Watts

Is there any advantage to using a 120watt adapter made for the 17" PowerBooks, instead of the 70watt, which should also be adequate on the 15"?
-oo0-GoldTrader-0oo- wrote on :

All of this time I thought Watts was a drummer in the greatest rock & roll show in the world.

I need a twelve-volt adapter to run my 15" Powerbook. THe last one I had was 45 watts. It worked OK until I started burning CD's. Then it started smoking and smelled of electrical mishap.

Is there any advantage to using a 120watt adapter made for the 17" PowerBooks, instead of the 70watt, which should also be adequate on the 15"?

What uses more juice?

Would the 120watt charge faster than the 70?

Would the 120watt put a greater demand on my 12-volt source?

As an aside, Is China in partnership with IBM on that Thinkpad thing?

zara replied on :

"-oo0-GoldTrader-0oo-" nomads_05@redacted.invalid wrote in message news:1122160440.016167.153240@redacted.invalid

All of this time I thought Watts was a drummer in the greatest rock & roll show in the world.

I need a twelve-volt adapter to run my 15" Powerbook. THe last one I had was 45 watts. It worked OK until I started burning CD's. Then it started smoking and smelled of electrical mishap.

Is there any advantage to using a 120watt adapter made for the 17" PowerBooks, instead of the 70watt, which should also be adequate on the 15"?

What uses more juice?

Would the 120watt charge faster than the 70?

Would the 120watt put a greater demand on my 12-volt source?

As an aside, Is China in partnership with IBM on that Thinkpad thing?

China is partners with everyone on everything. Learn Chinese

Matthew Kirkcaldie replied on :

In article 1122160440.016167.153240@redacted.invalid, "-oo0-GoldTrader-0oo-" nomads_05@redacted.invalid wrote:

As an aside, Is China in partnership with IBM on that Thinkpad thing?

No, IBM sold their PC business to a company called Lenovo. So ThinkPads now come from Lenovo, not IBM.

  MK.
Richard Johnson replied on :

"-oo0-GoldTrader-0oo-" nomads_05@redacted.invalid wrote in message news:1122160440.016167.153240@redacted.invalid

All of this time I thought Watts was a drummer in the greatest rock & roll show in the world.

I need a twelve-volt adapter to run my 15" Powerbook. THe last one I had was 45 watts. It worked OK until I started burning CD's. Then it started smoking and smelled of electrical mishap.

Is there any advantage to using a 120watt adapter made for the 17" PowerBooks, instead of the 70watt, which should also be adequate on the 15"?

What uses more juice?

Would the 120watt charge faster than the 70?

Would the 120watt put a greater demand on my 12-volt source?

As an aside, Is China in partnership with IBM on that Thinkpad thing?

Power (in Watts) is equal to current (Amperes) times the Voltage (Electric Force) or P = I X E. (I = Amperes) (E = Volts) (P = Watts)

All the rating in Watts says about a power source or supply is that is the maximum power it is able to deliver, reliably. That power is used by the battery charging system to charge the battery. The battery is only able to absorb power at a specific rate, otherwise it can burn or short out. The battery charging circuit handles that aspect of things and all it needs is the maximum power the battery can accept and stay whole. Putting a larger power source (Watts, not volts) will do nothing to increase the charge time. The normally delivered power supply will handle any surge required by the load (notebook) for normal operation. It is likely that if the notebook manufacturer has delivered a 70 Watt power supply, that is all that is needed. Using a more expensive 120 Watt unit will not likely aid the notebook in any way, nor charge the battery faster. The only gain is in the life of the supply as you will be under using it, thus keep the heat generated by the components down.

dave stanton replied on :

The only gain is in the

life of the supply as you will be under using it, thus keep the heat generated by the components down.

Which is not a bad thing at all.

