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I want to bench test a power supply. The Sam's computer facts troubleshooting manual says to use an #1129 on the +5v rail as a load. I would rather use a resistor. The lamp seems to have specs of max 6.4v 2.63A 16.83W. I can't find a resistance in a spec sheet.

How can I figure out which resistor to use in similar future circumstances?

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  • \$\begingroup\$ Are you sure they don't want the lamp for its tempco characteristic? \$\endgroup\$ Nov 17, 2017 at 23:28

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It probably draws a bit more than 2A from the 5V line. If the bulb acted as a resistor it would draw (5/6.4)*2.63 = 2.05A, but a bulb does not have a constant resistance- it drops as the voltage drops and the filament temperature changes.

I would think your goal here is to load the power supply near to its nameplate rating. If that is 2A you should load it to 2A. If that's 2.5A, load it to 2.5A.

In which case, the resistor you require is simply 5/I where I is the current. 2.5 ohms for a 2A current (if that's correct). Power dissipation is, of course, 25/R or 5V * I, so 10W typically for a 2A current. So you can use a resistor like the below 12.5W type and mount it to a heatsink.

enter image description here

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  • \$\begingroup\$ I have found a couple service manuals but I'm having trouble finding the exact max design power of the supply on a per-rail basis with which to do the calcs you suggest, thus I was hoping to infer a correct resistor value from the suggestion to use the bulb. \$\endgroup\$
    – nexus_2006
    May 9, 2017 at 18:35
  • \$\begingroup\$ A w.a.g. is that the bulb is 2.2A +/-10%. So the 2.5R resistor is probably close enough. \$\endgroup\$ May 9, 2017 at 19:18
  • \$\begingroup\$ The Commodore power supplies I've seen have the output power for the 5V rail molded into the bottom or on a label. If it says 8.5W @ 5.0V the resistor required is more like 3 ohms. \$\endgroup\$ May 10, 2017 at 17:14
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6.4V/2.6A= 2.5 ohms

but since bulbs are 10% R when at room temp

But reducing V+ @6.4V to 5V does not drop the current in a linear fashion due to the 10:1 PTC effect.

Light bulbs tend toward constant current sinks when V is reduced below rated due to PTC coefficient when hot.

R= 0.25 ohms cold then rises quickly to 2.5 ohms. How quick depends on size.

in series with a large C to simulate T rise time , which is probably 100ms or so.

  • an ultralow ESR tantalum of 100uF may get down below 1 Ohm ESR to simulate turning on this bulb with inrush current up to 26A from an ideal supply, but obviously limited in your case by it's current limiter or RdsOn+DCR(L)

You can also use a heatsunk power transistor darlington with a 100mV or o current shunt on emitter, as an active load with a pot to bias input current. Large caps aren't really necessary in a well designed Buck regulator. ...just really low ESR small caps.

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  • \$\begingroup\$ Commodore supplies were all linear types. Hence the dual outputs of regulated 5VDC and 9VAC @50/60Hz. \$\endgroup\$ May 10, 2017 at 17:13

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