Electronics Experts: I need a schematic for a simple power amplification circuit.

[Jeff7]

In this thread, I asked about LED's that run on low power, and Mark R suggeted a transistor-based circuit to boost the power to the LED. (The problem in the thread was that the power LED header on my Epox 8RDA+ is putting out too little power to light up anything but a red LED.)
So I need a circuit diagram that'll allow me to boost the power put out by the Power LED header, to something that a blue LED can use. Can anyone here provide a schematic, or else some information that could help? Thank you in advance.

 

[Demon-Xanth]

Use a LM393 (comparator)

Pin 1: tie a 100 ohm to 1k resistor between it and pin 8, attach the LED to this pin and ground
Pin 2: attach a 1k resistor going to pin 8, and another to pin 4
Pin 3: this is your input
Pin 4: ground
Pins 5, 6, and 7: tie it to pin 4
Pin 8: tie to +5V

The LM393 is a dual open collector comparator. Pin 1 is the output, 2 is the inverting input, and pin 3 is the non inverting input. Pins 5-7 are tied to ground to keep the other half from oscillating. Pin 8 is the positive supply and pin 4 is ground. It works by placing the inverting input at a place between what the input signal would be switching around. The resistor from pin 1 to 8 is a pull up, it can't go high on it's own. There are other variations to this that can be done .

 

[Marshallj]

One thing you need to keep in mind is that LED's are sensitive to current more than voltage.

That's why simple LED circuits use a current limiting resistor.

Unlike a light bulb whose current increases linearly with voltage, LED's are much different. With a light bulb, if you put twice the voltage in it, it will be about twice as bright. LED's may double brightness with only a small change in voltage. This is because they have such a low resistance.

For instance, I have a super bright 5 watt LED, and I have a datasheet for it here. It is typically rated for 6.84 volts and 700ma. If you do not limit the current, things can go out of hand very fast. At 6 volts, it's only drawing around 100 ma. But by 7 volts, it is drawing nearly 1000 ma. If you don't regulate the current going into the LED, it can burn out pretty easily. My LED can take 9 volts, but if you don't use a current limiting resistor, you can burn it out by pushing 7 volts.

 

[Jeff7]

Originally posted by: Marshallj
One thing you need to keep in mind is that LED's are sensitive to current more than voltage.

That's why simple LED circuits use a current limiting resistor.

Unlike a light bulb whose current increases linearly with voltage, LED's are much different. With a light bulb, if you put twice the voltage in it, it will be about twice as bright. LED's may double brightness with only a small change in voltage. This is because they have such a low resistance.

For instance, I have a super bright 5 watt LED, and I have a datasheet for it here. It is typically rated for 6.84 volts and 700ma. If you do not limit the current, things can go out of hand very fast. At 6 volts, it's only drawing around 100 ma. But by 7 volts, it is drawing nearly 1000 ma. If you don't regulate the current going into the LED, it can burn out pretty easily.

Yeah, I saw something about that in an article on new LED's; also that as their temperature increases, their resistance drops, allowing more current through, which further increases the temperature - too much of this too, and the LED will die.


Demon-Xanth - will your circuit increase both the available voltage and amperage, or just one? The LED header seems low on both - 2.58V @ 10.1mA is the output; the blue/white LED's I've seen are rated for over 3.2V, and 20mA if I remember correctly. Will it also keep the LED from killing itself due to temperature increase, and over-current?
I found two of those LM393 things, but one's labeled linear and the other's not. What's the difference there?

 

[Demon-Xanth]

You'll want to vary the pull up resistor accordingly. You'll want a 100 ohm. You may have to vary the bridge on pin 2 to get the proper level, such as use a 4.7k and 1k ohm to get about 0.9V

The open collector's nature is that it provides a very strong pull down, (all the way to ground) when it's not pulling down the resistor will pull it up, and run most of the current through the LED.

 

[Demon-Xanth]

The LM393 is a fairly standard part,
Here's a place to get them

Linear, TI, ST, National, and Fairchild are common sources. You'll probably want the DIP-8 package.

