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Series Regulation

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Then the LEDs instantly burn out because the peak current is too high.
The peak forward current is much higher than average current. Design will ensure that current will not exceed permissible limits (As i mentioned earlier LED is are connected in parallel serial combination with series resistance with each branch)

Of course the battery limits the current. It and its connecting wires have resistance.
The mosfet also has on-resistance that limits the current.
These current-limiting resistances either are too low so the current is too high and the LEDs burn out, or the resistances waste a lot of battery power like a linear regulator.

PWM reduces the heating of the Mosfet driver if its on-resistance is low enough. An external current-limiting device is needed and it wates power by making heat. Efficiency is not any better than a linear regulator and might be worse.

A buck converter increases efficiency.
It is an accepted fact that switching regulation is far more efficient than series regulation where efficiency comes down as input voltage increases. Here we have to conclude that only switched mode regulation will meet my needs.
Then about buck converter. What I am suggesting is also something with the same principle. Only difference is conventional buck converter uses and inductor to store energy and supply it to the load when the series transistor is OFF. In this case since the load is LED I am driving higher current (of course with in permissible peak current limits) when transistor is ON to maintain the average brightness and there is no current flow when it is OFF. I want to avoid inductor.
 
The only way you can use PWM and save power is to add an inductor and catch diode, but I don't think your MC is fast enough to implement a buck converter in software. The inductor would be quite large at low frequencies so you'd need to PWM at 20Khz or higher. An ADC on the MC would be needed also.
 
The only way you can use PWM and save power is to add an inductor and catch diode,
Pls note that here my load is LEDs. I have used LEDs earlier also for dynamic displays. Here it will be something like that. Higher voltage applied for lesser duration (duty cycle) to maintain same brightness.
but I don't think your MC is fast enough to implement a buck converter in software. The inductor would be quite large at low frequencies so you'd need to PWM at 20Khz or higher. An ADC on the MC would be needed also.
As I mentioned I plan to use LEDs (something like dynamic display) and avoid inductor. Instead of supplying DC to load a pulsed voltage will be supplied.
 
Ok, let me explain it to you this way. Let's say you have 12V, a LED string with a Vf of 10V, and a 100Ω resistor in series. This means that there will be 2V across the resistor and 20ma will flow. That will mean there is 40mW wasted in the resistor.
Now, let's say you have 14V, a LED string with a Vf of 10V, and a 100Ω resistor in series. This means that there will be 4V across the resistor and 40ma will flow. That will mean there is 160mW wasted in the resistor with no PWM. Now if we PWM it at 50% that means that an average of 80mW would be wasted in the resistor and an average of 20ma will flow.
If you have 14V, a LED string with a Vf of 10V, and a 200Ω resistor in series then there will be 4V across the resistor and 20ma will flow. That will mean there is 80mW wasted in the resistor with no PWM.
As you can see, there is no power saving using PWM without an inductor.
 
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Leds are less efficient at high peak currents. They waste the extra power as heat.
Your current-limiting resistors also waste the extra power as heat.

A buck regulator stores power, it doesn't waste power.

Philips Lumileds make buck and boost converters that have the inductor built-in and they are very efficient. They sense and regulate the average current.
 
Ok, let me explain it to you this way. Let's say you have 12V, a LED string with a Vf of 10V, and a 100Ω resistor in series. This means that there will be 2V across the resistor and 20ma will flow. That will mean there is 40mW wasted in the resistor.
Now, let's say you have 14V, a LED string with a Vf of 10V, and a 100Ω resistor in series. This means that there will be 4V across the resistor and 40ma will flow. That will mean there is 160mW wasted in the resistor with no PWM. Now if we PWM it at 50% that means that an average of 80mW would be wasted in the resistor and an average of 20ma will flow.
If you have 14V, a LED string with a Vf of 10V, and a 200Ω resistor in series then there will be 4V across the resistor and 20ma will flow. That will mean there is 80mW wasted in the resistor with no PWM.
As you can see, there is no power saving using PWM without an inductor.

I agree with all your explanations. Pls see the following explanation and correct it if you find necessary.
Take the same example.
At 12V input with out PWM current taken is 20mA. So power from 12V is 12x0.02 = 240mW.
At 14V with out PWM power = 14x0.04= 560mW
With 50% PWM power from 14V = 14x0.04/2 = 280mW only.(extra 40mW in the resistor. But total power has come down drastically)
Am I Right?
 
Leds are less efficient at high peak currents. They waste the extra power as heat.
Your current-limiting resistors also waste the extra power as heat.

A buck regulator stores power, it doesn't waste power.

Philips Lumileds make buck and boost converters that have the inductor built-in and they are very efficient. They sense and regulate the average current.
Yes. Here there is an important point. Whether these high brightness LEDs use for traffic lights are suitable for dynamic display. Their light O/P unlike some other type of LEDs may not increase much with increase in current from 20 to 40 mA. Then with this type of regulation, the lights will become dim when input voltage increases since duty cycle is is reduced.
In that case buck converter alone will be solution.:)
 
At 14V with out PWM power = 14x0.04= 560mW
With 50% PWM power from 14V = 14x0.04/2 = 280mW only.(extra 40mW in the resistor. But total power has come down drastically)
Am I Right?
Yes, but your LEDs will be 1/2 as bright because the average current is now 20ma with 50% PWM vs 40ma without it.
 
Yes, but your LEDs will be 1/2 as bright because the average current is now 20ma with 50% PWM vs 40ma without it.
With 12V also the average current was 20mA only. But it was continuous. Here it is 40mA with 50% duty cycle keeping the average as 20mA. So brightness should be same if the light O/P of LED varies proportionately with current (which may not be true in this case as audioguru pointed out).
Right?
 
Why not build a test circuit? A simple 555 timer and a variable resistor to limit current plus an LED and see how duty affects brightness.
 
Why not build a test circuit? A simple 555 timer and a variable resistor to limit current plus an LED and see how duty affects brightness.

Anyway I will be doing that with the MC and driver. But I put these doubts in the forum to get the opinion of people with practical experience so that I can save time and learn from their experience.
I had done dynamic display earlier and observed the brightness variation.
 
I use MV8191 red LEDs. The luminous intensity at 40mA is only 1.7 times the amount at 20mA so the brightness is not doubled. Your LEDs might be different.
 
I use MV8191 red LEDs. The luminous intensity at 40mA is only 1.7 times the amount at 20mA so the brightness is not doubled. Your LEDs might be different.

From the data sheet of MV8191, it seems to me that intensity Vs If characteristics is suitable for dynamic display, but light O/p is less for traffic light application.
 
The MV8191 is diffused and has a 45 degrees beam angle. The MV8141 is the same chip but is 40 degrees and is almost twice as bright since its package is clear.

Its curves show that it does not get very bright with high peak currents.
 
If you still want to use a voltage regulator, why not use a buck-boost switching regulator?
 
hi,
A very useful reference manual for SMPS.:)

Thank you very much:). I just had a glimpse. Seems to be very useful. I will definitely use this as a reference.
What is this 'buck boost' actually? Is it same as fly back converter where we can have the o/p voltage higher than input voltage?
 
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