I hooked a 12v battery up to a 1A 5v linear regulator at the input.
then hooked two amp meters in series with the input and output.
also measuring the voltage.
when I shorted the output. it output 1A at 0v (or very small voltage)
so only 1 watt on the output.. With 1A supplied at 5v on the output. then it would be outputting 5 watts maximum.
and the input current from the battery was less than 0.4A. so it has some form of PWM to increase efficiency? It was out of an old TV motherboard.
I hooked it up to my computers USB output. which was only 5.14v and it was still giving exactly 5.00v on the dot. so it has very very low dropout voltage.
then hooked two amp meters in series with the input and output.
also measuring the voltage.
when I shorted the output. it output 1A at 0v (or very small voltage)
so only 1 watt on the output.. With 1A supplied at 5v on the output. then it would be outputting 5 watts maximum.
and the input current from the battery was less than 0.4A. so it has some form of PWM to increase efficiency? It was out of an old TV motherboard.
I hooked it up to my computers USB output. which was only 5.14v and it was still giving exactly 5.00v on the dot. so it has very very low dropout voltage.
Have you looked up the device part number ?
Having 1 amp flowing in a short circuit dissipates zero wattage in the load (the short). Many regulators have 'foldback' protection which would reduce the current into a short.
0.4 amps flowing into the regulator under short circuit conditions would dissipate 4.8 watts in the regulator, and none in the short.
Having 1 amp flowing in a short circuit dissipates zero wattage in the load (the short). Many regulators have 'foldback' protection which would reduce the current into a short.
0.4 amps flowing into the regulator under short circuit conditions would dissipate 4.8 watts in the regulator, and none in the short.
Can you identify the regulator? It would be better to load it with a resistor than to short the output, anything could be happening under those circumstances, including release of the smoke.... did it take me 6mins to write that? I must have gone through a time warp...
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Well even if you have 1A and 0v it would be at least 1 watt.
if you had 0v. a short. but 5A were flowing in the 0v short. you would have 5 watts dissipated.
so 12v input and 5v1A output. if you had 0.5A input it'd be just over 5 watts dissipated in the regulator.. Somehow it will always dissipate around 5 watts no matter what the input voltage is. because it manages to reduce the input current in proportion to the voltage. so at 24v input. it might be around 0.2A input current? i havent tested that yet.
but i believe it might be true.
somehow its using PWM thats the only way you can get linear regulation to be any bit efficient at all..
its a KIA 7805A
it seems to have current limiting function built in rather than cutting off the output. so instead of cutting off the output. it allows the voltage to drop but only lets 1 amp flow on the output.
so it seems to only be 50% efficient dissipating half the power as heat. No matter what the circumstances.
Unless the voltage input is close to the output voltage. For example 6.5v input voltage. It would be more than 50% efficient. unless the voltage input raises above 2x the output voltage. then it becomes 50% efficient from that point on.
I find it facinating. but useful. I could use this as a phone charger from a solar panel. connected directly.. I think a 20W 12v solar panel with 1A max will be enough to charge a phone with it.
It would get pretty hot but having the massive heatsink i've got on it surely helps.. its attatched to a heatsink with a mounting clamp to clamp it to the heatsink. the mounting bracket also has a small section of fins to dissipate the heat as well.
so it has no chance of overheating in normal operation.
Whats the maximum input voltage? for safety margin.
if you had 0v. a short. but 5A were flowing in the 0v short. you would have 5 watts dissipated.
so 12v input and 5v1A output. if you had 0.5A input it'd be just over 5 watts dissipated in the regulator.. Somehow it will always dissipate around 5 watts no matter what the input voltage is. because it manages to reduce the input current in proportion to the voltage. so at 24v input. it might be around 0.2A input current? i havent tested that yet.
but i believe it might be true.
somehow its using PWM thats the only way you can get linear regulation to be any bit efficient at all..
its a KIA 7805A
it seems to have current limiting function built in rather than cutting off the output. so instead of cutting off the output. it allows the voltage to drop but only lets 1 amp flow on the output.
so it seems to only be 50% efficient dissipating half the power as heat. No matter what the circumstances.
