Transistor Or Resistor

Freight

Also it isn't adjustable, and therefor useless for my purpose.

Chris W

why not start with a discarded oil column heater, take out the element and re-wire it for low voltage operation.

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this can de done using a resistor sized for the largest load and pulse width modulation, but you may need input filtering so that the powersupply sees a more stable load.

possibly splitting the load into 4 or more parts and driveing them out-of-phase with each other cuould help too.

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Don't expect this tester to dissapate heat for any longer than it takes to test a battery. If you leave it connected as a permanent load you WILL have a fire.

Does anyone beleive you can get 100 amps through a small "battery clamp" with crimped wire terminations? I would not trust that rig for over 15 amps.

--- You're mistaken.

Let's look at your 14V supply like this, in Courier:

+------+ +------|+14 | +V | | | | [R2] | | [R1] | | | | G | | +---G NCH | | | S Q1 | | |O | | | |O S1 | | | | | | | +-----+------|0V | +------+

Note that with S1 closed the base of Q1 will be grounded, turning Q1 off.

When that happens there'll be no charge flowing through either R2 or Q1, with the result that R2 won't drop any voltage, but that it'll all be dropped across Q1.

Then, Q1 will dissipate:

P = IE = 0A * 14V = 0 watts

Now let's open S1:

+------+ +------|+14 | +V | | | | [R2] | | [R1] | | | | D | | +---G NCH | | | S Q1 | | | O | | | | O S1 | | | | | | | +-----+------|0V | +------+

Now Q1 will be enhanced and charge will flow.

Just for grins, assume Q1 has an Rds of zero ohms when it's on, so that it'll drop zero volts.

That means that 14V will be dropped across R2, and if the current through it is 50A, its resistance will be:

E 14V R = --- = ----- = 0.280 ohms, I 50A

and it'll dissipate:

P = IE = 50A * 14V = 700 watts.

Since there's no voltage dropped across Q1, it'll dissipate:

P = 50A * 0V = 0 watts

So with Q1 fully on or fully off it'll dissipate no power.

What about at other settings?

OK, let's replace S1 with a rheostat of such a value that at one end Q1 will be completely OFF and, at the other, completely ON.

Then let's call the supply voltage 'Vs', the current in the circuit 'It', the voltage across the resistor 'E1', the power it'll dissipate 'P1', the voltage across the MOSFET 'E2', its equivalent resistance 'R3', and the power it'll dissipate 'P2'.

If we start out with 10A, we'll get:

E1 = It R2 = 10A * 0.28R = 2.8 volts

P1 = It E1 = 10A * 2.8V = 28 watts

E2 = Vs - E1 = 14V - 2.8V = 11.2 volts E2 11.2V R3 = ---- = ------- = 1.12 ohm It 10A

P2 = E2 It = 11.2V * 10A = 112 watts

So...

If we do the math for four more instances separated by 10A each, we'll have:

It E1 E2 R3 P1 P2 A V V R W W

------+------+-------+-------+--------+-------- 0 0.0 14.0 OO 0 0 10 2.8 11.2 1.12 28 112 20 5.6 8.4 0.42 112 168 25 7.0 7.0 0.28 175 175 30 8.4 5.6 0.187 252 168 40 11.2 2.8 0.07 448 112 50 14.0 0.0 0.0 700 0

I stuck in an extra row at exactly half the output current, which is where the dissipation will peak in the MOSFET, in this case at 175 watts.

---

--- Huh???

---

--- Yeah... Explain the relationships between the three power supplies and how you plan to test them.

JF

whit3rd Inscribed thus:

I used to have a client that made and used a refrigerated water cooled heatsink on his PC CPU. He used a beer chiller and a pump to circulate the water.

Guess what eventually killed the MB...

Condensation !

--
Best Regards:
                Baron.

Wow! All these negative sounding posts. I'll address each reply below.

To Herman: Do you base your post on experience with the Harbor Freight unit, or is it speculation on your part? Are you aware that the battery clamps which you call small are the size typically found on automotive jumper cables? For the record, I own and have used the thing, and it works fine.

To Chris W: You dismissed it as useless for your purpose because it isn't adjustable. But the point was made that you could use the resistive element in the tester as the load - don't dismiss that idea out of hand, at least until you have investigated.

