Useful part

[Kristinsgrandpa]

Lin, I also am using the little voltage regulators from Radio Shack.
  They are great, you don't have to get out the calculator or the electronics manuals to figure out circuit specs and any voltage coming from your coach will be reduced to 12 V,

Ed

[Lin]

Pete,

I have suspected that the small light in the middle of the cluster was supposed to be the high beam indicator.  That whole cluster is miss-wired.  Right now that center light comes on when the main power switch is on.  I find that useful, since I might tend to leave the power switch on without that reminder.  Why does it have a main power toggle anyway?  It seems that this must also have been a modification.  The ignition switch, it would seem, should be able to turn on the power and energize the starter, but here it is like a system that uses an ignition key to turn on the power and then a start button to start it.  Only this is in reverse; a toggle turns on the power, and the ignition switch serves as a starter button.  I don't know anything about how it was originally, but I assume that this is not it.  At this point however, I do not see any reason to change it around.

[DavidInWilmNC]

Is this the item you guys are talking about: www.radioshack.com/product/index.jsp?productId=2062600&cp?  If so, how are you mounting it?

David

[Sojourner]

Keep-It-Simple.....24v red or whatever color led indicator light.....  http://web5.automationdirect.com/adc/Shopping/Catalog/Pushbuttons_-z-_Switches_-z-_Indicators/22mm_Metal/Non-Metallic_Monoblock_Indicators/LED

You can use 7812 voltage regulator but for only powering things that not already made for your requirement.
However if you still want to.....http://www.eidusa.com/Electronics_Kits_12_Voltage_Reg.htm
You can minus 3 diodes if your only input DC battery voltage.

FWIW

Sojourn for Christ, Jerry

[Tim Strommen]

A part like that may be mounted to a Printed-Circuit-Board (PCB), or in a pinch, all of the supporting components (stiffening caps on the output, decoupling caps on the input, and a return diode to protect the regulator from the charge stored in the stiffening caps) and the package itself can be hot glued to the back of an instrument cluster.

As that part says in its description - a heatsink may be needed in some cases.  This mostly depends on the draw of the load on the voltage regulator.  With a parts like the classic 7812 (for +12VDC), the 7809 (for +9VDC), or the 7805 (for +5VDC), in a TO-220 package the power limit is kept at about 15Watts.  The voltage regulator is considered "linear" since it is a transistor which is limiting the voltage that is present at the output based on an internal reference.  The transistor conducts more or less based on the circumstances of the output pin in relation to that internal reference.  Since the transistor is directly limiting the output power it will be exposed to the current which is being demanded of the device. 

[Edit]  The "dropout" of a 7812's transistor with a 1Amp load (its rating) is 2Volts.  This is the voltage difference from input to output which must exist before the part will correctly begin regulating its output voltage.  All of the rest of the difference is lost to the regulator as heat. To figure out the power dissapated by the voltage regulator use the following equation:

Power_Dissipated = ((Volts_Input - Volts_RegulatedOut) * Amps_Load) + (Volts_Input * Amps_Ground)

The power to ground in Linear regulators is negligable (less than 1/20Watt) so we can just omit that for simplicity.  Using the 12volt lightbulb example - we will assume a 5Watt light bulb at 12Volts.  If those of you who have read the post on diodes across a battery bank solenoid recall an equation:

Volts x Amps = Watts

We can work backwards too: 12volts / 5Watts = ~417mA (for the lightbulb)


So looking at the Linear power equation, we can calculate (I'm assuming a solid 28.8Volt alternator charge voltage):

((28.8Volts_Input - 12Volts_RegulatedOut) * .417Amps_Load = ~7Watts(Power_Dissipated)

Of course this all assumes a consistent ambient temperature of 25degC - if your temperature around the part is higer there is typically a "de-rating" (so it won't be able to support a full 1Amp ouput if the temperatures are high).  The bigger the voltage difference and higher the load curent, the harder the voltage regulator has to work to keep the voltage right.  A plastic part like this will typically have a maximum temperature of 150degC for operation at the silicon.  One must find if the cooling is enough for the temerature rise of the part.  There is a specification usually for this measurement "ndegC/Watt".

