Which is OK, because my voltage regulator would be dissipating less power then as well. Even with the diodes in line, my voltmeter read 4. I can free up some of the real estate on my board for another mode switch or dimming control or something.
Also, I can install a 5V charging jack or 10 so my wife and kids can charge their phones and MP3 players and such directly. That should give me some extra points. Motorcycle battery voltage is going all over the place, etc. The only way to get a stable Vin to the is to use a regulator to feed the regulator.
The scheme might be the best. The goal here is to distribute heat across several components. With series diodes however, the voltage drop is fixed and so is preferred in this context. I would stick with the but use a heatsink.
The can easily dissipate 2W with even a small heatsink, but without a heatsink it would exceed the maximum junction temperature rating of degC. All you guys are right I am sure. There are so many ways to get from A to B. One way may not be better than another. How about a combination. I think the goal here was to reduce the heat on the regulator.
You have a regulator that is designed to regulate. It is designed to have a heat sink. Just a little hint - it doesn't have to have lots of fins. Mount it on a piece of aluminium strap - preferably coloured. Mount it on the aluminium box you use for a case. Don't quote me with crap like that. My comment is in reference heat sharing with diode vs resistors where another joker seems to think a resistor is the way to go.
What the better solution is for the original poster is another story. Possibly, if it is a large enough ground plane. But then it will heat up all the other components too. Better to use a small finned heatsink.
I use one to supply a and with 50mA its fine but I have noticed that if the dragon supply comes through it instead of the USB it increases it to mA and it does then get too hot. But reading the OP surely using a zener diode will be the cheapest, easiest and arguably the most reliable method?
They are shunt regulators, so they work by diverting power from the load, wasting that power as heat. This may be OK when the power is low, but here it isn't all that low. In this case, using a zener diode would mean wasting around 3W or more in a series resistor, plus up to about 1W in the Zener diode.
Using a linear series regulator such as the , the OP has already established that there will be a maximum of 2W wasted in the regulator, with less wasted when fewer LED segments are illuminated. Yes and it would have very poor regulation. Yes but it wouldn't be any less heat, would it? Should I just get a big heat sink Yes and forget about it? Add the green and blue channels, and dissipation will tend to thrice that figure.
Over 1 watt is being dissipated by the linear regulator in this case. The regulator getting too hot to touch is not unexpected in this case. Adding the green and blue channels will make things much worse. The datasheet of the LM says:. Temperature rise is: 1. Yes, the LM can handle up to 1. There's no way it can do this without a substantial heatsink attached. Even at the mA your red LEDs are drawing, the regulator is dissipating well over 1 watt.
Its temperature will rise until the amount of heat being transferred to the air equals the amount of heat being generated. If there isn't much surface area over which that can happen, the temperature will quickly rise above the maximum rating of the device.
The coin adds a little area and a little thermal mass , but not nearly enough. The quoted maximum current the LM can supply requires that you provide a way for the chip to get rid of the heat it must produce to 'boil away' the voltage difference. You don't say what the output voltage of the LM is, I assume 5V. For 24 LED parts that is 3. That requires a decent heatsink for the LM A heatsink can be calculated much like a resistor.
You want to dissipate 4 Watt, over a temperature difference of let's say 40 degrees C. That is not a very big one. This will not reduce the total power dissipation, but it will distribute it over all resistors. Sign up to join this community. The best answers are voted up and rise to the top.
Stack Overflow for Teams — Collaborate and share knowledge with a private group. The following shows the proper connections:. Often times the bypass capacitors are left out of the regulator circuit. They must be in place in order to keep the regulator from oscillating and overheating.
At a minimum, a U10 0. This is all that is really necessary to stabilize the device. This does not have any bearing on the value of the filter capacitor on the input. This capacitor will smooth any ripple coming from a rectifier in an AC power supply. The value of the filter capacitor should be large enough to handle the load without ripple in the input. This web page will not go into detail with the calculations.
However, as a general rule, the larger the value, the better. Do not attempt to put a large value filter capacitor C3 on the output of the regulator. In fact, you really don't need any filter capacitor on the output of the regulator. It is not necessary and could damage the regulator if the input voltage drops quick and below the output voltage. If it is imperative that you use a large value on the output, or if sub circuitboards have filters on them, a 1N CR2 across the input and output of the regulator with the cathode to the input as shown below.
What does the diode CR2 do? If other loads on the rectifier side are heavier, the voltage on the input to the regulator will drop too quickly if there is a large capacitor C3 on the output.
This will backfeed into the output of the regulator and destroy it. The diode will pull the output voltage down to the input voltage if the input goes low quickly thus protecting the regular from being backfed.
Proper design of the regulator is vital when using the regulator.
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