There have been a number of posts addressing resistive elements of one sort or another - water tank heaters, hair dryers, toasters, headlights. Every one of the ideas mentioned have one problem or another associated with them, as well as benefits. For example, nichrome wire from toasters or hair dryers requires building a safe housing and determining the right means to connect to and the correct lengths of nichrome. On the plus side, nichrome makes a good element and you can't beat the price of a discarded toaster or hair dryer. Headlights are relatively expensive and relatively large and require building a mounting panel - but with reasonable care they can be used for long periods without overheating. The only one that would give you a load in a reasonable size already physically mounted in a safe configuration is the Harbor Freight tester. Sparky mentioned the problem with it. See my reply to him. Don't get me wrong - I am not pushing the tester as the best possible source for a resistive element. It is just one option.

To Sparky: Right. The tester is not intended to be connected to the source for an indefinite (and unattended) time. As you indicated, it is intended to be used for the time it takes to test a battery, less than a minute in all cases that I have used it. As I recall, the instruction manual addresses that, but I don't have it close to hand.

If he intends to hook up a long term load, he'll need to manage the (~700 watts) heat dissipation from whatever load he uses. If he uses it as designed (brief battery test) it's good to go in the case it comes in. He'll need to design a circuit to make the current variable.

Ed

s

Any auto-parts store has relays that take high current at low voltage, that will be less troublesome than transistors. Price scales nearly proportional to current-carrying capacity in most semiconductors.

In the vein of single-knob control, you MIGHT consider a DC motor (running a fan or brake, or stirring paint... any kind of energy- absorbing load). If the power source is wired to the rotor, and a variable current source to the stator, the load current (and motor speed) are controlled by that stator-current knob. An automobile starter motor might have considerable load capacity in such use; certainly a one-horsepower transistor is not feasible, a one-horsepower resistor is large, but a one-horsepower motor... ya see lots of those.

The fact that it isn't variable wasn't the only thing that led me to dismiss it. It is designed to draw 100 amps which is more than I want to draw. So the load it uses has far less resistance than what I need. I can't think of anyway to increase the resistance of it so I am at a loss as to how I could use it to make something that varies from as low as 3 to 5 to as much as 50 amps.

A bank 50 watt resistors, a large aluminum heat sink and a fan still seem like the easiest way to go. I just don't like having to switch various ones in and out to adjust the load but it is certainly doable just a less than perfect interface.

If I am missing something on how I could use the resistor in that thing, please enlighten me.

Chris W

BTW I very well may use it to do a full drain test of a battery but probably not at the full 50 amps so it could be under load for a long (2 or 3 hours) time.

Shortly after I posted that I realized there would be a point where the transistors would be dissipating the same power as the resistor. Since I am planing on using 50 watt resistors for the 12 V side of this, the transistor and resistor would both be dissipating 25 watts at that point? How realistic is it to find a transistor/heat sink combination to handle that? I was also thinking that the thermal runaway wouldn't be as much of an issue because the circuit could never have less resistance than the resistor so as the "effective" resistance of the transistor dropped with heat below half of the resistors resistance, it's load would start to get lower and therefor dissipate less power and cool off and start the cycle over. Surely it would equalize somewhere in there?

As for the 14V 5V and 3.3V.... those were all maximums and I currently don't see any situation where all three loads would be to that point at the same time. The only time all three loads would be used would be to test computer power supplies and even though it's not all that hard to find a power supply that is rated that high of current on each of those voltages, none of the ones I am talking about dealing with will put out that amount on all three at the same time. Also computer power supplies are of course 12 not 14V (really 13.8V give or take) like some of the other power supplies I use for other things.

In the end I think I will probably end up attaching a wire with powerpole connectors to each resistor and powerpole connectors to a common buss bar then plug in more and more resistors to change the load.

Chris W

Take a look at:

500 Amp Carbon Pile Load Tester

Item # 91129

Tests 12 volt batteries, alternators, regulators and starters by putting a load on the system to simulate working conditions.

• Adjustable load from 0 to 500 amps * Color-coded temperature compensation pass/fail chart * Color-coded separate volt and amp meters * Heavy duty 4 gauge solid copper wire

Overall dimensions: 10-1/2'' W x 5'' D x 10-1/4'' H Weight: 8.7 lbs.

$69.99

--
Anyone wanting to run for any political office in the US should have to
have a DD214, and a honorable discharge.

25W, dissipation that's the ballpark of a mid sized home sterio.

If you arrange it so that the loads come on one at a time (so the 125W setting sees 2 50W loads at full power and one at half power) then at any time only one transistor will producing heat and they can all share a single 25W sized heat sink. (if you can find room on such a small heatsink for 15 mosfets)

This can be made controlable with a single potentiometer with the addition of some small resitors and some cheap op-amps

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. . .

