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I want to illuminate a room. This will be done by two LED strips 60led/m, each 5m long, located along the tops of opposite walls.

In order to get full light power, they together consume 4A@12V. There's also additional 4m of cable needed to power each of these strips. In total, I get 18m of cable around the room.

Then, I want to have a dimmer. The simple way is to have uC switching a transistor with PWM. But having such a large antenna pulsing with 48W seems to be unwise.

What I learned, is that I could use an LC low-pass filter, to flatten the PWM on +12V line to a lower voltage.

EDIT: This is the latest diagram of the circuit. The only things left are uC and dimming potentiometer. C2 and L1 are calculated for 5kHz PWM.

LC low-pass filter

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  • \$\begingroup\$ emes - follow thge LED current path when Q1 is on and then off. There is NO path for L1 current with Q1 off. Inductor current CANNOT be suddenly stopped - Murphy and the laws of Physics do not allow it. L1 end attached to LD will "ring" to whatever it takes to maintain current. 1000 volts if necessary. \$\endgroup\$ – Russell McMahon Jul 5 '14 at 14:20
  • \$\begingroup\$ You don't need to low pass your pwm at all. Stick with the transistor only and you are good to go, don't forget a base resistor if you use a bjt. \$\endgroup\$ – Vladimir Cravero Jul 5 '14 at 14:30
  • \$\begingroup\$ @VladimirCravero - did you miss the part where he doesn't want to broadcast the PWM frequency? \$\endgroup\$ – WhatRoughBeast Jul 5 '14 at 15:32
  • \$\begingroup\$ Actually yes, but he can go as low as 1kHz, I don't think RF would be an issue \$\endgroup\$ – Vladimir Cravero Jul 5 '14 at 15:34
  • \$\begingroup\$ @RussellMcMahon - I think I get it, although I'm an electronics newbie. If I imagine current as flowing water and the inductor as a turbine, it makes sense :) \$\endgroup\$ – emesik Jul 5 '14 at 15:54
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This is a good idea. Theoretically you could use the pwm as is to dim the lights since an LED shines brighter the longer the on-periods are. Your concern about sending waves might be negligible as long you stay in the lower kHz-domain, yet no blinking will be seen for the human eye. So first I suggest you to make a roughly 5kHz pwm signal and apply it to your LEDs.

If you want to use a LowPass the calculation is easy: f=1/(2*pisqrt(CL)) This describes the -3dB frequency. So if you apply any signal, frequencies above f will be attenuated by 3dB at f, and falling with 40dB/decade. Example: choose C and L for a frequency of 100Hz, then apply a signal, of 1000 Hz and it will be 40dB less than the original signal, means devided by thousand. A 10kHz signal will be have -80dB, means there is nothing left of the signal.

So in your special case my suggestion is: make a 5kHz square signal from a controller and make an LC-LowPass for f=100 Hz (choose C e.g. 10 microF) and calculate L accordingly. It will produce perfect DC proportional to the pulsewidth.

Only your cicuit layout is wrong and won't work.

schematic

simulate this circuit – Schematic created using CircuitLab
Added: Diode D1 added. Inductor current cannot be interrupted. Current flows as per top arrow when Q2 is on and via D1 when Q2 is off.

I would try this, but I drew it quickly, so someone please look over it again! Hope this helps

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  • \$\begingroup\$ If you use a low pass filter, you won't be able to dim below the duty cycle equivalent to the forward voltage of the LEDs in the strip, but with unfiltered PWM you get a linear dimming down to nothing. \$\endgroup\$ – Pete Kirkham Jul 5 '14 at 13:37
  • \$\begingroup\$ Why should I drive the 12V line with PNP, not with NPN? Why the NPN would not work? \$\endgroup\$ – emesik Jul 5 '14 at 13:41
  • \$\begingroup\$ Circiuit is generally OK but: D1 added to provide a path for inductor current when Q2 is off. | R1 is too large. | R3 is too small. | Q2 needs to be much more capable - a P Channel MOSFET would be easier to use there. \$\endgroup\$ – Russell McMahon Jul 5 '14 at 14:18
  • \$\begingroup\$ Yes sorry, as I said, I drew it quickly. R3 should be 10k as well, and an input resistor must be inserted (!) about 10k. R1 could be reduced to 1k. The transistors are the defaults from the program, you should use something way bigger. A P-Channel Mosfet will probably the way to go. The two transistors are necessary since your controller will probably only allow for 5V and you need to control a voltage of 12V, thats why you have to use a voltage-level shifter circuit. If using BJTs use a darlington, or more adequately a Mosfet. \$\endgroup\$ – jjstcool Jul 5 '14 at 15:33
  • \$\begingroup\$ Ah, I just got your circuit right, that will work too:-) Sorry for sharing my misguided thoughts... \$\endgroup\$ – jjstcool Jul 5 '14 at 15:39

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