Mouse over the plugs for pinouts and voltages
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Keep in mind that power is measured in WATTS and that an increase in power consumption implies there has been an increase in WATTAGE.
If a resistor is used to add a load to a circuit, the power consumed will result in heat in the resistor and heat dissipation can become a problem.
If the power level becomes too high (resistors are manufactured with a stated resistance (OHMs) and WATTAGE capacity), it may result in an alteration of the
internal resistance or, in the worst case, the resistor burning out.
Power (watts) is calculated as P = V2 / R. Thus, increases in the power level
may be accomplished by increasing the voltage or by reducing the resistance. Higher voltage levels certainly would seem intuitively reasonable, but REDUCING THE RESISTANCE
to raise power comsumption??!!
AAAARRRRRRRGGGGGHHHHH -- This is too much like school, just tell me what to do!!!
If I wanted to increase the load in a circuit, wouldn't it seem more feasible to increase the resistance? This seeming paradox may be better explained by looking at the extremes.
By definition, a resistor restricts the electron flow through the circuit. If we can reduce the electron flow, we can also reduce the amount of work being performed over a period of time.
A high level of resistance allows only a small number of electrons to migrate -- materials with extremely high resistance levels are known simply as insulators -- glass, air, rubber, plastic, etc.
If I inserted a piece of glass into my circuit and attached the battery leads to each end, nothing would happen -- high resistance, no electron movement, no load and no heat. On the other hand, materials with very low resistance
are known as conductors; the lower the resistance, the better the conductor -- silver, copper, gold, aluminum. Placing a thin silver wire across my battery leads might result in the
wire beginning to glow or burning completely in two, depending of course on the battery capacity. Hence, low resistance leads to high electron flow, high load and possibly high heat generation.
The physics of heat generation is enmeshed in free electron flow, valence shell electrons and other factors that will remain outside this discussion.
It is suffice to say that using a resistor with a lower resistance rating will add a greater load to the circuit and produce increasingly higher levels of heat, either within the resistor itself
or in other parts of the circuit.
Many within the R/C community have been converting power supplies to drive their field chargers, but have not been satisfied with
the measured voltage on the +12 V output -- in several cases, the voltage levels have been lower than expected leading to long charge times, power supply shutdowns or charger
malfunctions. Relying on a little help from Mr. Ohm, Watt and Kirchhoff, we may be able to increase the amount of work expected from our PC power supply, and in turn, get it to step
up to the plate with an increased voltage level. The ATX switching mode power supplies require a static load to function and for many, a 10 ohm 10 watt resistor on the +5 V output
is sufficient, but voltage levels on the +12 V line may fall in the range of 11.5 to 11.75 volts, adequate for logic labs, but below desired levels for chargers. Five volt output generally
holds around 5.09 to 5.15 volts.
Returning to our power calculation, a 10 ohm load across 5.15 volts consumes about 2.65 watts of energy. P = V2 / R = 5.152 / 10 ~ 2.65 watts. By
virtue of being a regulated power supply, we cannot easily change the voltage levels -- the ATX power supply design guide specifies a +/- 5% variation for the 3.3, 5 and 12 volt outputs
and the internal circuitry is designed to maintain output within those specifications.
Consequently, to increase the perceived load, the most easily controlled variable is resistance. If we replace the 10 ohm resistor with a 2 ohm resistor, what change can we expect
in the load? As before, P = V2 / R = 5.152 / 2 ~ 13.26 watts. This substitution has certainly increased the load, but it has also introduced another
potentially destructive situation. The load resistor must now dissipate this increased energy and it does so in the form of heat. We can deal with heat to some extent, but when the load
exceeds the rating of the resistor, our best efforts may not save a 10 watt resistor carrying a continuous 13 watt load. My options are to find a whopper of a resistor with a high load rating or come up
with a workable solution using easily obtainable components.
Four laws will help us ramp up the perceived load and use off-the-shelf parts to successfully implement them.
