .33 ohm) with a bank of low voltage lamps. The mosfet current regulator provides the precision. Several small low voltage lamps.

While the lamp's resistance can increase dramatically as it lights, the mosfet would drive it harder by decreasing its own resistance. This would have the effect of transferring more heat from the Mosfet to the load resistor. And you can add and remove bulbs easily.

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Your last question imposes a totally new concept to your original post. You're now asking, if I understand your question, for a constant power load That will entail the use of an analog multiplier in the circuit. I saw a design for one on the web a while back.. have to search around for it. The circuit that I recommended originally won't do that without some additional circuitry. Look at the datasheet for the MC1495 (On Semi) and get an idea of how a multiplier works. You need the math function in the circuit to calculate the total power being dissipated so you can control it. An LM317 won't work here because it can't do the math. In this circuit, you already have the basics.. the voltage sense and current sense. Send the outputs of the IC1A and IC1B into the X and Y inputs of the multiplier. Scale the output of the multiplier with an opamp or voltage divider (depending on the gain requirements) and send that into the MOSFET gates. That should give you the ability to run the load in constant power mode.

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Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in 

you can control one at the load and the other at the supply, controling both at one end of the system is not possible - see Ohm's law.

Bye. Jasen

no. you use a milivontmeter and a low resistance (if you don't need isolation) or hall-effect cell.

el-cheapo multimeters work the same way (low resistance), the expensive ones (with the current clamp) use a hall effect cell.

Bye. Jasen