Dave

Obfus Kataa replied on :

Sun, 24 Jul 2005 (07:21 +0100 UTC) dave stanton wrote:

The only gain is in the

life of the supply as you will be under using it, thus keep the heat generated by the components down.

Which is not a bad thing at all.

Dave

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

Smitty replied on :

In article Pine.OSX.4.63.0507241016060.19458@redacted.invalid, Obfus Kataa vapaa@redacted.invalid wrote:

Sun, 24 Jul 2005 (07:21 +0100 UTC) dave stanton wrote:

The only gain is in the

life of the supply as you will be under using it, thus keep the heat generated by the components down.

Which is not a bad thing at all.

Dave

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

In general, I'd go along with that. I'd substitute the word "dissipate" for the word "withstand." Electronic devices generally produce heat as an unwanted byproduct, and use large (often finned to increase surface area) chunks of metal (heat sinks) to dissipate the heat to the outside world, as well as fans and ventilation slots.

The "efficiency" of the device in question determines the percentage of power that's wasted in heat generation. So if you're comparing two things with the same efficiency, I'd totally agree with you. The higher rated power supply would feel cooler to the touch not because it's producing less heat, but because it's dissipating it better.

Ross Bernheim replied on :

Smitty prestwhich@redacted.invalid wrote:

In article Pine.OSX.4.63.0507241016060.19458@redacted.invalid, Obfus Kataa vapaa@redacted.invalid wrote:

Sun, 24 Jul 2005 (07:21 +0100 UTC) dave stanton wrote:

The only gain is in the

life of the supply as you will be under using it, thus keep the heat generated by the components down.

Which is not a bad thing at all.

Dave

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

In general, I'd go along with that. I'd substitute the word "dissipate" for the word "withstand." Electronic devices generally produce heat as an unwanted byproduct, and use large (often finned to increase surface area) chunks of metal (heat sinks) to dissipate the heat to the outside world, as well as fans and ventilation slots.

The "efficiency" of the device in question determines the percentage of power that's wasted in heat generation. So if you're comparing two things with the same efficiency, I'd totally agree with you. The higher rated power supply would feel cooler to the touch not because it's producing less heat, but because it's dissipating it better.

Efficiency in a switching power supply such as used in this case varies with the load. Generally it is highest near the rated output and is less efficient at lower loads. The result is that the larger rated supply used at a lower power will quite often be less efficient and dissapate greater heat and run hotter!

Get the supply reccomended by the manufacturer.

Ross

Smitty replied on :

In article 1h08o9r.1t035px17cwjkyN%rossbernheim@redacted.invalid, rossbernheim@redacted.invalid (Ross Bernheim) wrote:

Efficiency in a switching power supply such as used in this case varies with the load. Generally it is highest near the rated output and is less efficient at lower loads. The result is that the larger rated supply used at a lower power will quite often be less efficient and dissapate greater heat and run hotter!

Get the supply reccomended by the manufacturer.

Ross

Hmm, I didn't know that about switching power supplies. I wonder whether that's theoretical or real-world, or both? In other words, if the efficiency of a larger supply drops because it's not running near capacity, would the increased heat dissipating capability built into it fall short of compensating for the extra heat that's being generated? If the efficiency drops from 70% to 65% when you go from 90% load to 50% load, let's say, then I wouldn't think the larger supply would really run hotter to the touch than a smaller supply running closer to max. Or does efficiency drop much more radically than that? (I realize we're talking about something with a whole lot of variables, here.)

However, the OP is talking about a "wal-wart" type adapter for a powerbook. Those aren't really switchers, are they? I thought they were unregulated, barely filtered linears. But, I'm often wrong about things...

Since the OP's previous adapter went up in smoke, I don't blame him for wanting to buy a king-daddy replacement.

dave stanton replied on :

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

I mean anything which contains electronic components should be run at the lowest temp you can to prevent premature failure.