 

[Marshallj]

Demon, you sure seem to know your electronics.

I have a question for you- I have this 5 watt LED and I want to make a regulator for it. I plan on using a 9 volt supply and the regulator to limit the current to around 800 ma. With the power involved, a simple resistor would create too much heat and be too inefficient (wasting batteries). What do you suggest I do for a regulator?

 

[Jeff7]

Thanks for the info so far; I'll draw a diagram of it for myself tomorrow, cause I'm getting a bit confused trying to visualize it - what goes where. Off to bed now (I work nightshift); there'll likely be more questions tomorrow though.

 

[Marshallj]

Originally posted by: Jeff7
[Demon-Xanth - will your circuit increase both the available voltage and amperage, or just one? The LED header seems low on both - 2.58V @ 10.1mA is the output; the blue/white LED's I've seen are rated for over 3.2V, and 20mA if I remember correctly. Will it also keep the LED from killing itself due to temperature increase, and over-current?

You can never get electricity for free in a circuit, so there's always a tradeoff. You can increase the voltage by decreaing the amperage (step up) or you can increase the amperage by decreasing the voltage (step down). But you can't increase both with a circuit because your output power must be equal to or less than your input power.

 

[Demon-Xanth]

Marshallj:
Unless you build a switching regulator, you're gonna waste the power anyways. But you'll need a 3 ohm, 3 watt resistor.

 

[Jeff7]

Originally posted by: Marshallj

Originally posted by: Jeff7
[Demon-Xanth - will your circuit increase both the available voltage and amperage, or just one? The LED header seems low on both - 2.58V @ 10.1mA is the output; the blue/white LED's I've seen are rated for over 3.2V, and 20mA if I remember correctly. Will it also keep the LED from killing itself due to temperature increase, and over-current?

You can never get electricity for free in a circuit, so there's always a tradeoff. You can increase the voltage by decreaing the amperage (step up) or you can increase the amperage by decreasing the voltage (step down). But you can't increase both with a circuit because your output power must be equal to or less than your input power.

(ok, ok, so I'm still not in bed)

Well yeah, I know that much. But it looks like this circuit is going to be plugged into the 5V line for its input - the voltage and amperage are considerably more than what the LED will need; I just need the right balance - the LED I'm going to use is rated for 3.2 - 4.3V forward voltage, and 20mA. When I connect it to the header by itself, it just barely lights up. But when I give it a 3V lithium coin battery, it is considerably brighter. I'm hoping to run it more in the higher end of its voltage range, around 4.0.

Would I be able to put a potentiometer on the circuit somewhere to adjust the output voltage? I have a multimeter, so I could test it first to make sure it is a safe voltage. I just don't want to get the circuit made up, then discover that the voltage is not much better than what it is currently from the header.

 

[Demon-Xanth]

You can add a pot where the 100 ohm pull up is, but I'd recommend atleast a 50 ohm in series with it.

 

[Jeff7]

Originally posted by: Demon-Xanth
You can add a pot where the 100 ohm pull up is, but I'd recommend atleast a 50 ohm in series with it.

Ok; how much resistance between pins 1 and 8 will yield what voltage? The reason for this question is that I need to know what rating of potentiometer I'll need before I order from Electronic Goldmine.

Edit: Ok, I drew up a little diagram.

Some questions:
"Pin3 - this is your input."
Is that the positive pin of the LED header on the motherboard?

"Pin4 - ground"
Is that the -5V line on the molex connector? What about the negative pin of the LED header?

 

[Demon-Xanth]

Almost 100% of the current will be flowing through the LED when the LED is to be lit. When it's not lit, the comparator will be taking 100% of the current (at 5V)

Voltage=Current*Resistance
Power=Voltage*Current

 

[Jeff7]

What do you think of the diagram? Is it at least accurate as far as the connections go? I've no experience at making up schematics. I can read simple ones, as far as designing a circuit board trace diagram is concerned; I imagine some sort of CAD-type software would be helpful for actual circuit design though.

