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Author Topic: Is LED brightness linearily proportional to supply current (dimming LEDs)?  (Read 5590 times)

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robmack

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I am making a circuit to dim a single 7-segment LED display. The display will show the selected gear on a motorcycle and will be mounted on the instrument cluster for the driver to see.  The behaviour I wish is for the display to have full brightness in daylight and much lower brightness at night.

Reference Dimming Circuit:  http://www.stephenhobley.com/blog/2010/09/17/light-controlled-pwm-generator/

I am using a simulator at the moment to refine the design.  Source voltage is 12V.  The output circuit uses a MOSFET, the gate of which is connected to the PWM output and pulled up to VDD using a 100K resistor.  The circuit shown below is a snapshot of the output stage in the simulator with the PWM output, MOSFET and 7-segment display shown. 



Will this driver circuit suitably provide the constant current required for driving the 7-segment LED display?

Is LED brightness linearily proportional to current?  That is, will the LED be half as bright if half the current is supplied to the display.  For reference, my circuit shows 280mA when the duty cycle of the PWM is 0% and 88mA with ~75% duty cycle.  This is about 70% less current so should I expect the LED to be 70% less bright?

Thanks.

- Robert

kam

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well, there is a big misunderstanding with the PWM in general. The pwm does not control the current neither the voltage, rather the power. what i mean i simple words is that, if a resistor of 1 ohm is connected to a source of 1v, it will draw 1A, everyone know this. Now if we PWM this source with 50% duty cycle, will the resistor draw half the amps of half the voltage? Neither! It will still draw 1A at 1V, since the PWM does not change the voltage. BUT, in time, the PWM pulses delivers HALF the power on the LED, since 50% duty cycle means hat half of the time the LED is off, and half it is on. These for the PWM.

Now for the current -> The LEDs are absolutely proportional to the current. Half the current means half the brightness. BUT, half the current does not mean half the voltage at all. For example, the current may change from 0 to full in a voltage range of 2.8 to 3.2, and this region is absolutely NOT linear (it is exponential).

To your question if this circuit will work, yes it will work(the pull-up resistor is too high - make it 1Kohm or less). To your question if this circuit will be linear, no it will be not. But do you really want it to be linear? I am not sure if linear is best for your application, this is something that you'd rather test (and share with us  :D :D ;))












robmack

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Thank you for the quick reply and for the hint about the pull-up. I just copied an existing resistor from the design without much thought about its value.  I'll correct that.

To be specific about my quoted currents, the simulator probe showed instantaneous, RMS and DC current values.  Your explanation was validated by what I saw in the probe -- the P-P voltage and current were the same as the steady-state current and voltage.  I quoted in my first post was the displayed DC current from the probe (88mA).  The RMS current was slightly higher, at 112mA.

As for the linearity, I don't require it for the application.  What I do require is that the display dim to a level that is "acceptable" for the given light conditions.  Since the human eye is non-linear in its response to light and colour, this "level of acceptability" could be wide given different 7-segment displays and different LED colours.  So, you're right...I'll have to experiment to see what works for the display that I have chosen.  That display is a Kingbright SC39-11SRWA, chosen for its visibility in daylight conditions.

When the circuit is completed, I'll share the results and design.  It is intended for particular models of BMW motorcycles (early K-series and R-series) to allow for substituting an aftermarket instrument cluster.

Cheers,

kam

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so you own such a bike? which one?

robmack

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I have three motorcycles at the moment.  Two are BMW K-bikes from the 1980's -- a K100 and a K75.  The third is a late model Moto Guzzi.  The BMWs have the gear indicator in the instrument cluster and, when I customize one of the K75 which will involve removing and replacing the cluster, I will need to manufacture a circuit to simulate the lost functionality.  Part of that circuit is this 7-segment dimmer.