12V Regulator Max Input...

[Degrom]

Hi.

I am busy with a small power supply and I am adding a few regulators to get a 5V and 12V supply.

My question is : My power supply is putting out about 6.4A@36V and regulators max out at 36V. What size resister can I use to bring down the volts a bit(Just to be save)?

Would also appreciate if you explain how you came up with the calculation so I can use it in the future as well...

Thanks.

[Al_The_Man]

Originally Posted by Degrom My power supply is putting out about 6.4A@36V and regulators max out at 36V. What size resister can I use to bring down the volts a bit(Just to be save)? .

Generally not a good idea, you can lose regulation.
If a linear supply, look at adding an extra Tfmr winding or two, much more efficient.
Al.

[Degrom]

Originally Posted by Al_The_Man Generally not a good idea, you can lose regulation. If a linear supply, look at adding an extra Tfmr winding or two, much more efficient. Al.

Hi Al,

Why do you say regulation can be lost?

Is it cause the Amps are not fixed all the time...

[Al_The_Man]

Resistor is rather waste full, heat dissipation, and also if current varies or peaks, it could cause the regulator to 'drop out' .
Al.

[Degrom]

So how about using a 24 regulator(Can go up to 40v) and stepping it down a bit and then use the 5 and 12 volt regulators of that?

I am not going to use the 5 and 12 volt regulators to drive heavy stuff.. I need the 12v for a fan and the 5 volt for signaling relays...

[gar]

071117-2200 EST USA

Degrom:

Even if you are within the voltage limits you must evaluate your power dissipation in the regulator. You are not going to supply a 7805 with 35 V and load it to 1 A because this will dissipate (35-5) * 1 = 30 W.

A voltage source with a minimum of 8 V at low line voltage and a maximum of 11.4 V at 1 amp will have a maximum dissipation of (11.4 - 5) = 6.4 W. Much more reasonable.

.

[sswitaj]

First, Gar's absolutely right, if you're expecting to pull more than a few milliamps out of your regulator, you're going to dealing with power dissipation issues.

Linear regulators do what they do by acting like a variable resistor between the supply and load. They essentially "burn off" the excess voltage, dissipating it as heat.

In your case, your power supply is 7 times the voltage of your load. That means for every 1 watt the regulator delivers as 200 mA of current into the load, it needs to throw away 6 more watts as waste heat.

In an ideal world you would probably be much better served by using a small transformer, bridge, and cap to make a bulk DC supply closer to 16 volts (for a 12 volt regulator) or 9 volts (for a 5 volt regulator). That would give you two volts of ripple before you needed to worry about drop-out.

In a system that uses a lot of 5V but only a little 12V, you might want to feed 16 volts raw DC into the 12 volt part, then take the 12 volt output and feed it to the 5 volt part, thus spreading the dissipation.

Alternately, instead of a linear supply, you might want to look into the excellent series of small switchers. Many of them are just as easy to use as the LM78/79 series, but since they run at 70% efficiency they deliver amp-sized outputs with MUCH less heat.

I design a lot of equipment that runs off 24 volt busses, and I use the PT78ST105H, 5V regulator form TI a lot. Yes, it's $14, versus a .65 cent 7805, but I don't have to cool it.

But back to your original question, resistors are a bad choice for dropping voltage, since when your load goes to zero, the dropping effect goes away, and you can damage voltage-sensitive components.

They are, however, really useful for dissipating *power*. Imagine the situation where you have a 7805 delivering 1A at 5V from an 12V source.

The '05 is dissapating 7 watts, a lot for a TO220 case. Insert a judiciously chosen dropping resistor, say 5 ohms, before the regulator, now the *resistor* is taking the beating, dissipating 5 watts, and delivering 7 volts to the regulator, which now only has to dissipate 2 watts.

You do have to worry about peak load (or you can get into dropout problems) but for stable loads, this technique can be a lifesaver. When the load goes away, btw, there's no problem, the resistor just drops less voltage, and the regulator input goes *up*, away from dropout.

And *finally*, answering your first question, "what if I really *am* only using 20mA, power *isn't* a problem and I just need to drop a couple of volts?"

Use a string of a few diodes. Each diode will drop .6 to .7v, regardless of load. Just run a 20K resistor to ground so the string is always biased "on" by a milliamp or two so that even when the regulator load goes to zero, some small current still flows and the drop still works.

[Degrom]

Thanks for all the input everyone...

I have built my main power supply really good, but did not think it would turn out to be a real issue
just to get power from it for the fans.

I think it would be better to just get a small transformer to power the fan's and keep the main power supply
uninterrupted and up to max amps for the Stepper to use.

I have a few of those transformers that plug into the wall that are rated at 12v. If I can use one of those I
would not even need to get a extra rectifier and capacitor for the smaller power supply.

[Eclipze]

You would be lucky to make a 12V 1A supply for the same cost as picking up a plugpack from the shop. I get 12V 1A plugpacks for $10. I cannot beat that cost building my own

[NC Cams]

Why ANYONE would still use a 7805 in place of an "automotive grade" LM2940? The 2940 and its derivatives offer MUCH better protection and comparable regulation while offering superior overvoltage and reverse polarity and voltage spike protection - especially in battery powered and LDO use and/or extreme environments - car underhoods are DEFINITETLY extreme environments both environmentally and electrically.

Eclipze is right: The availability of cheap wall bug transformers these days makes one think twice about the need to do add on complexity of pre-regulation and/or finding other ways to find 5 or 12vdc - life is too short. A call to Digikey or Mouser is a quick, easy fix for a prompt low voltage add on wall bug.

[gar]

071122-0947 EST USA

NC:

Good device, but it does not solve the power dissipation problem other than you can work with a slightly lower source voltage. Therefore reducing power dissipation at high AC line voltage. Above I had assumed a range of 95 to 135 V which translated to 8 to 13.4 V power supply output, as input to the regulator.

It is no solution for a 36 V input at any substantial current.

A Wakefield 623K (4.75 x 3 x 0.5) has a normal convection air thermal resistance of about 3.5 deg C/watt, or 6.3 deg F/watt. At 6 W that is about a rise of 38 deg F. If you stayed at the 6 watt level of dissipation in the regulator, then a 5 V output and 36 V in you would need to stay below about I = 6/31 = 0.19 A.

.

[Eclipze]

LM2940 is only rated to max 26V. The 7805 and 7812 etc... might have an absolute rating of 35..40V, but it doesn't have enough margin to operate with a 36V input.

You cannot put a resistor on the front of it to fix it. You could try a zener regulator combo to reduce the voltage prior to the regulator (transistor + zener) which would help burn some of the power off... but it's just the wrong approach.

Using a massive heatsink has limited benefits. Your creating a lot of heat and stressing the silicon with high temperatures.

The answer is... it's not a good solution to derive 5 and 12V from a 36V power supply with a linear regulator. Perhaps with a switch mode regulator, but the complexity of choosing the right part and associated components , wiring configuration and making sure it works across the desired current range is not something people want to mess with for hobby CNC. Plugpacks are cheap, reliable, easy to get and inexpensive.

It would be realistic to have a 5V regulator running from the 12V plugpack... otherwise, just get another plugpack for the 5V.