PE/424CR/ February 9, 2006 Light-dimming Circuit 1

Light Dimmer Circuit

One of the main applications of single-phase ac-to-ac converter is the dimming of the

lights. The circuit used for this application especially for the incandescent filaments

lamps. One of the important attributes of the incandescent filament lap is its low cold

resistance and high hot resistance. For preliminary calculations, the cold resistance may

be neglected.

Figure 1: Light dimmer circuit using a Triac

We can replace the two SCRs connected in anti-parallel by a bidirectional TRIAC

(Triode AC) as shown in Figure 1. Since anode of one SCR and the cathode of the other

are connected together when two SCRs are connected in anti-parallel. Therefore, it does

not make sense to call the two main terminals of the triac as anode and cathode. These

terminals are usually called the main-terminal-1 and main-terminal-2. The gating pulse is

applied between the gate and the main-terminal-1. The triac can be triggered by applying

a positive or a negative gating current regardless of the polarity of the main terminal-1.

The other new device in this circuit is called the DIAC. It is a 4-layer p-n-p-n

semiconductor device that acts as a bidirectional diode. It is designed to breakdown at the

same voltage in either direction. Its voltage-current characteristic is given in Figure 2. It

behaves like an on/off switch. When the magnitude of the voltage across the DIAC is less

that its forward-breakdown voltage FBOV , the device is open and acts as an open circuit.

As soon the magnitude of the voltage across the device tends to exceed its forward-

breakdown voltage, the device breakdown and begins to conduct. The voltage across the

device once its start conducting is usually less than its breakdown voltage. Theoretically,

we would prefer it to be zero. Practically, it varies from one device to the other.

PE/424CR/ February 9, 2006 Light-dimming Circuit 2

Figure 2: Voltage-current characteristic of a DIAC

We now explain the operation of the dimmer circuit of Figure 1 when the applied voltage

is given as

)tsin(V)t(v Ms ω= V

As the input voltage begins its positive half cycle, the TRIAC is off and the

capacitor begins to charge toward the maximum value of the input voltage. The charging

circuit consists of 1R and the cold resistance LR of the incandescent lamp. Let us denote

the total resistance in the charging circuit as R such that

L1 RRR += (1)

For a voltage of )t(vC across the circuit for π≤ω≤ t0 , the differential equation

for the RC circuit is

)tsin(V)t(vdt

)t(dvRC MC

C ω=+ (2)

By now we should be able to write the solution in the general form of the above

differential equation as

φω−+φ−ωω+

= tan/t

2

MC eK)tsin(

)RC(1

V)t(v (3)

where

RCtan ω=φ (4)

When we apply the initial condition, i.e. the voltage across the capacitor is zero at

t = 0, we obtain the constant of integration K as

)sin()RC(1

VK

2

M φω+

=

Thus, the voltage across the capacitor is

PE/424CR/ February 9, 2006 Light-dimming Circuit 3

( )φω−φ+φ−ωω+

= tan/t

2

MC e)sin()tsin(

)RC(1

V)t(v

The DIAC will breakdown at α=ωt when the voltage across the capacitor is equal to

FBOV . In other words,

2

M

FBOtan/ )RC(1V

Ve)sin()sin( ω+=φ+φ−α φα− (5)

This is a nonlinear equation that can be solved using computer or a programmable

calculator.

Example: _____________________________________________________________

A 120-V, 60-Hz single-phase voltage source is available in the laboratory. A TRIAC

dimmer circuit is used to adjust the light intensity of a 120-V, 100-W incandescent

filament lamp. A 5-kΩ potentiometer is used in along with a 1-µF capacitor. If the

breakdown voltage of the DIAC is 60 V and the potentiometer is adjusted to 4.5 kΩ,

determine the angle at which the TRIAC will begin conduction. Sketch the output voltage

for one time period. What is the new power rating of the lamp?

Solution:

Since no information is provided for the cold resistance of the incandescent filament

lamp, let us neglect it because it will be very small in comparison with 4.5-kΩ setting of

the potentiometer.

Hence, 5.4R = kΩ, C = 1 µF, 60VFBO = V, and 7.1692120VM == V.

=×××π=ω=φ −61014500120RCtan 1.696

038.1)(tantan 1 =φ=φ − rad (or o48.59 )

Equation (5) can now be written as

2696.1/ 696.117.169

60e)038.1sin()038.1sin( +=+−α α−

We can solve this equation numerically or graphically. The numerical solution yields

351.1=α rad (or )41.77 o

PE/424CR/ February 9, 2006 Light-dimming Circuit 4

To solve the nonlinear equation graphically, the voltage across the capacitor as

]e)038.1sin()038.1t[sin(696.11

V)t(v 696.1/t

2

MC

ω−+−ω+

=ω

After couple of adjustments for the angle and the magnitude, the plot of voltage as a

function of tω (degrees) is shown below.

60 62.5 65 67.5 70 72.5 75 77.5 80 82.5 85 87.5 900

20

40

60

80

100

120

140

160

180

200Angle at which TRIAC turns on

Angle in Degrees

Cap

acit

or

volt

age

( V

)

From the graph, it is evident that the capacitor voltage reaches a value of 60 V when

o5.77t =ω . This is in agreement with the numerically obtained result.

We can now write an expression for the output voltage across the lamp as

π≤ω≤α+πω

π≤ω≤αω=

2t,V)tsin(V

t,V)tsin(V)t(v

M

M

o

The input and the output waveforms are sketched below.

0 30 60 90 120 150 180 210 240 270 300 330 360200

160

120

80

40

0

40

80

120

160

200Input and OUtput Voltages

Angle in Degrees

Volt

age

axis