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Basic Electronics » Voltage Regulator

May 25, 2010
by kgb
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if I input a 10V to a 5V voltage regulator, there would be a 5V extra. Where would it go? So, does the voltage regulator dissipate the extra 5V?

-KGB

May 25, 2010
by mcai8sh4
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The extra energy is dissipated as heat (you may notice the metal part getting warm).

-Steve

May 25, 2010
by wayward
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Also, a part of the current is used to drive the regulator itself.

-Zoran

June 05, 2010
by devinsbusiness
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If you are trying to push much amperage through it you may want to try an LM7805 regulator. It is made to handle up to 1 amp. Devin

June 05, 2010
by mongo
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Another little trick is to use the 7805 to drive a 2N3055 transistor and get even more current. But the 7805 can usually handle 1500 mA. It does get hot, especially if your supply is higher and a heat sink is recommended.

June 08, 2010
by Ralphxyz
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So with a 10v supply would the LM7805 be more efficient with a 750 mA current just because it didn't "heat up" as much?

There still is the original question of the missing 5 volts, it is still missing, where is it going now if not in heat?

Ralph

June 08, 2010
by bretm
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Heat isn't the only option. Switching converters using active components (inductors, capacitors, transistors) are more efficient than using resistors and generate less waste heat. As an illustration, imagine converting the 10V DC to AC, running it through a 2-to-1 step-down transformer, and then rectifying it again.

June 08, 2010
by BobaMosfet
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Voltage is not dissipated in heat. Only the flow of current generates heat.

BM

June 08, 2010
by mongo
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OK, picture this...

Using a 7805 regulator and drawing 1 amp at the output. You are using 5 watts of power as the load. Now, if you are supplying it with 7.5 volts, your input power will be just over 7.5 watts. (Including the internal circuit draw). If you are supplying it with 15 volts, you would have an input power of 15+ watts. Still 1 amp or so but now you are changing voltage to watts, in turn, it becomes heat. It is more efficient to drive the circuit with the lowest possible source simply based on the watts dissipated. The higher the supply voltage, the higher power dissipation and waste heat.

June 08, 2010
by Ralphxyz
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Ok, so if I have a solar generated power source with each volt at a premium.

Would it be best to store power at 6 ,12 ,18 or 24 volt for a 5 volt application?

Ralph

June 08, 2010
by mongo
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Well, since most anything you find out there commercially starts at 12V, it kinda answers it for you. Changing DC to DC is less efficient that AC to AC but most systems are designed around these losses. There are ways of converting that are more efficient, such as actual DC to DC converters but they are not cheap.

My solar/wind system is 12V but I don't use it for the low voltage applications. I use an inverter to create house current so I can operate appliances and standard lighting.

If you want to build a dedicated system for 5V applications,with what is readily available for batteries, an 8 volt system would be the most efficient but not by much over a 12 volt system. 2V cells are harder to find, as are the 4V batteries, (Though you can find them in those small emergency lights that hang on the walls in places like office buildings)

I would stick with the 12 volt solar. Custom voltages would be at a higher premium than it would be worth.

Something to keep in mind with solar... even though my system is 12V, the solar panels generate up to 36 volts at no load. Loaded, they run around 18 volts but the controller does the work from there. It in itself is not all that efficient but it's as good as it can get with current technology. (well, 2004 technology anyway) My panels net me about 250 watts to charge the batteries and the wind turbine can generate up to 400 watts at 23 MPH winds. (Less at lower wind speeds) So I have the capacity of about 10 KWH per day before taxing the system.

June 09, 2010
by BobaMosfet
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mongo-

Ahhhh grasshopper. You must meditate on my statement to truly discern it's meaning. And your eyes will be opened. Your quick answer shows that you do not yet see with your inner electronics eye.

BM

June 09, 2010
by mongo
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Boba... Huh?

True, voltage is not dissipated as heat but wattage is. You still have to deal with that part of it. Input current will pretty much be the same as output current but any current at excess voltage is watts wasted. That does go up in heat.

June 10, 2010
by Ralphxyz
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Well if "voltage is not dissipated as heat" it still disappears.

