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You are here: MacNN Forums > Hardware - Troubleshooting and Discussion > Mac Notebooks > MPC7447 - 1 volt x 8 amps is 8 watts

MPC7447 - 1 volt x 8 amps is 8 watts
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aehaas
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Jul 31, 2003, 03:45 PM
 
Nobody ever discusses the fact that these little powerbook chips are drawing 8 to 12 amps. How do they get that out of those 14 volt batteries? Are they running individual cells in parallel?

aehaas
     
Eug
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Jul 31, 2003, 03:48 PM
 
You could as the same question about 5/12 V power supplies, connected to 115 V AC outlets, and 1.5 V desktop CPUs.

Anyways, the MPC7447/7457 is 15.8 Watts at 1 GHz, at 1.3 V I think. The MPC7457RX1000NB is 8.3 Watts at 1 GHz, at 1.1 V.
     
crispinwilliams
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Jul 31, 2003, 06:23 PM
 
If you have a 14V battery but only need 1 V - then you need a circuit to convert the 14V to 1V.. A simple way would be just to use a resistor - drop 13V across the resistor.. but thos is grossly stupid way to do it - you would waste all your power in heat in the resistor.. so what you do is have a switch that you turm on for an instance - and then turn off again... and use this current pulse to charge up a capacitor. By monitoring the voltage on this capacitor - the circuit decides when to open the switch and give another current pulse. Assuming that you design this circuit efficiently -(ie no power wasted in the switch) then you would end up drawing a current of 600 mA at 14 V to give you 8A at 1V... still a lot of power - and the circuitry and battery has to be able to deal with spikes of large current ..
     
neutrino23
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Jul 31, 2003, 07:38 PM
 
Originally posted by crispinwilliams:
If you have a 14V battery but only need 1 V - then you need a circuit to convert the 14V to 1V.. A simple way would be just to use a resistor - drop 13V across the resistor.. but thos is grossly stupid way to do it - you would waste all your power in heat in the resistor.. so what you do is have a switch that you turm on for an instance - and then turn off again... and use this current pulse to charge up a capacitor. By monitoring the voltage on this capacitor - the circuit decides when to open the switch and give another current pulse. Assuming that you design this circuit efficiently -(ie no power wasted in the switch) then you would end up drawing a current of 600 mA at 14 V to give you 8A at 1V... still a lot of power - and the circuitry and battery has to be able to deal with spikes of large current ..
A switch and a capacitor might be used to step down voltage but it won't step up the current.

A DC-DC converter is needed. This will send pulses of electricity (by an oscillator or the device you mention) through one side of a transformer. This device can change Voltage and Current. The product of V * I on the input is the same as V * I at the output. The ratio of input coils to output coils determines how these change from input to output. Other circuits at the output change the AC signal to DC and filter it.

Engineers have been working on this sort of power converter for battery operated equipment for a long time. It is a very sophisticated technology.
Happy owner of a new 15" Al PB.
     
Gametes
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Aug 1, 2003, 12:51 AM
 
Ya, my powerbook burns about a watt a minute. And it certainly feels like it too
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