Calling all Electronic Engineers
13 February 2008, 08:26 PM
#1
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Calling all Electronic Engineers
If I have a power source of 55w (controlled by a Computer) and a device that has a power consumption of 35w how can I trick the computer to think the power consumption is still 55w instead of 35w?
been recommended add a capacitor wired in parallel with the device consuming 35w
if a capacitor is required do i use a 1000uf or 4700uf? (those numbers mean nothing to me
)
Don't want a huge capacitor
been recommended add a capacitor wired in parallel with the device consuming 35w
if a capacitor is required do i use a 1000uf or 4700uf? (those numbers mean nothing to me
)Don't want a huge capacitor
13 February 2008, 08:42 PM
#2
If the power source is regulated and doesnt require a feedback from the load then it doesnt matter. the 55w power source will supply as much power as the load needs eg 35w
Andy
Btw adding a capacitor in parallel will only draw a current whilst it charges to the supply voltage assuming its DC, this will take a few milliseconds
Andy
Btw adding a capacitor in parallel will only draw a current whilst it charges to the supply voltage assuming its DC, this will take a few milliseconds
13 February 2008, 08:53 PM
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eeemmmmm
The problem that i'm gonna have is that computer detects less power is being use so it automatically thinks the device is broke (original device was 55w which is the same as the power output)
The problem that i'm gonna have is that computer detects less power is being use so it automatically thinks the device is broke (original device was 55w which is the same as the power output)
13 February 2008, 08:54 PM
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13 February 2008, 09:02 PM
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It sounds like the power supply requires some form of feedback or the tolerances on the power supply start up are very carefully monitored. You would need to obtain a circuit diagram to see what is happening and then have someone devise a solution. Or just get the original 55w load replaced.
13 February 2008, 10:30 PM
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More information needed I'm afraid 
We need to know what the load is and how the computer is monitoring the power being delivered to it. For example, if the load is resistive, and the power supply delivers a constant voltage, then most likely the computer is (somehow) measuring the current in the circuit.
If that's the case then the easiest fix would be to wire a dummy load in parallel with the real one to draw the extra 20W. Measure the voltage across the load, and then add a resistor R in parallel, where the value of R is given by V^2 / 20 (where V is the voltage across the load). Make sure you use a resistor rated for at least 20W dissipation, providing a heatsink if necessary. It will get hot.
This works for both ac and dc; use RMS values for voltage and current if it's an ac circuit.
If you know how the computer is measuring the load current, you could try and recalibrate it to over-read by a factor of 55/35 (11/7). For example, if it's sensing the voltage across a low value series resistor R, change R for one with a value 11R/7. But how practical that actually is depends on how the monitoring is done, whether or not you have or can trace a schematic, and what else that sensor is connected to. Without knowing exactly what the equipment is, I wouldn't suggest tinkering unless you're 100% you know what you're doing.
We need to know what the load is and how the computer is monitoring the power being delivered to it. For example, if the load is resistive, and the power supply delivers a constant voltage, then most likely the computer is (somehow) measuring the current in the circuit.
If that's the case then the easiest fix would be to wire a dummy load in parallel with the real one to draw the extra 20W. Measure the voltage across the load, and then add a resistor R in parallel, where the value of R is given by V^2 / 20 (where V is the voltage across the load). Make sure you use a resistor rated for at least 20W dissipation, providing a heatsink if necessary. It will get hot.
This works for both ac and dc; use RMS values for voltage and current if it's an ac circuit.
If you know how the computer is measuring the load current, you could try and recalibrate it to over-read by a factor of 55/35 (11/7). For example, if it's sensing the voltage across a low value series resistor R, change R for one with a value 11R/7. But how practical that actually is depends on how the monitoring is done, whether or not you have or can trace a schematic, and what else that sensor is connected to. Without knowing exactly what the equipment is, I wouldn't suggest tinkering unless you're 100% you know what you're doing.
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13 February 2008, 11:21 PM
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If that's the case then the easiest fix would be to wire a dummy load in parallel with the real one to draw the extra 20W. Measure the voltage across the load, and then add a resistor R in parallel, where the value of R is given by V^2 / 20 (where V is the voltage across the load). Make sure you use a resistor rated for at least 20W dissipation, providing a heatsink if necessary. It will get hot.
.
.
This seems to be the easiest option however I wouldnt have a clue how to spec a resister
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