Multiple Regulators to step down - Page 1
Forum section: electronics / beginners

author: PDW
author_rank: Newbie
timestamp: July 17, 2017, 10:19:50 am
content: I am stepping down a 6-7 volt DC to 3.3 volt to power ESP8266 board. The regulator (LD33V) is hot, as expected i suppose. I am asking if I can use an LM3805 to get down to 5 V, then push that through the LD33V to drop from 5 V to 3.3 V.Thanks, new to the forum, hope this is enough info.

author: veryevil
author_rank: Unknown
timestamp: July 17, 2017, 10:24:53 am
content: You sould be able to however all it will achieve is to spread the heat over two devices. The same overall power will still be consumed (slightly more) so you will just get two hot spots instead of one.The same could be achieved with a power resistor before but you would need to calulate a sensible value based on maximum current and voltage drop. You would still however end up heating two things up instead of one.

author: capt bullshot
author_rank: Super Contributor
timestamp: July 17, 2017, 10:27:34 am
content: You can also use a suitable zener diode or some silicon diodes in series to drop the voltage and dissipate heat.

author: wraper
author_rank: Unknown
timestamp: July 17, 2017, 10:47:28 am
content: Two regulators in series make sense if you need both 5V and 3.3V in your circuit and want to evenly spread heat dissipation between those 2 regulators. Otherwise you might improve the cooling of the regulator (larger ground pad/via stitching, heatsink) or using buck converter instead.

author: Zero999
author_rank: Super Contributor
timestamp: July 17, 2017, 01:19:34 pm
content: Another option is just to add a resistor in series with the linear regulator. The resistor should drop enough voltage to keep the regulator cool, yet not so much it operates in drop-out, causing the output voltage to fall. You may need to increase the size of the regulator's input decoupling capacitor, to keep the AC impedance low.

author: Pack34
author_rank: Frequent Contributor
timestamp: July 17, 2017, 01:23:06 pm
content: Why not move to a switching power supply?

author: MrAl
author_rank: Super Contributor
timestamp: July 17, 2017, 03:34:31 pm
content: Hi,Just to note, if you have N regulators each dropping the same voltage then the power (and thus heat) distribution is approximately P/N where P is the total power lost by going from the higher voltage to the lower voltage.For example, if we go from 9v to 6v and then to 3v, that's a voltage drop of 3v for each of the two regulators, so if the current is 1 amp then we loose 3 watts in each regulator and a total of 6 watts.If instead we went from 9v to 8v to 7v to 6v to 5v to 4v to 3v that would require 6 regulators, each dissipating 1 watt each for again a total of 6 watts.To contrast, if we went from 9v to 4v and then to 3v (two regulators) the first one would dissipate 5 watts while the second would only dissipate 1 watt, so with similar device packages the heat would not be distributed very well because one would get much hotter.So the idea is to try to distribute the heat as evenly as possible.This actually would apply to any type of regulator even a buck, but with the buck the power lost would be much less and often we dont need more than one regulator which is much nicer.

author: Teledog
author_rank: Regular Contributor
timestamp: July 17, 2017, 06:54:29 pm
content: I agree with Pack34.I tried the same thing with the LD33V & ESP8266..got wayy too hot!Decided to go with an eBay "lm2596 module"..bigger package size, but input anywhere from ~5-32VDC ..without worries.Get the units with the blue 10 turn pot though - for 3.3V "accuracy"..they're ~99 cents or less each.I typically order a pile of them for assorted projects.G' Luck!

author: KL27x
author_rank: Super Contributor
timestamp: July 18, 2017, 06:07:40 am
content: Your regulator needs 1V of overhead. That leaves about 2V extra. If you were actually drawing 800mA, the series power resistor should be 2V/0.8A = 2.5 ohms. Wattage is 2V x 0.8A = 1.6W. So you would want a 2.5 ohm 2W resistor.At full 800mA load, the resistor will be dissipating 2/3 of the heat. The regulator will be dissipating only 0.8W. You just cut the thermal load of the regulator by 66%.But at only 400mA, the series resistor will only drop 1V. So the regulator will be dropping 2V at 400mA. It is still dissipating 0.8W. Which, without a heatsink, is still Hot. If your expected max current draw is only 400mA, you can use a 5 ohm resistor with suitable rating. If you have intermittent burst draw over 400mA, you can beef up the output caps as a compromise. This will get you down to 0.4W.Basically, choose the lowest max load that will work for you, to get the highest resistor possible. This will probably work out better than the 5V regulator... which takes roughly 1/3 of the load off the LDO over any current draw. (A bit more under some circumstances, but not much).

author: digsys
author_rank: Unknown
timestamp: July 18, 2017, 06:39:52 am
content: Or just use one of these -https://www.digikey.com/product-detail/en/recom-power/R-78E3.3-0.5/945-1661-5-ND/3593412For 3 bucks, I wouldn't fool around with adding extra bits, PLUS it goes to 30V+ Input

author: Codebird
author_rank: Regular Contributor
timestamp: July 18, 2017, 07:34:39 am
content: Short answer: Yes, this would work, but it might not be the most compact or lowest cost way to achieve your aim, and it certainly won't be very power-efficient.