PWM lamp dimmer.

A simple and efficient PWM lamp dimmer using timer IC NE555 is discussed in this article. Yesterdays linear regulator based dimmers can only attain a maximum efficiency  of 50% and are far inferior when compared to the PWM based dimmers which can hit well over 90% efficiency. Since less amount of power is wasted as heat, the switching elements of PWM dimmers require a smaller heat sink and this saves a lot of size and weight. In simple words, the most outstanding features of the PWM based lamp dimmers are high efficiency and low physical size. The circuit diagram of a 12V PWM lamp dimmer is shown below.

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12V PWM lamp dimmer
Fig 1 : PWM lamp dimmer using NE555

As you can see, NE555 timer IC which is wired as an astable multivibrator operating at 2.8KHz forms the heart of this circuit. Resistors R1,R2, POT R3 and capacitor C1 are the timing components. Duty cycle of the IC’s output can be adjusted using the POT R3. higher the duty cycle means higher the lamp brightness and lower the duty cycle means lower the lamp brightness. Diode D1 by-passes the lower half of the POT R3 during the charging cycle of the astable multivibrator. This is done in order to keep the output frequency constant irrespective of the duty cycle. Transistors Q1 and Q2 forms a darlington driver stage for the 12V lamp. Resistor R4 limits the base current of transistor Q1.

Understanding the variable duty cycle astable multivibrator.

As I have said earlier, the variable duty cycle astable multi vibrator based on NE555 forms the foundation of this circuit and a good knowledge on it is essential for designing projects like this. For the ease of explanation the timing side of the astable multivibrator is redrawn in the figure below.

astable multivibrator variable duty cycle
Fig 2: Astable multivibrator with variable duty cycle

Upper and lower halves of the POT R3 are denoted as Rx and Ry respectively. Consider the output of the astable multivibrator to be high at the starting instant. Now the capacitor C1 charges through the path R1, Rx, and R2. The lower half of POT R3 ie; Ry is out of the scene because the diode D1 by-passes it. When the voltage across the capacitor reaches 2/3 Vcc, the internal upper comparator flips its output which makes the internal flip flop to toggle its output. As a result the output of the astable multivibrator goes low. In simple words, the output of the astable multivibrator remains high until the charge across C1 becomes equal to 2/3 Vcc and here it is according to the equation Ton =0.67(R1+Rx+R2)C1.

Since the internal flip flop is set now, the capacitor starts discharging through the path R2,Ry into the discharge pin. When the voltage across the capacitor C1 becomes 1/3 Vcc, the lower comparator flips its output and this in turn makes the internal flip flop to toggle its output again. This makes the output of the astable multivibrator high. To be simple, the output of the astable multivibrator remains low until the voltage across the capacitor C1 becomes 1/3 Vcc and it is according to the equation Toff = 0.67(R2+Ry)C1. Have a look at the internal block diagram of NE555 timer shown below for better understanding.

555 timer internal block diagram
Fig3: NE555 internal block diagram
How does the frequency remain constant irrespective of the position of POT3 knob?.

What ever may be the position of  POT3 knob, the total resistance across it remains the same (50K here). If anything decreases in the upper side (Rx) the same amount will be increased in the lower (Ry) and the same thing gets applied to the higher(Ton) and lower(Toff) time periods. The derivation shown below will help you to grasp the matter easily.

With reference to Fig 2, we have:

Ton = 0.67(R1+Rx+R2)C1

Toff= 0.67(R2+Ry)C1

Total time period of the output waveform “T” is according to the equation :

T = Ton + Toff

There fore, T = 0.67(R1+Rx+R2+R2+Ry)C1

                        T= 0.67(R1+2R2+Rx+Ry)C1

We know that Rx+Ry = R3

There fore T = 0.67(R1+2R2+R3)C1

Therefore frequency F = 1/(0.67(R1+2R2+R3)C1) 

From the above equation its is clear that the frequency depends only on the value of the components C1, R1, R2  and the over all value of R3 and it has nothing to do with the position of R3 knob.



  1. Colin Mitchell

    A badly designed circuit.
    The output of the 555 can deliver 100mA and the first transistor is not needed.
    When the transistor is removed, the lamp will achieve much higher brightness.
    The Darlington arrangement is a DISASTER.

  2. plz tell me substitute of 2n2222 and bd 139
    can i use bc 547 instead of 2n2222 and sl 100 instead of bd139???

    • Seetharaman

      Q1 can be BC148B and for Q2 you can use SL100 or recent S8050

  3. Can we use a 9 V DC battery instead of 12 V DC supply…and do we hav to use a regulator or is it automatic?

    • Seetharaman

      Use regulated supply of 12 volt DC. if you use 9 volt regulated supply the bulb rating also should be reduced to 9 volt. please note that this circuit will be effective for incandescent lamps only.

  4. Nice circuit. I would like to once test this circuit in my home as it would be more useful to get experience for my project in the next month. Thanks for sharing this circuit.

  5. how is this work as a dimmer bcze if the dutycycle is 65% and time period
    (T) is 10 sec then for 6.5 sec the output of 555 tmr is high so the lamp will light brightly and for 3.5 sec lamp will be dim pls can anyone explain this to me

  6. Dear sir,
    Thanks for the PWM circuit and as the explanation is very clear and simple.
    What is the ideal frequency to run a motor with a PWM circuit.
    Thanks and Regards

    • Safoor, that depends on how much you want from your motor (like speed, torque)

  7. sir, can i use 12v 20w bulb instead of the above mentioned lamp?

    • Seetharaman

      Hi Shamanth BD 139 can only handle just 1amp current. hence replace it with a power transistor like 2N3055 or BD182 or replace both Q1 Q2 with a single power darlington transistor like TIP122 with a good heat sink.

  8. Quixilver

    I’d like to use this on a 12v 50Watt spotlight. do I need to change any components.

  9. Raveesh H P

    Please inform me whether this circuit can be modified to control the 230 volts ac soldering Iron (ceramic heater) using Triac and if so what are the modification required.

    Please reply.