Unless the voltage input is close to the output voltage. For example 6.5v input voltage. It would be more than 50% efficient. unless the voltage input raises above 2x the output voltage. then it becomes 50% efficient from that point on.
I find it facinating. but useful. I could use this as a phone charger from a solar panel. connected directly.. I think a 20W 12v solar panel with 1A max will be enough to charge a phone with it.
It would get pretty hot but having the massive heatsink i've got on it surely helps.. its attatched to a heatsink with a mounting clamp to clamp it to the heatsink. the mounting bracket also has a small section of fins to dissipate the heat as well.
so it has no chance of overheating in normal operation.
Whats the maximum input voltage? for safety margin.
http://www.kec.co.kr/data/databook/pdf/KIA/Eng/KIA7805AP~KIA7824AP.pdf
You might get low drop out with no load
You might get low drop out with no load
You must be using a very unusual arithmetic system if 1 times 0 >= 1.
Anyway, the KIA7805A is a simple series regulator (see http://www.kec.co.kr/data/databook/pdf/KIA/Eng/KIA7805AP~KIA7824AP.pdf ), so it can't possibly output more current than goes in. Did you perhaps measure the output current directly after applying the short and the input current a bit later, after the regulator had heated up and the thermal protection had kicked in?
Anyway, the KIA7805A is a simple series regulator (see http://www.kec.co.kr/data/databook/pdf/KIA/Eng/KIA7805AP~KIA7824AP.pdf ), so it can't possibly output more current than goes in. Did you perhaps measure the output current directly after applying the short and the input current a bit later, after the regulator had heated up and the thermal protection had kicked in?
The power dissipated in a DC circuit like this is the product of the voltage and the current. So 0 v (as measured across the short) and a current of 1 amp gives 0*1 which is 0.
Well if it was 1v and 1 amp then it would be 1 watt
but no wire is perfect. so even if it measures 0v. there would still be a very tiny voltage across the "short" or "wire" so it would have a small amount of wattage dissipated.
Im correcting myself because I made a mistake.
The thermal protection had not kicked in. it was still just barely warm to the touch. the heatsink was barely warm.
It must be some sort of special PWM linear regulator that adjusts the output to get exactly 5v through some sort of PWM.
because it would dissipate an enormous amount of power if you fed it 15v and 1A. it'd be dissipating 10 watts when outputting 5v 1A!!! a more efficient way would be very high frequency simple PWM so the maximum dissipated power would be more closer to 5 watts.
I gave it a load of 0.5A on the output at 5v. so 2.5W
and at the input of 12v it was just between 0.2A and 0.3A when I checked with my multimeter.
but no wire is perfect. so even if it measures 0v. there would still be a very tiny voltage across the "short" or "wire" so it would have a small amount of wattage dissipated.
Im correcting myself because I made a mistake.
The thermal protection had not kicked in. it was still just barely warm to the touch. the heatsink was barely warm.
It must be some sort of special PWM linear regulator that adjusts the output to get exactly 5v through some sort of PWM.
because it would dissipate an enormous amount of power if you fed it 15v and 1A. it'd be dissipating 10 watts when outputting 5v 1A!!! a more efficient way would be very high frequency simple PWM so the maximum dissipated power would be more closer to 5 watts.
I gave it a load of 0.5A on the output at 5v. so 2.5W
and at the input of 12v it was just between 0.2A and 0.3A when I checked with my multimeter.
