Transistor Or Resistor

[Chris W]

sparky wrote:

On Apr 29, 11:21 am, ehsjr <eh...@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.

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

Chris W

 

[sparky]

On Apr 29, 11:21 am, ehsjr <eh...@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.

 

[Jasen Betts]

On 2010-04-29, Baron <baron.nospam@linuxmaniac.nospam.net> wrote:

gearhead Inscribed thus:

On Apr 29, 8:10 am, baron <baron.nos...@linuxmaniac.nospam.net> wrote:

Which ever way you go you are going to have to get rid of around 700
watts of heat.  Water cooling is starting to look good.



Absotively posolutely.
I used to have a dummy load for tuning the output tubes on my old Swan
350.
It was a fat-ass resistor in a paint can that you fill with oil.
OP can kluge up his own liquid-cooled load, but speaking from
experience the water gets nasty.

I agree it does if its static. Actually I was thinking along the lines
of flowing water through a tank or tube containing the resistors.

Which is probably why people prefer oil, even though thermally it
probably doesn't quite have the performance of water. Plus oil
doesn't evaporate. But for something just to bang together for a
quick test, I just dunk a power resistor in a yogurt container or
something.
And as others have pointed out, there are plenty of other options,
like incandescent lamps, hair dryers and such.
OP should be thinking about how to get rid of heat, not how to hook up
a lot of transistors. In this situation, the only sensible use of
transistors is as switches in full saturation (for bjt's) or triode
mode (mosfets). He doesn't even know how to interpret the thermal
info on a datasheet properly.

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



--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[Jasen Betts]

tOn 2010-04-29, Chris W <1qazse4@cox.net> wrote:

The ideal situation would be to have a single pot that I could use to
adjust the load.

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.


--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[Herman]

"sparky" <sparky12x@yahoo.com> wrote in message
news:6109a09c-0ab8-448b-8fc3-2fe0944b7967@w36g2000yqw.googlegroups.com...
On Apr 29, 11:21 am, ehsjr <eh...@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.

 

[John Fields]

On Thu, 29 Apr 2010 18:24:43 -0500, Chris W <1qazse4@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

 

[baron]

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.

 

[ehsjr]

Herman wrote:

"sparky" <sparky12x@yahoo.com> wrote in message
news:6109a09c-0ab8-448b-8fc3-2fe0944b7967@w36g2000yqw.googlegroups.com...
On Apr 29, 11:21 am, ehsjr <eh...@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

 

[whit3rd]

On Apr 29, 4:24 pm, Chris W <1qaz...@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.

 

[Chris W]

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.

 

[Chris W]

John Fields wrote:

On Thu, 29 Apr 2010 18:24:43 -0500, Chris W <1qazse4@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

 

[Michael A. Terrell]

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.

 

[Jasen Betts]

On 2010-04-30, Chris W <1qazse4@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@netfront.net ---

 

[John Fields]

On Fri, 30 Apr 2010 18:18:57 -0500, Chris W <1qazse4@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

 

[John Fields]

On Sat, 01 May 2010 08:07:41 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Fri, 30 Apr 2010 18:18:57 -0500, Chris W <1qazse4@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.

---
Slow Saturday:

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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
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SYMATTR Value 2.4
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SYMATTR InstName R9
SYMATTR Value 2.4
TEXT -1186 450 Left 0 !.tran 2

JF

 

[ehsjr]

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.

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<--[330R]--| | | |
T \e | | |
| | /c /c /c
| +-----------|---|---}}--|
| \e \e \e
| | | |
| [R] [R] [R]
| | | |
Gnd -+--------------------------+---+--}}---+

2N3055's were mentioned in the thread, so we'll discuss that.
Say you use 10 2N3055's. At max, each would provide 5
amps to draw 50 amps through the element, but the voltage
drop across the transistor would be small, around 1 volt.
So the transistors would each dissipate about 5 watts,
the emitter resistors would each dissipate about 2.5 watts,
while the element dissipated the other ~650 watts. The
element could be the 2 series Harbor Freight testers,
wirewound resistors, nicrome wire, whatever.

For 50A you'll need about 4 amps to drive the transistors' base
circuit (all 10 2N3055 bases connected together), so a
total of 11 2N3055 power transistors. And they will need
to be properly heatsinked. Note that the driver transistor -
the left hand 2N3055 - does not have an emitter resistor. Its
emitter is connected to drive the bases of the other 2N3055's.
You can drive the base of the driver transistor through a
330 ohm resistor connected to the wiper of a 500 ohm 1watt pot,
connected across the supply, or circuit of your choice that
can provide up to 40 mA to the base of the driver transistor.

You need that or similar power transistor circuit to get the
1 potentiometer variability you wanted. Note that at lower
current, heat in the transistors will increase. For example,
say you want to test at ten amps. Each load transistor will
supply 1 amp. The voltage drop in the element will be 2.8
volts, and the drop in the emitter resistors will be ~.5
volts, leaving roughly 10.7 volts across each transistor.
Ignoring the driver transistor, that's about 10.7 watts in
each of the 10 2N3055 load transistors or a total of ~112
watts to get rid of. The element would dissipate only ~28
watts in that case. The point here is that there will be a
lot of heat in the transistors. Your design must take that
into account. Also, 2.5 watts (worst case) in each emitter
resistor is nothing to sneeze at - they'll be plenty hot
when you run the thing at max. Touch one, and you'll get
burned. So, you need a big heatsink, forced air cooling
(or better) and a cage to prevent accidently touching
any of the hot components.

Your mention (below) of a 2 or 3 hour test at reduced current
implies unattended testing. That calls for more than the
simple circuit described so far. Already shown is the change
in the location of maximum heat. At max, the majority of the
heat is generated in the element while at minimum the majority
of the heat is generated in the transistor circuit.

Any "non-brief" testing implies unattended testing, and means
you need to add some form of safety to the design, because no
one's there to keep things under control. For example, say the
driver transistor shorts out during the long term test. What
was supposed to be a 10 amp (for example) load suddenly becomes
a 50 amp load. That's not good news for the "home team" and in
fact could burn down the home.

That's the primary reason I recommended testing using the HF
testing device as designed to be used in my initial post.
You're talking a 50 amp current. That is not to be trifled
with. What is simple at low current becomes far less trivial
at high currents.

Aside from destroying the circuit itself and/or burning down
the house, an unattended high current load test can destroy the
battery or supply being tested, if things go awry or if you
overlook something.

Example: you set up your load to draw 10 amps from a battery.
Something happens, and you don't get back to the thing in
2-3 hours. You could permanently damage the battery that way
by discharging too far. That means you need an automatic
shut off designed into the load circuit. Obviously, you need
some form of thermal shut down if it gets too hot. And, what
happens if one transistor shorts or opens? One opening could
cause a cascade effect where others open or emitter resistors
die. Nothing in the simple circuit is protected.

Ed

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.

 

[John Fields]

On Sat, 01 May 2010 08:07:41 -0500, John Fields
jfields@austininstruments.com> wrote:

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.

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


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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

 

[John Fields]

On Fri, 30 Apr 2010 22:33:07 -0400, "Michael A. Terrell"
<mike.terrell@earthlink.net> wrote:

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

---
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

 

[Michael A. Terrell]

John Fields wrote:

On Fri, 30 Apr 2010 22:33:07 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


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

---
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."

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.


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