+15V to +5V via 78L05 is getting too hot, should I use 78L12->78L08->78L05 or? - Page 1
Forum section: electronics / beginners

author: dusan
author_rank: Regular Contributor
timestamp: April 06, 2024, 05:41:47 pm
content: I need to power bluepill (~30mA) from 15V source. When I use just 78L05 it's getting really hot. If I use chain of 78L12->78L08->78L05 then they are only warm but fine. I could also use diodes as a droppers (they remain only warm, not hot). Which is better? Is there a better solution? I don't have switching/buck regulator.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 06, 2024, 05:48:59 pm
content: The regulators in series are not that good. They need extra current that flows to ground. Another alternative would be using the larger 7805 in a TO220 case that gets rid of the heat better.Instead of the choice of diodes one can also use a zener diode - these also come with different power ratings / size.

author: schmitt trigger
author_rank: Super Contributor
timestamp: April 06, 2024, 05:50:01 pm
content: Google the term 7805 switch mode replacement.Universal-solder dot ca has a device for US$2.78, I am sure you will find other vendors.

author: Siwastaja
author_rank: Super Contributor
timestamp: April 06, 2024, 05:53:50 pm
content: The diode string solution has the following advantages:* Probably cheaper* Less total current consumption by getting rid of the ground pin currents of the 7812 and 7808* Added reverse polarity protectionBoth solutions will of course work.One obvious solution is to just use 7805 in TO220 package. It can easily handle the 10V*30mA = 300mW, even without a heatsink. For example, a random datasheet says Rth_J-A = 19degC/W, mere 6degC temperature increase at 300mW. Compare this to 78L05 in TO92 package, 159degC/W, so 48degC temperature increase. Which is not usually too much, but if you have any risk of +15V actually being a tad more, and working in hot environment, then it does become a problem.Speaking of which, maybe you don't have a problem. If you are really sure that 30mA is your worst case current and 15V input not exceeded by much, and there are no other hot components in tight enclosure close to the 78L05, maybe you don't have any problem. Touch can be deceiving - say 70-80degC case temperature feels very hot but isn't a problem.

author: wasedadoc
author_rank: Super Contributor
timestamp: April 06, 2024, 06:04:05 pm
content: Inserting a series 150 Ohm 0.5W resistor will halve the heat produced in the regulator IC.

author: Zero999
author_rank: Super Contributor
timestamp: April 06, 2024, 06:28:34 pm
content: Quote from: Siwastaja on April 06, 2024, 05:53:50 pmI doubt it's too hot. If it's outputting 5V and isn't shutting down, then it's fine. The maximum operating temperature rating is 125°C, so it's unlikely 48°C temperature rise is not an issue, unless it's in a high temperature environment.My guess is it's about 70°C, which is a bit too hot to handle, which has concerned the original poster, when in reality it's fine.

author: iMo
author_rank: Super Contributor
timestamp: April 06, 2024, 06:30:42 pm
content: I vote for a 6V8 500mW zener in series (cathode to +15V)6.8V*(30mA+5mA)=0.238W power loss at the zener..(8.2V-5V)*30mA + 8.2V*5mA =0.137mW power loss at 78L05 (==Tamb+21.7C)

author: shapirus
author_rank: Super Contributor
timestamp: April 06, 2024, 06:45:43 pm
content: Quote from: dusan on April 06, 2024, 05:41:47 pmYou can simply use a series resistor to drop some of the voltage before the regulator.For example, you want to drop, say, 15V - 5V - 4V = 6V on the resistor (15V in, 5V out, 4V on the 70L05).Then the resistor will have to be 6V / 30mA = 200 Ohm. It will dissipate (30 mA)2*200 Ohm = 0.18W, so a generic 1/4W THT or a 1206 SMD resistor will work just fine (but it'll get somewhat hot, so don't place heat-sensitive stuff near it).You then also get an opportunity for an additional power supply noise filtering for free: just add a cap after the resistor, and they'll form an RC filter.This of course works only if you dont expect significant increases in current, which will cause an extra voltage drop on the resistor, which may not leave enough headroom for the 78L05 to operate properly. But even if there are current spikes, but they are short enough, an adequately sized capacitor on the output of 78L05 will take care of it.Otherwise, using a zener, as suggested earlier, will be the way to go.

author: Siwastaja
author_rank: Super Contributor
timestamp: April 06, 2024, 07:01:03 pm
content: Series resistor requires true knowledge about worst case PEAK currents, including start up (capacitors charging). Power-up might not even be monotonic; a sudden increase in current might cause the input voltage to sag too much, triggering brown-out detection on the circuit and make it oscillate and never start properly. Therefore, I don't recommend using a resistor - linear current dependency is too much for many/most circuits which expect stable voltage source (and if they didn't, you wouldn't use a regulator).Resistor is a safe choice only if the value is small enough not to help much under average current consumption.Just use diodes or a zener diode.

author: tridac
author_rank: Regular Contributor
timestamp: April 06, 2024, 07:03:21 pm
content: Assuming 15 volts @ 30 ma into the regulator = 10 volts drop = 300mW. Running that hot may work, but will reduce reliability margin. For example, what happens if the reg shorts and drops 15 v onto the pill ?. If you must run that close to margin, then fit a clamp zener, say 6.2v 5w, across the output, to protect the load.Only use 78Lxx series for a very small few mA loads here. TO220 7805 are a stock part here, and tend to use them, as they are cheap and get the job done.

author: iMo
author_rank: Super Contributor
timestamp: April 06, 2024, 07:33:01 pm
content: Quote from: iMo on April 06, 2024, 06:30:42 pm(8.2V-5V)*30mA + 8.2V*5mA =0.137mW power loss at 7805 (==Tamb+2.6C)PS: EEs (Electronics Enthusiast) do communicate via schematics, graphs, math, simulations in LTspice..

author: Xena E
author_rank: Frequent Contributor
timestamp: April 06, 2024, 07:43:13 pm
content: My vote is to use a 500mA or 1A version as noted by others, or buy/make a TO92 heat sink clip for the 78LO5.At 30mA it will be fine anyhow, however the heatsink or uprated part solutions will give it a better safety margin.

author: Zero999
author_rank: Super Contributor
timestamp: April 06, 2024, 07:51:47 pm
content: One downside with a larger regulator is it will have a higher current limit, which might not be desirable, as you won't get the same level of protection.I say, just leave it as it is. It's not too hot, just fine.

author: kripton2035
author_rank: Super Contributor
timestamp: April 06, 2024, 09:26:30 pm
content: https://www.romanblack.com/smps/smps.htmorhttps://www.aliexpress.com/item/4000549890103.html

author: Doctorandus_P
author_rank: Super Contributor
timestamp: April 06, 2024, 10:50:21 pm
content: Those SMPS boards are a good solution, but a bit overkill fror 30mA.TO220 is much better at heat dissipation then TO92.I assume you have a bunch of 78L05 chips. You can stack them on top of each other, where you put the GND pin of the first to the +5V output of the "next one". This way each only sees a 5V difference, and thus a lower power dissipation. But I see this mostly as a "quick fix" when you've already built half the circuit and do not have the TO220 variant on hand.