There is some thermal resistance to the ambient air and some to the metal tab, with a TO-220, you can expect between 4 to 7degC/Watt to the tab, and about 50 to 100degC/Watt to the ambient air (through the plastic package).

Using those values, we can use another simple equation:

Power_Dissipated * Thermal_Resistance = Temperature_Silicon

or 7Watts * 75C/W = 525degC_Silicon temperature rise (or "poof" in the silicon world... since the temperature limit for that part is about 150degC).  You'd need to run this part in a -375degC ambient to avoid burning out the part!


Before looking for a heatsink it's a good idea to find out how much temperature rise your part can allow in its environment.  If you have a part that's rated for 150degC and an ambient temperature of 70degC - the allowable rise of the heatsink is:

T_RiseTotal = T_PartMaxTemp - T_AmbientMax

or in this case: 150 - 70 = 80degC

All of the thermal resistances must be accounted for - tab resistance, glue or thermal paste resistance, and heatsink resistance.  The way thermal resistances are summed is really just as easy as series resistors:

T_R"tab" + T_R"adhesive" + T_R"heatsink" = T_R"Total"

If you have an thermal paste which has a thermal resistance of .0045DegC/Watt (like "Arctic Silver") and a heatsink which has a thermal resistance to ambient of 3degC/Watt with no moving air you'd get roughly:

5.5C/W + .0045C/W + 3C/W = 8.545C/W

Running the heat-rise numbers again, we see:

7Watts * 8.545C/W = 59.815degC of heat rise to the heatsink

Since we know that the maximum allowable heat rise is 80C, 60C is a good 20deg margin of safety, and the part should run fine for long periods of time.


We can also look at the efficiency of this solution with respect to power consumption.  The bulb I used for this example is 5Watts.  The V-Reg in this example will dissapte 7Watts as heat as a "cost" of the regulation.  The total power consumed to light up this bulb is 12Watts - with only 42% of the power going to the bulb.  If you are boondocking - notice of his type of detail can help squeeze the most run-time out of your batteries.  Replacing an incandescent bulb drawing 5Watts with an LED which only draws 0.58Watts (28.8volt charging voltage - 3.5Volt @ 20mA blue LED with a current limiting resistor) is a savings of 11.42Watts.  Even with an LED, running one 3.5volt part from a 28.8volt supply with a current limiting resistor means that about 88% of that power is still being wasted as heat by the resistor.  Keep in mind the total loss of 510mW to heat is very small, but the useful work is relatively low as well - the LED is simply more efficient at outputting light at the specified spectrum than an incandescent bulb is - where upwards of 90% of that 5Watts is being emitted as IR "heat" whereas an LED may only emit about 10% of its 70mW as IR "heat".  While the v-reg to light-bulb solution has moer of the power going to the load than the "regulator" the overall power draw is higher than that of the LED solution (and you can get more usefull work out of the LED option by simply putting more LEDs in series - for instance 6 LEDs of the above specs gives you 73% useful work and only about 27% waste as heat in the resistor).

Thanks to "pvcces" (Tom Caffrey) for pointing out my absurd miss-posting of info earlier today (this is the second time I've done that... )[/Edit]


For applications which require more than 1Amp of output - linear regulators are not likely to be the most efficient way to go (due to the drop-out there is a lot of waste once the power goes up).  For things like running 110Watts of 12Volt headlights off of 24Volts, there are switching regulators (like those found in common PC power supplies) which are more efficient (>60%).

Cheers!

-Tim

[Lin]

David

Yes, that looks like it.  I did not mount it at all.  There is no weight to it, so it is supported by the wires I soldered to the leads.  I did make sure exposed wires were taped though>

[FloridaCliff]

Keep-It-Simple.....24v red or whatever color led indicator light.....  http://web5.automationdirect.com/adc/Shopping/Catalog/Pushbuttons_-z-_Switches_-z-_Indicators/22mm_Metal/Non-Metallic_Monoblock_Indicators/LED

FWIW

Sojourn for Christ, Jerry

Jerry,

Thanks for the link.