The original question is still of interest to me, how efficient is a voltage regulator? Apparently just using a voltage regulator means voltage loss rather as heat or ether or whatever.

I am picturing a multi-use solar powered system with different voltage requirements so one project uses 5 volts another might use 12 volts or more. It seems as if one is powering a motor a higher voltage is used (depending on the load of course).

Thanks for the discussion so far it has been interesting.

Ralph

June 10, 2010
by mongo
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This goes back to the reduction in DC voltage. 12V into a 5V regulator decreases it by 7V. If you are dropping the voltage at 7V and one Amp, it equates to 7 watts dissipated as heat. similarly, if your current draw is only 500mA, then the dissipated wattage is cut to 3.5 watts. It's pretty linear in that respect.

Since the regulators typically require a minimum of 7.5V to give you a regulated 5V output, you still lose 2.5Watts as heat at one Amp draw. The lower the current draw, naturally, the wattage decreases with it. That is why I mentioned the 8V system for efficiency. 12V systems are just more practical, though there is a higher loss overall.

June 10, 2010
by BobaMosfet
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Ralphxyz,

Voltage is not something you can dissipate. It isn't a solid, material. Voltage is not energy. Voltage is a term that describes an aspect of energy. It is literally a relative measure of charge.

A linear voltage regulator, like the LM7805 is not as efficient as a switching regulator, because it is using a transistor like a resistor. That 6mA quiescent current is used by the regulator to control a voltage-mirror and and transistor or transistor array within the regulator. The voltage mirror usually generates a 1-volt reference, so the regulator has something to compare against, and then it adjusts the relative charge disparity between the ground and output pin so that there is a 5VDC difference. It is built robust enough to handle up to 7.5 Watts of energy (5VDC across it at 1.5 Amps). If the downstream load tries to draw more than that, the LM7805 protects itself. But if it didn't, it would begin to break down and fail. A Linear regulator usually generates a very smooth voltage and it is very responsive to changes in the input side.

A switching regulator (usually) uses a transistor like an on/off switch. It's output while much more efficient, tends to be noisier (not smooth), and it is not necessarily as fast to respond as the linear regulator to voltage changes on the upstream side. However, depending on model, it may have the ability to output more voltage than is being input-- a very handy feature.

Voltage doesn't disappear, it's right there. Hook the LM7805 to a 12V 800Amp car battery and leave the output pin disconnected. The regulator will never get warm, yet seems like it's a controlled short with 9600 Watts of energy potentially bearing down on it! It doesn't fry, because it's only utilizing 6mA at approximately 2VDC - in other words it's only dealing with approximately 12mW.

And here's the magic no one understands: Current resisted is current not dissipated and therefore cannot get warm. Only the current that is NOT resisted (that is allowed to flow through the component) can generate heat.

Doesn't matter if its an LM7805, a resistor, an IC, a lightbulb, a 22kV transmission line, or a Lighting Bolt-- it works the same way.

Hope that helps

BM

June 10, 2010
by BobaMosfet
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I amend my statement "And here's the magic no one understands". I'm sure many people do, but the problem is no one had stated it, and this is integral to helping less knowledgeable folks understand this stuff.

BM

June 10, 2010
by BobaMosfet
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I also add that I may come across as pompous, conceited, arrogant, or condescending. This has more to do with my weird, slightly informal way of writing than with how I feel, except sometimes I'm frustrated because I don't have a whiteboard so I can draw pictures (which I find much more pleasing that endless amounts of text) to share stuff.

I offer apologies to anyone who may feel I was being PCAC, it was not meant at all.

Steps off soapbox

BM

June 10, 2010
by mcai8sh4
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BM : haha, don't worry about it, your explanations are always appreciated (at least by me) you never insult people on a personal level, so in my book, there can be no problem.

Thanks for all the information, I read practically every post in these forums, and have learnt a lot. Now I want you to have a whiteboard.

Keep it up!

Steve

June 10, 2010
by Ralphxyz
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Really, I understand pictures a lot better than text. Especially those rambling explanations usually seen on the web. Boba your post are always appreciated, by this one of little learning.

Ralph

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