Dave

Fred Moore replied on :

In article pan.2005.07.25.18.13.01.273661@redacted.invalid, dave stanton me@redacted.invalid wrote:

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

I mean anything which contains electronic components should be run at the lowest temp you can to prevent premature failure.

And if I remember correctly, integrated circuit failure increases with the fourth power of absolute temperature.

--Fred

Barry Watzman replied on :

Using a higher power adapter is always fine as long as it has the correct voltage, connector and polarity.

The adapter does not determine the amount of power used, the laptop does. The adapter should be an "all you can eat buffet" for the laptop .... "here is 15 volts, take as much as you want, UP TO {maximum amount}.

With regard to {maximum amount}, watts = volts * amps

So, a 60 watt 15-volt adapter is capable of supplying any amount of current UP TO 4 amps. What it does, however, is offer the laptop 15 volts, and the laptops decides how much it will take, again, UP TO the limit of the adapter (4 amps for a 60 watt, 15-volt adapter). If the laptop only takes 3 amps, then the adapter, fully capable of operating at 60 watts, is only actually operating at 45 watts.

-oo0-GoldTrader-0oo- wrote:

All of this time I thought Watts was a drummer in the greatest rock & roll show in the world.

I need a twelve-volt adapter to run my 15" Powerbook. THe last one I had was 45 watts. It worked OK until I started burning CD's. Then it started smoking and smelled of electrical mishap.

Is there any advantage to using a 120watt adapter made for the 17" PowerBooks, instead of the 70watt, which should also be adequate on the 15"?

What uses more juice?

Would the 120watt charge faster than the 70?

Would the 120watt put a greater demand on my 12-volt source?

As an aside, Is China in partnership with IBM on that Thinkpad thing?

Ross Bernheim replied on :

Smitty prestwhich@redacted.invalid wrote:

However, the OP is talking about a "wal-wart" type adapter for a powerbook. Those aren't really switchers, are they? I thought they were unregulated, barely filtered linears. But, I'm often wrong about things...

Many of the wall wart supplies for computers and even external drives are small switchers these days. The majority of the electronics have been shrunk to a single inexpensive chip and the requisite inductor and switching transistor are relatively small and inexpensive.

The old transformer based linear supplies are rapidly disappearing due to the high cost of the transfomers and the shipping weight. Also the switchers can relatively easily be made 110/220 auto sensing.

Ross

I've been a pedestrian ever since I could walk. (Graffiti: Gene Mora)

Clark Martin replied on :

In article 1h08o9r.1t035px17cwjkyN%rossbernheim@redacted.invalid, rossbernheim@redacted.invalid (Ross Bernheim) wrote:

Smitty prestwhich@redacted.invalid wrote:

In article Pine.OSX.4.63.0507241016060.19458@redacted.invalid, Obfus Kataa vapaa@redacted.invalid wrote:

Sun, 24 Jul 2005 (07:21 +0100 UTC) dave stanton wrote:

The only gain is in the

life of the supply as you will be under using it, thus keep the heat generated by the components down.

Which is not a bad thing at all.

Dave

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

In general, I'd go along with that. I'd substitute the word "dissipate" for the word "withstand." Electronic devices generally produce heat as an unwanted byproduct, and use large (often finned to increase surface area) chunks of metal (heat sinks) to dissipate the heat to the outside world, as well as fans and ventilation slots.

The "efficiency" of the device in question determines the percentage of power that's wasted in heat generation. So if you're comparing two things with the same efficiency, I'd totally agree with you. The higher rated power supply would feel cooler to the touch not because it's producing less heat, but because it's dissipating it better.

Efficiency in a switching power supply such as used in this case varies with the load. Generally it is highest near the rated output and is less efficient at lower loads. The result is that the larger rated supply used at a lower power will quite often be less efficient and dissapate greater heat and run hotter!

It will typically be less efficient and dissapate more heat but it may run cooler since it's designed to handle a higher overall power dissipation.

