Linear Voltage Regulators (LM7805) | AO #17 

+codedemonelectronics Appreciate you being honest. ;) (Seriously, thanks.)

+script0r inside Thank you.

this was a perfect video with zero unusefull time . yes there must be sick haters everywhere .

@@tolgadabbagh1877 "Sick haters" That's quite an extreme view of someone who just pressed a button that was made for pressing.

I wish I could make them full time, but I can't. So I'm shooting for one a month or at least more than one a quarter.

By the way, if I put a 0.47uF capacitor in the input (instead of .33uF, as manafucter says) there is no problem, right? :) Thanks

anything less than 10 uF is fine.

Hello Poland! Thanks for watching.

I think anything over 450°C is safe to say, you need a heat sink. ;)

Thanks, should I have more questions I know who to ask. LOL.

It is 2V above the output voltage.

Because our other videos aren't high quality?! Just kidding. Thanks for the feedback.

"Electricity" does not flow. "Electricity" is a construct. Current flows when a Voltage is applied. In the case of a linear regulator the current into it is equal to the current out of it (minus some tiny loss.) The regulator, effectively, changes the voltage dropped across itself to maintain the required output voltage. So heat disspiated is equal to the voltage drop across the regulator times the current through it. A regulator that switches "on and off" is a switching regulator.

For low-frequency or bulk decoupling they work fine. But you'll want to put a low-value ceramic in parallel with it for higher frequency noise. While called a decoupling or bypass capacitor, the ceramic is really a filter cap in that case. The caveat here is if you use polymer aluminum (or polymer tantalum) capacitors instead of traditional wet aluminum electrolytics. Polymers have significantly lower ESR.

That's what most designs do. The bulk electrolytic cap is *decoupling* the circuit from the power supply. Hence the name, "decoupling capacitor."

It helps but does not eliminate it. It’s just a positive thermal coefficient.

@@AddOhms How protective is that action - thermal runaway usually happens so quickly - how would I calculate a safe margin?

It's not something I would ever have a design that relies on [for a linear regulator.] If it is "thermally running away" there is a serious problem in the design.

You. cannot. Series makes no sense and parallel they would try to regulate each other.

Yes, but you shouldn't. A decouple capacitor DECOUPLES a device from the power supply. So you should place the larger capacitor close to your load, not the regulator.

+Power Max Another issue with large capacitors on the output is when the regulator first turns on, the initial inrush current of that large capacitance can cause the regulator to go into thermal shutdown. So you end up with a linear regulator which a huge oscillation, while trying to charge up the output capacitor--the same cap that was put there to help it!

AddOhms Yikes! I do notice that when I attempt to design a opamp based LDO from scratch in LTspice, if I use ideal capacitors (ESR

AddOhms Something that I hope to do is to design my own accurate, arduino controlled lab power supply. One that will be at least 0-15V with 2 isolated channels, at at least 0-5A, which will allow full voltage and current range adjustment. That is proving to be a challenge with things like power dissipation, loop stability, transformer / preregulator stages, dropout voltage of the pass transistor element, etc. It is all getting quite complex! I have a dual-rail design in LTspice that seems to work well, I can give you a link to my old instructables question about it.

+Power Max For the longest time I wanted to design my own: power supply and oscilloscope. Eventually I bought them instead. Both are going to be "someday" projects for me.

Voltage regulators actively (and accurately) regulate voltage. Zeners are passive, so they drift under various conditions. Watch: Voltage Dividers as Regulators ?! | AO #22 ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE--kEh0TYjYYE.html

You probably want a switching regulator. Or at the very least, an LDO with an enable pin. Either way, not a circuit I can help with.

I have zero experience with this kind of system. But if the output voltage changes with RPM, it isn’t a regulator.

@@AddOhms thanks for the reply ... any idea what I can use to regulate the dc? Something like dc to dc that's adjustable? I just don't want it to go over 14.8vdc that's all ... thanks again 👍

+TheProCactus Nope

Would it be something relevant to you?

No. I'm not an electrician and I don't know electrical codes well enough to talk about them.

Well thanks for your thought, I look forward to watching more of your videos.

+Chetan Amrao I converted to nF. 0.33uF is 330nF.

okay i didn't notice that! LOL

Without a smoothing capacitor and a regulator, you won’t have DC.

Regulators usually have a minimum load specification. It's the minimum current they need to regulate the output. If there is no load, there is nothing for the regulator to, well, regulate. If you're using a LM7805, it's minimum load is probably 5 milliamps. Add a 1 kohm resistor for testing.

+MrLookslikeMani Hi Germany! Yeah 100-470uF get used pretty often. I just want to make the point they aren't always necessary. If you go with much larger values than 470uF, you may actually cause the regulator to oscillate when it first powers-on. Capacitors are a short circuit when they first start to charge. So too much capacitance can cause too much current, activating the thermal shutdown. Looking back, I should have added that in this video. I may make a separate one to explain better.

+AddOhms Thanks a lot. :)

My $0.02 - I've been using 10 mF caps for these for years. Never had a problem. I had always assumed that this was the norm.

Measure the input and output voltage with a multimeter, while there is a load on the regulator. Unloaded regulators, like batteries, may not output the right voltage.

+Matt Hunter You can't just wire regulators in parallel with each other. They will attempt to regulate each other. There are techniques to isolate them, but generally that's not efficient. The best approach is to get a single supply that sufficient for the load. This could mean, splitting the load up. Example, half the LEDs on one and the other half on the other.

