Transistor Or Resistor
in [Basics]
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From: Herman on 30 Apr 2010 07:23 "sparky" <sparky12x(a)yahoo.com> wrote in message news:6109a09c-0ab8-448b-8fc3-2fe0944b7967(a)w36g2000yqw.googlegroups.com... On Apr 29, 11:21 am, ehsjr <eh...(a)nospamverizon.net> wrote: > Chris W wrote: > > I want to make a load center to test power supplies and batteries. I > > was thinking of using 50 Watt 4 ohm resistors for 12V loads but I will > > need 15 of them to get the current drain I want. I would also like to > > load 5V and 3.3V lines and of course that would require different > > resistors. > > > I was wondering if this wouldn't be a lot easier with a power > > transistor. The 50 Watt resistors are going to cost a little over $3 > > each and I will probably need 30 of them to get the loads I want. > > > The goal is to have a variable load of about 3 to 50 amps on as much as > > 14V and from about 1 to 25 amps on 5V and 3.3V. Can someone recommend a > > specific transistor that would work good? I am hoping I can do it with > > fewer transistors. I do plan on using a large heat sink and fan to keep > > this cool. > > > Thanks, > > Chris W > > 100 Amp 6 Volt/12 Volt Battery Load Tester > > Item # 90636 at Harbor > Freighthttp://www.harborfreight.com/100-amp-6-volt-12-volt-battery-load-test... > > On sale now for $19.99 > > Use as is, or use the element as a load resistor in whatever > circuit you design. Using it as is will save you $$, burned > out power transistors, large heat sinks etc - and the need > for Joerg to provide sound effects for circuit demise. > > Ed- Hide quoted text - > > - Show quoted text - 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.
From: John Fields on 30 Apr 2010 09:01 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
From: baron on 30 Apr 2010 09:20 whit3rd Inscribed thus: > I've seen (for laboratory magnets) arrays of dozens of 2N3055 > transistors, on copper plate heatsink with a soldered-on tube > for cooling water... NOT pretty. 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.
From: ehsjr on 30 Apr 2010 12:49 Herman wrote: > "sparky" <sparky12x(a)yahoo.com> wrote in message > news:6109a09c-0ab8-448b-8fc3-2fe0944b7967(a)w36g2000yqw.googlegroups.com... > On Apr 29, 11:21 am, ehsjr <eh...(a)nospamverizon.net> wrote: > >>Chris W wrote: >> >>>I want to make a load center to test power supplies and batteries. I >>>was thinking of using 50 Watt 4 ohm resistors for 12V loads but I will >>>need 15 of them to get the current drain I want. I would also like to >>>load 5V and 3.3V lines and of course that would require different >>>resistors. >> >>>I was wondering if this wouldn't be a lot easier with a power >>>transistor. The 50 Watt resistors are going to cost a little over $3 >>>each and I will probably need 30 of them to get the loads I want. >> >>>The goal is to have a variable load of about 3 to 50 amps on as much as >>>14V and from about 1 to 25 amps on 5V and 3.3V. Can someone recommend a >>>specific transistor that would work good? I am hoping I can do it with >>>fewer transistors. I do plan on using a large heat sink and fan to keep >>>this cool. >> >>>Thanks, >>>Chris W >> >>100 Amp 6 Volt/12 Volt Battery Load Tester >> >>Item # 90636 at Harbor >>Freighthttp://www.harborfreight.com/100-amp-6-volt-12-volt-battery-load-test... >> >>On sale now for $19.99 >> >>Use as is, or use the element as a load resistor in whatever >>circuit you design. Using it as is will save you $$, burned >>out power transistors, large heat sinks etc - and the need >>for Joerg to provide sound effects for circuit demise. >> >>Ed- Hide quoted text - >> >>- Show quoted text - > > > 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. > > 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
From: whit3rd on 30 Apr 2010 13:47
On Apr 29, 4:24 pm, Chris W <1qaz...(a)cox.net> wrote: > 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. 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. |