No one calc. what you have offer without specs of LEDs and the LM324 has inadequate slew rate to drive a square wave unless using a low frequency such as 1~10kHz. The best way to measure it is with an Ammeter before a huge capacitor across V+ to 0V. Or use a current sense R of 50m ohms and scope the current.
Hi...
Yes you are right it's operating at 200Hz not Khz... My mistake.
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Hi...
I have a bench power supply, I configured the Voltage and current at 6V and 1 Amp. The LED strip glowing perfect on 100% duty cycle. I measured the Voltage and current at 100 % Duty cycle is 5.96V and current is 0.93A...
I don't know how to calculate the wattage 10-90% duty cycle.
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Hi...
Thanks for your reply.
I purchased the led strip from Local Market.
What power loss can one expect?
* loss in the Mosfet
* loss in the driving circuitry.
Loss in the Mosfet can be switching loss and conductive loss. Switching loss at 200Hz should be negligible. Conductive loss is the voltage across the Mosfet in ON state (scope) times voltage times duty cycle.
Loss in the rest of the circuitry may come from a power supply circuit including smoothing bulk capacitor. It will be shown as DC current when the PWM is in OFF state. This power loss will be current x voltage.
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Masuring PWM'd current with a DVM is somehow tricky.
There are meters that show the average current. On a square wave this is: peak_current x duty_cycle.
There are meters that show true RMS_including DC. They show peak_current x sqrt(duty_cycle)
There are meters that show true RMS of a high pass filtered current. I don't know the exact formula for this.
And there are meters that show some value that I call "fake RMS". They show only correct RMS values of clean sine shaped currents.
They may use "peak value" method or "average of a rectified value" method. The exact formula depend on the meter and usually is unknown.
--> Either use a meter with known method and known formula or use a scope.
Modern scopes often have mathematical functions .... they can exactely calculate the power from random current and voltage waveforms.
What power loss can one expect?
* loss in the Mosfet
* loss in the driving circuitry.
Loss in the Mosfet can be switching loss and conductive loss. Switching loss at 200Hz should be negligible. Conductive loss is the voltage across the Mosfet in ON state (scope) times voltage times duty cycle.
Loss in the rest of the circuitry may come from a power supply circuit including smoothing bulk capacitor. It will be shown as DC current when the PWM is in OFF state. This power loss will be current x voltage.
****
Masuring PWM'd current with a DVM is somehow tricky.
There are meters that show the average current. On a square wave this is: peak_current x duty_cycle.
There are meters that show true RMS_including DC. They show peak_current x sqrt(duty_cycle)
There are meters that show true RMS of a high pass filtered current. I don't know the exact formula for this.
And there are meters that show some value that I call "fake RMS". They show only correct RMS values of clean sine shaped currents.
They may use "peak value" method or "average of a rectified value" method. The exact formula depend on the meter and usually is unknown.
--> Either use a meter with known method and known formula or use a scope.
Modern scopes often have mathematical functions .... they can exactely calculate the power from random current and voltage waveforms.
I measured RMS of output Voltage and Output Current and calculated the wattage depending on that. In 10% Duty cycle it's showing 19% Efficient.
Let me Demonstrate what actually I want:-
I Want to Drive A LED Strip of 3W 6V from a 7.4-8.4V 2S2P 4000mAH Lipo battery. I want to calculate the time backup if I use analog PWM using op-Amp.
Is there any better way to make an efficient dimmer? Should I use Integrated Circuit which generate pwm which will be efficient?
RMS voltage and RMS current is not usefull for this, because you don't have an ohmic load with constant V to I relationship.
An indication for low efficiency is that the switching circuit becomes warm.
With 19% it will be not warm, but hot. Imagine: 19% means that 81% is heat. This is about four times the power of the LED.
RMS voltage and RMS current is not usefull for this, because you don't have an ohmic load with constant V to I relationship.
An indication for low efficiency is that the switching circuit becomes warm.
With 19% it will be not warm, but hot. Imagine: 19% means that 81% is heat. This is about four times the power of the LED.
Hmmm... you are right...
I don't have proper DMM which will show me the proper Wattage of a chopped signal, nor I have a proper oscope who will calculate the power output mathematically. My oscope is more than 5-6 years old and no option to calculate the wattage.
I am also confused about how to calculate the effective wattage of the PWM out.
we gave you several methods and calculation formulas.
--> averaging methods are good, RMS not.
--> any scope (without mathematics function) picture can give a good estimation. (btw: what scope do you have?)
--> calculate the loss is good because: input_power = output_power + loss
I'm under the impression that your "efficiency" consideration are flawed from the start.
Power is average of instantaneous voltage multiply instantaneous current. No chance to determine the input or output power of a pulsed signal with DMM voltage and current measurements. That's only possible for DC signals without current or voltage ripple.
Applying PWM to a load with constant resistance (e.g. a resistive heater, also roughly applicable to LED stripes with series resistor) gives rated power multiply duty cycle.
For best efficiency you need to use PWM with an proper sized inductor in series with the load.
Otherwise the PWM efficiency is no better than a linear regulator for the same average LED current.