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Hi. You might want to replace the LM317T with a simple current-limiting resistor. Diodes are current-driven devices. Your white LED may have 3.5V across it if you put 20 mA through it, and it will only have 3.7V across it if it has a lethal 50 mA. By the time it gets to 3.9V, you might just have a smoking blob instead of an LED. Just use a 560 ohm, 1/2 watt resistor in series with the LED, and you should be fine. That will give you about 20mA through the LED.

Using an LM7812 as a series regulator for the fans is a good idea. Your schematic doesn't show a connect between the GND pin of the 7812 and GND -- I'd guess that's just an oversight. You should place a small (10uF) cap from input to GND, and from output to GND, to avoid oscillations which might cause problems.

The part that's bothersome, though, is the peltier cooler. You say you've got a cooler that needs 15V at up to 15A, but your power supply is only capable of 15V at 4.5A. If that's not a typo, you're in trouble. Choose a power supply that will provide the voltage you want, but make sure it can provide enough current.

If you've got a voltage source, it's rated to provide a certain amount of current. That means you can't exceed that amount, or something won't work right. It might be automatic shutdown, current limiting (the power supply drops down to a lower voltage to keep current below maximum), or the power supply might be damaged or even smoke. A good power supply will be able to provide a steady DC voltage over a wide range of current, as long as it's less than the maximum. As an extra note, you should know that some switching power supplies require a minimum load current, too. It's best to ask before specifying.

Good luck Chris

Good luck Chris

With the fans, you're OK - they'll "draw as many amps as they need," but you need a current-limiting resistor in series with LED. You would be better-off connecting the LED, in series with a resistor, across the 12V or

15V supply. Forget about VREG1 altogether. Calculate the LED current using Ohm's law:

I = (Vs - Vf) / Rs

Vs = Supply voltage Vf = Approximate forward voltage drop across LED (3.5V ish) Rs = Series resistor

You need decoupling capacitors (try 100nF) on the input and output of the other (remaining) voltage regulator.

Here is the revised version of my circuit:

What do decoupling capacitors do? I know that a capacitor stores a charge, but what purpose does the decoupling capacitor serve?

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Yup, things that are designated as 12v 15v etc will only take what current they need. An LED needs to have the current limited, your schematic would likely 'terminate' the LED. While LED's have a forward voltage you have to deduct that from the supply voltage and then calculate a resistance that will limit the current. Say you have taken 3.3v as the LED forward voltage, 15v - 3.3v = 11.7v . So to limit the current to 20mA a resistor of 11.7v / .02mA is required, which would be 585ohms (use 680ohms). No need for the voltreg here just the resistor. Note that you can use a lower current for very little loss of LED output, if power is an issue 10mA or less will still give a good output (particularly with high intensity LED')s.

-- Regards ..... Rheilly Phoull

When digital circuits change states, there is a very brief time when both output devices are conducting. That creates a spike on the power rails which you can easily see on a scope. The caps absorb much of this by providing a local source of power.

Since it contains a feedback loop, the regulator has the potential to oscillate - depending on the load impedance. It is standard practice to decouple the regulator inputs and outputs. Look at the typical application circuit on the datasheet. The capacitors must be mounted close to the regulator to be effective.

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In this situation the caps restrain the regulator from oscillating.

Once more, switching on the Peltier will switch off or blow your power supply.

petrus bitbyter

How do I keep the peltier from trying pull more amps than the powersupply can handle with out droping the voltage?

Thanks

petrus bitbyter wrote:

power

If you built a switching buck regulator, you could lower the supply voltage to the peltier so it would draw less current (and move less heat). The current demand is roughly proportional to the supply voltage. So, for instance, if you lowered the 15 volt supply to 10 volts, the peltier current would fall from 15 amps to 10 for a total power to the peltier of 10 volts * 10 amperes = 100 watts.

Assuming the efficiency of the buck regulator was 90%, its input current would be (100 watts * 100/90%)/ 15 volts = 7.4 amps.

You don't! You make sure your power supply has the necessary head room and let the peltier take what it needs.

Are those the maximum ratings for the Peltier? In practice, their efficiency drops very low as you approach the maximum voltage and current, as they will internally generate more heat than they are pumping.

Best to run it at roughly half of its voltage and current rating, or less.

Mark

Would I have to build a buck regulator like John Popelish described in order to limit the current.

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in

A buck regulator is just one way to do this. You could also try getting another power supply.

Mark