I looked in the circuitry diagram and read some of the datasheet info on it
it seems there might be some form of minimal PWM to keep the efficiency reasonable.
that works with the current limiting.
so its still better than a zener diode and transistor with a current limiting circuit added on.
because if it was just that simple. then it would be taking 1A input no matter what the input voltage is. and it'd be dissipating an enormous amount of power!
it seems there might be some form of minimal PWM to keep the efficiency reasonable.
that works with the current limiting.
so its still better than a zener diode and transistor with a current limiting circuit added on.
because if it was just that simple. then it would be taking 1A input no matter what the input voltage is. and it'd be dissipating an enormous amount of power!
The internal circuits are in the data sheets. There are only transistors inside, no clock circuits, I have never seen a linear regulator IC with a pulse modulation circuit.
Why couldn't it be done then?
It shouldn't be that hard or take that much circuitry?
if the output voltage is fixed. then all the components can be built into the one chip even the capacitors..
and it would have very good efficiency if it was PWM with current limiting.
It still doesn't explain how the input and output current were different!!
It must be to do with the current limiting somehow creating a false PWM effect and making it more efficient.
It shouldn't be that hard or take that much circuitry?
if the output voltage is fixed. then all the components can be built into the one chip even the capacitors..
and it would have very good efficiency if it was PWM with current limiting.
It still doesn't explain how the input and output current were different!!
It must be to do with the current limiting somehow creating a false PWM effect and making it more efficient.
Errr......didya see DUG's link?.....that's the real deal....expensive though. Sorry, wasn't a lik exactly. but you get my drift
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It isn't hard to make PWM regulators in a single IC, they already do it. But sometimes you just want a linear regulator. Look up switching regulator ICs.
A linear three-leg regulator is stone simple to implement, and it has no noise issues. As soon as you get into switching regulators. you have serious noise considerations. Yes, good designs minimizes those issues, but they cannot be ignored.
A linear three-leg regulator is stone simple to implement, and it has no noise issues. As soon as you get into switching regulators. you have serious noise considerations. Yes, good designs minimizes those issues, but they cannot be ignored.
if its integrated into an IC and has a very high switching frequency with good built in filtering capacitors integrated into the chip you wouldn't even need a single external component.
but I guess no one wants to make a single-chip highly efficient PWM dedicated regulator
if they do exist they probably would be too expensive to be feasible.
but I guess no one wants to make a single-chip highly efficient PWM dedicated regulator
if they do exist they probably would be too expensive to be feasible.
There is a bit more to it than that I'm afraid.
The whole point of a switching supply is efficiency and that means having the control element (transistor) either fully on or fully off. Some means of storing the energy needs to be used, and for efficiency that means magnetic components such as an inductor.
The inductor can be configured and used in ways that a capacitor can not when it comes to energy storage.
The whole point of a switching supply is efficiency and that means having the control element (transistor) either fully on or fully off. Some means of storing the energy needs to be used, and for efficiency that means magnetic components such as an inductor.
The inductor can be configured and used in ways that a capacitor can not when it comes to energy storage.
P = I*V = I^2*R = V^2/R
If you know that V= 0, then it follows that the power is 0 (zero).
well the thing is you can have a basic PWM that just adjust the output voltage from PWM frequency and a smoothing capacitor with feedback and a resistor.
since its not a buck converter it doesn't need an inductor when its basic PWM
which would still be more efficient than linear regulation.
thats how solar charge controllers work without an inductor. they just switch the solar panel on and off with PWM.
if you did it at high enough frequency with good enough regulation and smoothing on the output and feedback you could possibly still do it in one chip might draw a few milliamps of current but that shouldn't affect the efficiency too much.
since its not a buck converter it doesn't need an inductor when its basic PWM
which would still be more efficient than linear regulation.
thats how solar charge controllers work without an inductor. they just switch the solar panel on and off with PWM.
if you did it at high enough frequency with good enough regulation and smoothing on the output and feedback you could possibly still do it in one chip might draw a few milliamps of current but that shouldn't affect the efficiency too much.
If the series pass element is a MOSFET with very low RDSon and you have a proper gate driver (bootstrapped voltage over the output), the voltage drop may be very little.
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