Transistor Or Resistor
in [Basics]
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From: Chris W on 30 Apr 2010 19:01 ehsjr wrote: > > 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. > 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. > 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. >
From: Chris W on 30 Apr 2010 19:18 John Fields wrote: > On Thu, 29 Apr 2010 18:24:43 -0500, Chris W <1qazse4(a)cox.net> wrote: > >> Thanks for all the replies. I still have a few questions. >> First it seams that using only transistors is not a good idea. The main >> reason I was hoping to get away from using all the resistors is the >> cumbersome way of adjusting the load by switching in various numbers of >> resistors and the fact that the resistors are only going to be able to >> be used to dissipate the maximum amount of energy at one voltage. >> >> Some one suggested using transistors as switches to the resistors. This >> could make it a bit easier because I could then use a single small >> switch to add several resistors to the load. However that doesn't >> really do much to make the interface to adjust the load any more elegant. >> >> Using that method the best idea I have come up with to adjust the load >> is to configure it so my first switch added 1 resistor to the load, the >> second switch added 2, the third, 4 and so on. Then I would treat the >> row of switches like a binary number to increment the load. >> >> The ideal situation would be to have a single pot that I could use to >> adjust the load. Alternatively having 4 or 5 pots where I would use the >> first one to turn the load up to it's max then the second one to add in >> that load, etc. How hard would it be to use transistors to control the >> current the resistors are sinking? That seems like it might be the best >> solution. If I'm not mistaken the transistors wouldn't have to sink >> very many watts at all. > > --- > 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. > --- > >> As for dissipating 700 watts, it is closer 1000 watts >> 14V * 50A + 5V * 25A + 3.3V * 25A = 907.5 watts > > --- > Huh??? > --- > >> That is the worst case, and will likely rarely see that high of a load >> except for a very short time (30 second or less). The most it will ever >> see for an extend time will be 600 watts. I have a 100 watt RF dummy >> load. A heat sink 10 times the size of that dummy load would be big, >> but still manageable. A few good fans would make it even more manageable. >> >> Any thoughts? > > --- > Yeah... > Explain the relationships between the three power supplies and how you > plan to test them. > > JF 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
From: Michael A. Terrell on 30 Apr 2010 22:33 Chris W wrote: > > ehsjr wrote: > > > > 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. > > > > 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. Take a look at: <http://www.harborfreight.com/500-amp-carbon-pile-load-tester-91129.html> 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.
From: Jasen Betts on 1 May 2010 00:56 On 2010-04-30, Chris W <1qazse4(a)cox.net> wrote: > 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? 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 --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: John Fields on 1 May 2010 09:07
On Fri, 30 Apr 2010 18:18:57 -0500, Chris W <1qazse4(a)cox.net> wrote: .. .. .. >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. --- 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 |