 

[Demon-Xanth]

Pin 3 is going to be whatever pin on the MB header changes, there is no "-5V" on the molex connectors, just ground and +5 (and +12), the LED will go from the pot to ground. I would recommend adding a resistor in series with the pot (towards the VCC pin) so there no way to tie the pin permenantly high, as that will fry the chip REAL quick.

 

[PrincessGuard]

Just a little side track, but is there any benefit for using the LM393 rather than a driver IC like the ULN2003?

 

[Demon-Xanth]

The LM393 is a VERY common IC. Also, the threashold of the LM393 can adjusted by the resistor bridge on the inverting input (pin 2). The LM393 also features an open collector output that allows it to easily drive a LED.

...and I just pulled it off the top of my head.

 

[Jeff7]

Originally posted by: Demon-Xanth
Pin 3 is going to be whatever pin on the MB header changes, there is no "-5V" on the molex connectors, just ground and +5 (and +12), the LED will go from the pot to ground. I would recommend adding a resistor in series with the pot (towards the VCC pin) so there no way to tie the pin permenantly high, as that will fry the chip REAL quick.

So I'll water cool it.

Ok, so a 100 ohm pot, and a 100 ohm resistor? The resistors you are prescribing, they are all just common 1/4W resistors?

 

[Demon-Xanth]

Yeah, 1/4W should be pleanty, though you might want 1/2W on that 50 ohm.

 

[Mark R]

With the power involved, a simple resistor would create too much heat and be too inefficient (wasting batteries). What do you suggest I do for a regulator?

Nat semi LM2575

A lot more expensive than a resistor though, plus needs a sizeable inductor. You'll need a current sense amplifier also to run it in constant current mode.

 

[Demon-Xanth]

Originally posted by: Mark R

With the power involved, a simple resistor would create too much heat and be too inefficient (wasting batteries). What do you suggest I do for a regulator?

Nat semi LM2575

A lot more expensive than a resistor though, plus needs a sizeable inductor. You'll need a current sense amplifier also to run it in constant current mode.

Also, there is not THAT much power involved. The worst case scenario is only about half a watt. And that is in a failure state (the output is shorted to ground). Also, a linear regulator STILL would dissipate the same amount of heat as a resistor and just add complexity. And you'd still need a way of turning it on and off.

In the end, it's 4x the complexity with 0 power savings, and most likely more power lost. And you're talking about a system w/ a CPU that takes 50W of power. That's 100x what this would ever pull.

Don't over think the problem.

Edit:
A real life example of overthinking the problem (true story):
An engineer was appointed to make a battery charger. He made one with all of the bells and whistles to make sure the batteries didn't get overcharged, and sense when they're charged enough. The people using it were upset because the batteries weren't lasting very long. Another engineer came up with a solution: replace the entire board with a resistor. The batteries stayed charged much longer, there were no problems. Everyone was happy. (I've worked with both of these engineers, but this was at a different company)

 

[Marshallj]

Originally posted by: Demon-Xanth

In the end, it's 4x the complexity with 0 power savings, and most likely more power lost. And you're talking about a system w/ a CPU that takes 50W of power. That's 100x what this would ever pull.

Don't over think the problem.

Edit

The problem with my LED is that it's 5 watts and the battery voltage will drop when I'm using it. If I make a current regulator, I can make sure that the LED gets the same amount of current (and stay the same brightness) even as the voltage on the batteries drop.

 

[Jeff7]

Ok, I'm back here. I've updated the "diagram" - here.

I updated your instructions, Demon-Xanth, to include the potentiometer for adjustable output.

Pin 1: tie a 50 ohm (1/2W) resistor and 100 ohm trimmer between it and pin 8, attach the LED to this pin and ground
Pin 2: attach a 1k resistor going to pin 8, and another 1K to pin 4
Pin 3: this is your input
Pin 4: ground
Pins 5, 6, and 7: tie it to pin 4
Pin 8: tie to +5V

Now I know that the pin 3 is the input - the positive of the LED header, but I couldn't find in the thread where the negative input goes.

 

[dmcowen674]

Radioshack used to have a great starter basic Electronics book series. Hopefully they still have it and you could have everything you need.