Ah, I see. They will be in pairs, might it be ok to have one regulator at each pair of lights? (11W total with the regulator supplying 12v at 1A) That way the main wiring is unregulated but each module is safe

Yes! That is the point of a regulator. If is constantly changing between 7 and 9 volts, you might need to add some more capacitance on the input to help smooth it out. Watch the output to see how much it varies, if at all.

If yes How can it be achieved.

With an external circuit yes. Do a search for "LM317 Current Limiter."

Thankyou sir Got the idea :-)

I don't know what you mean by "what you are doing". I have an Electrical Engineering degree.

Overall you're okay. 220uF on the input is fine. You're decoupling the regulator from the rest of the circuit. 220uF directly on the output might be a bit high. On an older linear like the 7805, it'll be fine. But it's better to put a small capacitance (1-10uF) on the output for filtering while putting the big caps near large loads, like a motor or LED strip.

LTSpice, Falstad, and ngspice.

@@AddOhms ill try them thanks a lot!

these days I have been studying converters and regulators, like buck and boost, and I do not understand what the output current depends on

+Mohannad SMARTHERO The load determine the current drawn. The regulator's inductor and capacitor determine the current it can provide.

almost clear enough, I still need to make some tests, thanks alot

Resistance. Junction (to) Ambient. What's the thermal resistance between the junction (the silicon) and outside air.

Linear regulators are horribly inefficient for high current applications. Almost all high current / high power supplies are switching supplies.

huh? They do. “Max current.” or “Load regulation”.

Without voltages and a power level, it’s impossible to say. Modern switching mode converters are exceptionally small. The real decision comes down to cost.

shubham raj aonkimonki YESS! To start, your ripple frequency are almost some 100MHz so you objectif is to pass only the DC! Well that’s great bc DC’s frequencies are almost nul in simple terms you have to find the stability, then you know that the formula that gives the capacitance impedance is: 1/(2pi*f*C) otherwise when f=0(DC) the impedance is like too too high so current can’t flow through the capacitor and if f= almost 100MHz then you’ll find your 330nF ( behind that a complicated maths and a big knowledge of impedance at different frequency)

I don't provide one-on-one help.

You can't put regulators, switching or linear, in parallel. They'll try to regulate each other, until they destroy each other.

@@AddOhms Really? Oh. Thanks!

Datasheet clearly says max is 2W without a fin (heatsink). And 50W with one.

@@AddOhms thanks. So if I can use 2amp than 50w÷2A=25volt deference? So I can have 2amp 5v output with a heatsink?

@@AddOhms50w at what ambient or junction temperature? I did some calculation of heatsink value for 45c ambient temperature and 125c maximum junction temperature, with 14w power dissipation, 2c/w junction-case heat resistance,2c/w case to sink resistance,not sure if it was correct or not but I ended up with 1.714c/w heatsink ambient resistance value. Is there a heatsink with such value?

The conditions are stated in the datasheet.

The total heat generated across the regulators is the same. In fact, slightly more. If they are too close to each other, you would not get any benefit. And multiple regulators would likely be more expensive than a heat sink.

AddOhms 24 volt to 12v 100 mA will produce about 1.3 watt * 65 = 84.5 C toooo much for a heat sink

Well, that is just wrong. 84.5°C (temp rise) + 25°C (ambient) = 109.5°C. That's still below the maximum junction temperature of 125°C or 150°C. So at 100 mA, an LM7812 in a SOT-223 package would not even need a heat sink for proper operation. Use a TO-220 and the junction is only going to be at 56.2°C.

Neither has a “turn-on” voltage. They drop their dropout voltage, which is about 2 V. So they both need at least 7V in to guarantee 5V out. I don’t know where you’re getting your misinformation. There is no functional difference. In this case, LM and ML are the (original) vendor’s initials.

@@AddOhms Sorry I meant no disrespect, I was just confused with the (LM7805) in the description of the video but yet you show the ML7805, and yes there are specific differences from different vendors. As for the TOV I'm not exactly sure where I found that but I guess that should be a minimum voltage and even that is different per the different spec's I've been looking at. Sorry if I sounded disrespectful, I wasn't trying to be. Also, the LM7805 seems to have a steady output voltage of 5v where the ML7805 has an output range from 4.8-5.2v, and the ML7805 has a higher maximum input voltage.

+Ryan Smyth Sure... use this account: 1M4eteJeAzfmcYCnTtLg7rVqB41Nx6n1be. Thanks.

I don't have any suggestions for this project.

In 7 years, you’re the first person to claim I said that. So, it’s more likely you misheard the pronunciation.

The 555 is cool and has enabled many cool applications. However nearly every digital system of the 1980s and most through the 1990s and even today, the 7805 and its variants are present. The 555 might be number 2, but it isn’t as popular as the voltage reg.

@@AddOhms I will agree, but it's limited purpose is why I'd go with the NE555. Longevity, 7805 wins hands down, as you said, it has been used in nearly everything, even with efficiency and switching supplies, some things still use them. The NE55% however, was phased out quickly when uC's took over. Thanks for responding :) You have some great content!

To provide decoupling for the stuff that gets powered by the board.

@@AddOhms Sure, but why 47uf, not the ones in the datasheet of the regulator? Thanks for your reply