Driving a LED

[audioguru]

They arent in series with each other, it seems there wasnt enough voltage to drive them in series.

You don't need the 555. If you have a 9V battery then two 3.3V LEDs in series need 6.6V and a current-limiting resistor for 15mA is 160 ohms. Two of these results in four LEDs operating at 15mA each. The same as nine in parallel operating at 6.67mA each. My current is 30mA. Your current is 66mA. Mine and yours make the same amount of light. My battery lasts a lot longer than yours. My circuit is very simple.

So each LED is getting Pulses of 66mA (They are in parallel of each other..)

No, your LEDs are dim.
9 identical LEDs in parallel share the total current. So each LED gets only 66mA/9= 7.33mA.

My LED flashlights flash to save power.
They flash five 33% duty-cycle pulses at about 10Hz then a pause of about 1/2 a second. The pulse current is 100mA and the blue or white ordinary LEDs are about 4.5V at such a high current. They are extremely bright with just a single LED and the average current is 100mA x 0.33 x 0.5= 16.5mA. I use a low-dropout 5V regulator to keep the brightness the same as the 9V battery voltage drops to 6V as it runs down. The little 9V battery lasts a long time.

 

[Overclocked]

You don't need the 555. If you have a 9V battery then two 3.3V LEDs in series need 6.6V and a current-limiting resistor for 15mA is 160 ohms. Two of these results in four LEDs operating at 15mA each. The same as nine in parallel operating at 6.67mA each. My current is 30mA. Your current is 66mA. Mine and yours make the same amount of light. My battery lasts a lot longer than yours. My circuit is very simple.


No, your LEDs are dim.
9 identical LEDs in parallel share the total current. So each LED gets only 66mA/9= 7.33mA.

My LED flashlights flash to save power.
They flash five 33% duty-cycle pulses at about 10Hz then a pause of about 1/2 a second. The pulse current is 100mA and the blue or white ordinary LEDs are about 4.5V at such a high current. They are extremely bright with just a single LED and the average current is 100mA x 0.33 x 0.5= 16.5mA. I use a low-dropout 5V regulator to keep the brightness the same as the 9V battery voltage drops to 6V as it runs down. The little 9V battery lasts a long time.

So your saying If I used JUST a 9V and hooked 2 LEDs in series with a resistor, I would have more effciency Vs if I used a 555 Timer? Impossible! Arent you wasting current through the resistor? Isnt power Wasted through the LDR?

Err, I should have been more clearer, Those Numbers Are TOTAL Current going through the whole circuit, but if the 555 timer consumes only 1 to 2mA, then its pretty close.

Now Im reading that Pulsing the LEDs ruins the yellow phospher coating on the white LEDs, Is this true?

Speaking of LEDs, did anyone catch last nights Mythbusters? They found out LEDs have the longest Life time and Only use 1W per hour. I wonder how the power company would charge you if you used LEDs throughout your house, since they charge by the kW/hour. Id bet they would round up somehow...Wonder if the meters could even read in watts...

 

[eng1]

Now Im reading that Pulsing the LEDs ruins the yellow phospher coating on the white LEDs, Is this true?

I think that they couldn't be damaged. But if the frequency is high enough, you wouldn't see a white light.


Can you post the schematic of you circuit?

 

[audioguru]

If you pulse an LED with 20mA and a 50% duty cycle, it would look exactly as bright as the same LED with 10mA continuous. The battery power used is also the same, so the pulsing doesn't save anything.

An LED's datasheet has a max continuous current amount and a max pulsed amount that is higher and states the pulse-width duration. If you operate LEDs at their max rating then some will die early and most will be at half brightness in about 10 years.

 

[Overclocked]

I think that they couldn't be damaged. But if the frequency is high enough, you wouldn't see a white light.


Can you post the schematic of you circuit?

Can Do. I didnt show the different emitter resistors with the switch.

 

[audioguru]

Your 555 circuit makes a 67% on, 33% off duty-cylcle. So the LEDs are dimmed 33%. Why? Your switched resistors should do the dimming.

 

[Overclocked]

Your 555 circuit makes a 67% on, 33% off duty-cylcle. So the LEDs are dimmed 33%. Why? Your switched resistors should do the dimming.

So If I understand Correctly, I should have eliminated the emitter resistor all together? Or are you saying I should Eliminate the 555 Timer all together and Just use the transistor?

 

[audioguru]

If you eliminate the current-limiting resistor then the LEDs and transistor blow up.

The 555 and the transistor are doing nothing but dimming. An ordinary resistor can do dimming just as well all by itself.

 

[Overclocked]

If you eliminate the current-limiting resistor then the LEDs and transistor blow up.

The 555 and the transistor are doing nothing but dimming. An ordinary resistor can do dimming just as well all by itself.