I am all about keeping it simple.   And at that price! 

Cliff

[Kristinsgrandpa]

  I mounted mine to the birdcage frame of the bus with a sheetmetal screw.

I sure hope the heat doesn't mess up the paint on the side of the coach.

Ed

[Tim Strommen]

I don't think that'll be a problem - the inverse might be true if you get moisture on the inside of the sheetmetal when it's cold outside and warm inside (moisture might eventually corrode that part).

-Tim

[pvcces]

Tim, why did you add the load to the drop out power? I wouldn't think that the heat that is dissipated in the load could show up as heat in the regulator.

Tom Caffrey

[Tim Strommen]

Because I was stupid...  I'm revising the above equations. (that's what I get for doing it over lunch...)

-Tim 

[pvcces]

Thanks, Tim, for the clarification.

Tom Caffrey

[gumpy]

Gumpy,

Please school me.  If you are putting a resistor in line with the led anyway, why can't you match the resistance the bulb would have provided?  Expiring minds want to know........


Thanks

Frank

Sorry Frank. I wasn't following this post.

The resistor value is calculated by the LED voltage and current. In the case of reds and greens, it's roughtly 20 milliamps at 2 volts. I use a 1500 ohm resistor, whcih is figured for 17 milliamps on a 28 volt system ((28 -2) / 0.017). I clean out the bulb's metal base of any residual glass and adhesive, and drill a tiny hole through the lead point on the base. One end of resistor is soldered to the positive side of the LED, as short as possible. The other resistor lead is put through the hole and soldered to the lead point. File smooth. The negative lead of the LED is bent up and over the lip of the metal base, and soldered there.

Just make sure your leads don't short together or touch the base where they're not supposed to. You
can fill the base with epoxy, if you want, but I've found if I get everything positioned carefully, I can
cut the center leads very short and they won't touch anything, and are quite secure once soldered to
the base. I usually bend the positive LED lead up along the LED, and solder it there, so the resistor sort of
shields it from touching the case.

craig

[Tim Strommen]

To find a current limiting resistor for is not horribly tricky - filament bulbs are positive-temperature-coefficient devices.  This means the lower the temperature, the lower the resistance they have.  A light bulb is designed for a specific operating temperature, power and lifetime and current by picking the filament material, filament wire gauge, and filament wire length.

You already know the design voltage for the bulb (in this case 12-14.4Volts), and you know the power (I used an example 5Watt bulb which is common for indicators - check your bulb thought as YMMV).

As shown in an above post, you can work backwards from power and voltage to find current (this will be the operating current):

Power / Voltage = Current

or

5Watts / 13.2Volts = 0.379Amps (379mA)

This 379mA is what we want to limit our lamp's current to.  If you have a system voltage of 28.8Volts, you simply subtract the lamp design voltage from this voltage to find the dropping voltage required of the current limiting resistor.

28.8Volts_System - 13.2Volts_BulbDesign = 15.6Volts_ResistorDrop

With this figured out, we can use the equation:

V
___
IxR

Using some basic math now, we can figure out a resistor value which will "consume" 15.6Volts at 379mA:

15.6Volts / 0.379Amps (379mA) = ~41.1Ohms

This is a value which is only available with 1% tolerance or better parts, for 5% the next best resistor is 43Ohms. The other specification we need to find is the power dissapation the resistor will need to avoid over-heating.  This is easy, going back to the first equation in this post, but this time solving for "power" by:

Voltage x Current = Power

or

15.6Volts * 0.379Amps = 5.9124Watts

You MUST pick the next size above this value or higher (you cannot pick a lower value because its closer - like 5Watts) so a comon standard value is 10Watts.

Cheers!

-Tim

[DrivingMissLazy]

Tim, would not a small zener diode be the best way to go? I do not think there would be any lost energy like with a resistor.

That is how I dropped the 12 volts to 6 volts for my 46 ford instrument panel. No heat sink or anything.


Richard