J. Clarke replied on :

Obfus Kataa wrote:

Sun, 24 Jul 2005 (07:21 +0100 UTC) dave stanton wrote:

The only gain is in the

life of the supply as you will be under using it, thus keep the heat generated by the components down.

Which is not a bad thing at all.

Dave

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

The thermal energy dissipated would be the same, but the 2500 watt heater, if it is a typical design, would be running at a lower temperature than the 1500 watt when both were set for 900 watts.

It is not heat, per se, that is bad for electronics, it is temperature sustained over time. A 200 watt power supply that is being used to power a device drawing 20 watts will typically operate at a much lower temperature than a 20 watt supply being used to power the same device, and so may be expected to last longer.

Barry Watzman replied on :

Your statement is not necessarily true. A 2500 watt heater and a 1500 watt heater both set to 900 watts will produce the same amount of heat. Now IF the 2500 watt model is physically larger, it's temperature may be lower. Maybe. But the problem doesn't address that and, in fact, both heaters may well be the same size, or the 1500 watt heater might even be bigger. Your conclusion "A 200 watt power supply that is being used to power a device drawing 20 watts will typically operate at a much lower temperature than a 20 watt supply being used to power the same device" may be true, but it isn't necessarily or automatically true in all cases.

J. Clarke wrote:

Obfus Kataa wrote:

Sun, 24 Jul 2005 (07:21 +0100 UTC) dave stanton wrote:

The only gain is in the

life of the supply as you will be under using it, thus keep the heat generated by the components down.

Which is not a bad thing at all.

Dave

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

The thermal energy dissipated would be the same, but the 2500 watt heater, if it is a typical design, would be running at a lower temperature than the 1500 watt when both were set for 900 watts.

It is not heat, per se, that is bad for electronics, it is temperature sustained over time. A 200 watt power supply that is being used to power a device drawing 20 watts will typically operate at a much lower temperature than a 20 watt supply being used to power the same device, and so may be expected to last longer.

J. Clarke replied on :

Barry Watzman wrote:

Your statement is not necessarily true. A 2500 watt heater and a 1500 watt heater both set to 900 watts will produce the same amount of heat. Now IF the 2500 watt model is physically larger, it's temperature may be lower. Maybe. But the problem doesn't address that and, in fact, both heaters may well be the same size, or the 1500 watt heater might even be bigger. Your conclusion "A 200 watt power supply that is being used to power a device drawing 20 watts will typically operate at a much lower temperature than a 20 watt supply being used to power the same device" may be true, but it isn't necessarily or automatically true in all cases.

Barry, I mean you no offense, but you're being pedantic.

And I'm not going to write a treatise on heater design or power supply design here just to satisfy your need for minute cataloging of caveats.

Bottom line, unless there's something really wonky about the design, for a given output a big power supply will run at a lower temperature than a little power supply.

Yes, there are exceptions, but you are not likely to find them on the shelf at CompUSA.

J. Clarke wrote:

Obfus Kataa wrote:

Sun, 24 Jul 2005 (07:21 +0100 UTC) dave stanton wrote:

The only gain is in the

life of the supply as you will be under using it, thus keep the heat generated by the components down.

Which is not a bad thing at all.

Dave

The heat generated to produce a given wattage is dependent on the capacity of the device producing the wattage? Does that mean that a heater capable of 2500 W. set at 900 W. produces less heat than does a 1500 W. heater set for 900 W.?

I think the heat generated would be the same, but the higher-rated device would be designed to withstand greater heat than would the lower-rated device.

The thermal energy dissipated would be the same, but the 2500 watt heater, if it is a typical design, would be running at a lower temperature than the 1500 watt when both were set for 900 watts.

It is not heat, per se, that is bad for electronics, it is temperature sustained over time. A 200 watt power supply that is being used to power a device drawing 20 watts will typically operate at a much lower temperature than a 20 watt supply being used to power the same device, and so may be expected to last longer.