Yes but how long will a battery last when using resistive circuits or my circuit?

 

[freeskier89]

If you pulse an LED with 20mA and a 50% duty cycle, it would look exactly as bright as the same LED with 10mA continuous. The battery power used is also the same, so the pulsing doesn't save anything.

What he has been trying to say along is that it will last the same amount of time

The power consumption of the two setups are identical.

BTW: I did see the mythbusters the other night... I loved the explosives

 

[Sceadwian]

If the circuit will naturally provide 20ma's but the LED will only handle 10 continous, then using a resistor will let you run it, but the resistor will drop the voltage (as heat and wasted power) If you properly adjust the duty cycle of an LED and don't violate it's maximum junction temperature (even instantly) it can be run off of the same battery without the resistors hence it's more power efficient the battery will last longer, because the resistors aren't being used to waste power, the PWM duty cycle is used to moderate the total average power dissapated in the LED itself. With LED's the big picture is actually the junction temperature, which is difficult to manage, and very difficult to find pulse derateings for.

 

[audioguru]

The max continuous current for most ordinary LEDs is 30mA. The max peak current is 100mA. You must limit the peak current to 100mA or the LED will blow up.
PWM limits the power, not the current.

The voltage of a battery drops as it is being used up. To keep the brightness of an LED flashlight from dimming too much, a voltage regulator can be used as an automatic series resistor, or a fixed series resistor that absorbs some of the extra power.

 

[Overclocked]

The max continuous current for most ordinary LEDs is 30mA. The max peak current is 100mA. You must limit the peak current to 100mA or the LED will blow up.
PWM limits the power, not the current.

The voltage of a battery drops as it is being used up. To keep the brightness of an LED flashlight from dimming too much, a voltage regulator can be used as an automatic series resistor, or a fixed series resistor that absorbs some of the extra power.

So a LDO Operating in Constant Current mode would work?

 

[Hero999]

The problem with that is there's always a voltage loss even if it has a dropout of 0V. For example a perfect 5V regulator will loose 5V working in constant current mode.

 

[audioguru]

A two-transistors current regulator is good enough to mintain the brightness of LEDs as the battery voltage runs down.

 

[Overclocked]

A two-transistors current regulator is good enough to mintain the brightness of LEDs as the battery voltage runs down.

But what about efficiency?

I Just Built that Circuit and experimented for a half hour. Looks like My battery is Dieing, I measured it to be around 6.67V. This maybe a good thing, since it would give me an Idea of How the circuit would perform when the battery is near its death.

Interestingly enough, I used 8 Blue LEDs (Each is Rated at 3.6V 20mA) And Got around 22mA Draw Total. Each LED Got 3V. I used Blue since I dont have any White LEDs LEDs left.

Heres what I used for values:

 

[audioguru]

Why did you use 4 strings of LEDs? With a total of only 22mA then each string draws only 22/4= 5.5mA. One string of two LEDs in series would produce the same amount of light at 22mA and the battery would last as long.

Why is the value of the pot only 1k? It is wasting power. The current gain of the transistor is typically 150 so if the battery voltage drops to 7V and the LED current is 22mA then the value of the pot is (7V-6V)/(22ma/150)= 6.8k. Use a 10k pot.

Replace your dead battery.

 

[Overclocked]

Why did you use 4 strings of LEDs? With a total of only 22mA then each string draws only 22/4= 5.5mA. One string of two LEDs in series would produce the same amount of light at 22mA and the battery would last as long.

Why is the value of the pot only 1k? It is wasting power. The current gain of the transistor is typically 150 so if the battery voltage drops to 7V and the LED current is 22mA then the value of the pot is (7V-6V)/(22ma/150)= 6.8k. Use a 10k pot.

Replace your dead battery.

To maximize brightness.(or atleast I thought it did...Maybe all it did was increase the angle?)

I used 1K for experimentation, I do know 6.8k is a standard value, but I'll use 10k for experimentation.

 

[audioguru]

Your transistors have a very wide range of current gain and you don't know the amount for the transistors you have, so you need the trimpot to adjust for the amount of output current.

 

[Overclocked]

Your transistors have a very wide range of current gain and you don't know the amount for the transistors you have, so you need the trimpot to adjust for the amount of output current.

I Fully Optimized the circuit today. I used a 12V source but I used a pot to adjust the voltage to appr 8.99V.

I used 2 LEDs in series. The total current being consumed is 21.20mA, and Each LEDs Gets 3V. I also noticed that when I adjusted the voltage That the LEDs remain at the same brightness no matter what the Input voltage (but this is to be expected of a constant current source). The LEDs start to get dim below 6.65V.

What would happen if I changed the Transistor to say, a darlington transistor?