1. The total resistance in a SERIES circuit equals the sum of the individual resistances: Rtotal = R1 + R2 + ... + Rn
2. The sum of voltage drops in a series circuit will equal the voltage source: Vs = Vd1 + Vd2 + ... + Vdn
3. The voltage drop across a resistor in a series circuit is directly proportional to the size (Ohm rating) of the resistor: Vd1 = Vs x R1 / Rtotal
4. The total power consumed in a series circuit is equal to the sum of the individual powers used by each circuit component: Ptotal = P1 + P2 + ... + Pn
Using (1) above, I can connect two 1 Ohm resistors in a series to get a total resistance of 2 Ohms. Using (2) and (3), I know the total voltage drop across both resistors will equal 5.15 volts
and since each resistor comprises half the total resistance, each will drop 2.575 volts. Using (4) and applying my power formula, each resistor will dissipate 6.63 watts of power, i.e.
P1 = V2 / R1 = 2.5752 / 1 ~ 6.63 watts. Total power = 6.63 + 6.63 = 13.26 watts.
By using two easily obtained 1 Ohm 10 Watt resistors (Radio Shack), we can wire them in series across the +5 volt (red/black) output and increase the load on our power supply with an attendant increase in
output voltage on the 12 volt line. Both resistors will be running at about 65% of their rated wattage and will not be damaged by overload. However, they will get very hot -- the single 10 Ohm
resistor was dumping about 2.65 watts while each of the 1 Ohm resistors will generate nearly 2.5 times that. To keep them cooled down, it is strongly suggested that both be attached
to the PS case with heatsink compound to help reduce heat buildup.
On the power supplies I have tested, all produced higher voltage levels, with increases of .15 to .2 volts and total output of 11.85 to 12.06 volts.
If you are converting your PC power supply with the intention of driving a field charger, you might consider the substitution indicated above to get a little more voltage. Be mindful that
the conversion comes with a tradeoff in the form of more heat. However, a little care in heatsinking your resistors should provide a reliable and long lasting source for regulated DC power.
Unexpected Shutdown -- I've made the recommended changes, but when I connect my charger, the PS still shuts down. Now what?? The usage of a PC powersupply as
a substitute for a field charger power source falls far outside the intent of the original design specifications. Once the PS is running and stable, the overload circuitry is tuned to
detect high current sinks and shut the PS down -- under normal usage, these sinks would be indicative of an internal short in the PC. When a microcomputer is running, powersupply
demands change, but these are minimal and are usually associated with optical, hard or floppy drive usage or USB devices being attached.
Some field chargers produce a high current sink when first attached and generate a latch into the overload state -- as designed. The specifications state that the PS will remain latched
until the load is removed from the rail -- the PS will either automatically reset or may require at least one PS_ON cycle. Even though your PS may have sufficient wattage to
effectively drive the charger, the load change is the problem. One method that has been effective is to attach the charger before powering up -- to the PS, the charger
now appears as a high draw motherboard and not as a potential short occurring after the PS is stable.
Doesn't work!!! -- When I tested my PS before making the conversion, it worked fine, but now it won't power up at all. What should I try??
If the power supply were functioning before being modified, there are two prime suspects to check out. A failure to correctly reconnect the remote
sense wires can keep some PS from latching on. See the Design Guide Update for remote sensing information. The
other likely culprit is a short causing the overload circuitry to capture. Occasionally, this will cause a fan bump -- the fan will attempt to spin up, then stop
and there will be no voltage on the output rails. Cycling the ON/OFF switch produces the same result: a bump, then Latch_Off. Check your internal
wiring to ensure none of the clipped ends are shorted to the case or another rail. A second potential source for shorting is binding posts. The majority
of binding posts have either a plastic bushing that passes thru the case or a shoulder on the post and retaining washer that keeps the post centered.
Occasionally, the second style will not be correctly centered and the bare threaded post will come into contact with the PS case.
With the PS unplugged and using your VOM or DMM, test each post in turn for a short to the case -- none of the DC rails, including GROUND, should
have case contact. The case is grounded on the AC side; the DC side is isolated. If one appears grounded to the case, realign the post and test again.
Should none of modifications prove workable, your only remaining solution may be to purchase a standard
benchtop powersupply from a retail vendor or one of the on-line auction sites -- perhaps not as satisfying as a DIY project, but it will get you back into the air.
Updated: March 13, 2009