--- That's probably best.

If you want a simple, less klugie way to do it, though, you might want to try this:

1. Determine the maximum current into the load box.
2. Determine what resolution you want out of the load box.

Let's say you want 50 amps max into the box and you'd like to switch the load in 5 amp steps.

Since 50A/5A per step = 10, you know you're going to need 10 resistors and 10 transistors, and with a little bit of work you can figure out that if you've got a 12V supply and you want to pull 5 amps out of it, the load resistance required to do that will be:

E 12V R = --- = ----- = 2.4 ohms, I 5A

the resistor will dissipate:

P = IE = 5A * 12V = 60 watts,

and the finished circuit will look like this, on the ends, with eight identical stages in between. +-----+ +-----------------------|+12 | | R1 Q1 | | +--[2R4]--D S------+--|GND | | G | +-----+ | | | DUT S1>---|-----------+--[1k]--+ | | . . . . . . | | | | | | | R10 Q10 | +--[2R4]--D S------+ G | | | S10>--------------+--[1k]--+

The advantages?

If you use logic level MOSFETS with an Rds(on) of 20mV at 5A Id, that's a dissipation of 100 milliwatts per device, which means _no_ heat sink at all.

Plus, you'll be able to drive the gates with 5V CMOS if you want/need to.

JF

On Sat, 01 May 2010 08:07:41 -0500, John Fields wrote:

--- Slow Saturday:

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R0 SYMATTR InstName M1 SYMATTR Value IRL3915 SYMBOL voltage 240 288 R0 WINDOW 0 -40 3 Left 0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value PULSE(0 5 0 1e-6 1e-6 1 2) SYMBOL nmos 80 160 R0 SYMATTR InstName M2 SYMATTR Value IRL3915 SYMBOL voltage 80 288 R0 WINDOW 0 -42 1 Left 0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V3 SYMATTR Value PULSE(0 5 .1 1e-6 1e-6 1 2) SYMBOL nmos -80 160 R0 SYMATTR InstName M3 SYMATTR Value IRL3915 SYMBOL nmos -240 160 R0 SYMATTR InstName M4 SYMATTR Value IRL3915 SYMBOL nmos -400 160 R0 SYMATTR InstName M5 SYMATTR Value IRL3915 SYMBOL nmos -560 160 R0 SYMATTR InstName M6 SYMATTR Value IRL3915 SYMBOL nmos -720 160 R0 SYMATTR InstName M7 SYMATTR Value IRL3915 SYMBOL nmos -880 160 R0 SYMATTR InstName M8 SYMATTR Value IRL3915 SYMBOL nmos -1040 160 R0 SYMATTR InstName M9 SYMATTR Value IRL3915 SYMBOL nmos -1200 160 R0 SYMATTR InstName M10 SYMATTR Value IRL3915 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WINDOW 0 -42 1 Left 0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V9 SYMATTR Value PULSE(0 5 .7 1e-6 1e-6 1 2) SYMBOL voltage -1040 288 R0 WINDOW 0 -57 -3 Left 0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V10 SYMATTR Value PULSE(0 5 .8 1e-6 1e-6 1 2) SYMBOL voltage -1200 288 R0 WINDOW 0 -57 1 Left 0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V11 SYMATTR Value PULSE(0 5 .9 1e-6 1e-6 1 2) SYMBOL res -1008 32 R0 SYMATTR InstName R1 SYMATTR Value 2.4 SYMBOL res -848 32 R0 SYMATTR InstName R2 SYMATTR Value 2.4 SYMBOL res -688 32 R0 SYMATTR InstName R3 SYMATTR Value 2.4 SYMBOL res -528 32 R0 SYMATTR InstName R4 SYMATTR Value 2.4 SYMBOL res -368 32 R0 SYMATTR InstName R5 SYMATTR Value 2.4 SYMBOL res -208 32 R0 SYMATTR InstName R6 SYMATTR Value 2.4 SYMBOL res -48 32 R0 SYMATTR InstName R7 SYMATTR Value 2.4 SYMBOL res 112 32 R0 SYMATTR InstName R8 SYMATTR Value 2.4 SYMBOL res 272 32 R0 SYMATTR InstName R9 SYMATTR Value 2.4 TEXT -1186 450 Left 0 !.tran 2

JF

This is a long post. If you understand the guts of it, you'll understand the warning at the end.

2 in series gives you .28 ohms and a 50 watt load. $40 dollars, no heat sink needed (for brief tests), already assembled and less than the $90 dollars you'd spend for 30 resistors at $3.00 each. HF also has a 50 amp unit which could be used, I think. (Not certain about that one)

For variability, you'd still need a bunch of power transistors in parallel, but with individual .5 ohm 5 watt emitter resistors. Something like this:

• ---+-------------+---ELEMENT---+---+--}}---+ | | | | | P /c | | | 0

--- Sunday fun; measure the current out of the 12V source:

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-2256 1408 -2256 1328 FLAG -2256 1408 0 SYMBOL voltage -2160 640 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value 12 SYMBOL nmos -1936 656 R0 SYMATTR InstName M1 SYMATTR Value IRL3915 SYMBOL nmos -1696 656 R0 SYMATTR InstName M2 SYMATTR Value IRL3915 SYMBOL nmos -1456 656 R0 SYMATTR InstName M3 SYMATTR Value IRL3915 SYMBOL nmos -1216 656 R0 SYMATTR InstName M4 SYMATTR Value IRL3915 SYMBOL nmos -976 656 R0 SYMATTR InstName M5 SYMATTR Value IRL3915 SYMBOL nmos -736 656 R0 SYMATTR InstName M6 SYMATTR Value IRL3915 SYMBOL nmos -496 656 R0 SYMATTR InstName M7 SYMATTR Value IRL3915 SYMBOL nmos -256 656 R0 SYMATTR InstName M8 SYMATTR Value IRL3915 SYMBOL nmos -16 656 R0 SYMATTR InstName M9 SYMATTR Value IRL3915 SYMBOL nmos 224 656 R0 SYMATTR InstName M10 SYMATTR Value IRL3915 SYMBOL res 256 528 R0 SYMATTR InstName R10 SYMATTR Value 2.4 SYMBOL res -1904 528 R0 SYMATTR InstName R1 SYMATTR Value 2.4 SYMBOL res -1664 528 R0 SYMATTR InstName R2 SYMATTR Value 2.4 SYMBOL res -1424 528 R0 SYMATTR InstName R3 SYMATTR Value 2.4 SYMBOL res -1184 528 R0 SYMATTR InstName R4 SYMATTR Value 2.4 SYMBOL res -944 528 R0 SYMATTR InstName R5 SYMATTR Value 2.4 SYMBOL res -704 528 R0 SYMATTR InstName R6 SYMATTR Value 2.4 SYMBOL res -464 528 R0 SYMATTR InstName R7 SYMATTR Value 2.4 SYMBOL res -224 528 R0 SYMATTR InstName R8 SYMATTR Value 2.4 SYMBOL res 16 528 R0 SYMATTR InstName R9 SYMATTR Value 2.4 SYMBOL Digital\\dflop -2048 928 R0 SYMATTR InstName A1 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL voltage -2048 1184 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value PULSE(0 5 0 0 0 .001 0 1) SYMATTR InstName V12 SYMBOL voltage -2160 1184 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value PULSE(0 5 1 0 0 .5 1) SYMATTR InstName V13 SYMBOL Digital\\dflop -1808 928 R0 SYMATTR InstName A2 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL Digital\\dflop -1568 928 R0 SYMATTR InstName A3 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL Digital\\dflop -1328 928 R0 SYMATTR InstName A4 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL Digital\\dflop -1088 928 R0 SYMATTR InstName A5 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL Digital\\dflop -848 928 R0 SYMATTR InstName A6 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL Digital\\dflop -608 928 R0 SYMATTR InstName A7 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL Digital\\dflop -368 928 R0 SYMATTR InstName A8 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL Digital\\dflop -128 928 R0 SYMATTR InstName A9 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 SYMBOL Digital\\dflop 112 928 R0 SYMATTR InstName A10 SYMATTR SpiceLine Td=10n tripdt=10n trise=30n vhigh=5 TEXT -2248 1360 Left 0 !.tran 0 42 .001 uic

JF

--
Caveat...

From the manual, at:

http://manuals.harborfreight.com/manuals/91000-91999/91129.pdf

"15 seconds per test with 1 minute cooldown; 3 tests in 5 minutes."

JF

I was thinking of using it in parallel with fixed resistors to trim the load current. I've used the parallel 2N3055 transistors for a variable load before. There was an article in a ham radio magazine years ago, "Power Supply Checker Outer" or something in that vein that had a simple design. At that time the surplus marked was flooded with heatsinks with three to five house numbered 2N3055 transistors for a couple dollars.

The magazine was most likely '73', or 'Ham Radio' and in the late '70s or early '80s.

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