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 VFBO , the the dev device ice is open open and and acts acts as as an an ope open n circ circuit uit.. As soon the magnitude of the voltage across the device tends to exceed its forwardbreakdown 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.
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Light-dimming Circuit
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 v s ( t ) = VM sin( ωt ) 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 R 1 and the cold resistance R L of the incandescent lamp. Let us denote the total resistance in the charging circuit as R such that R = R1 + R L
(1)
For a voltage of v C ( t) across the circuit for 0 ≤ ωt ≤ π , the differential equation for the RC circuit is
RC
dv C ( t ) dt
+ v C ( t ) = VM sin(ωt )
(2)
By now we should be able to write the solution in the general form of the above differential equation as v C (t) =
VM 1 + ( ωRC)
2
sin( ωt − φ) + K e − ωt / tan φ
(3)
where tan φ = ωRC
(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 K=
VM 1 + (ωRC)
2
sin( φ)
Thus, the voltage across the capacitor is
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Light-dimming Circuit
v C (t) =
VM 1 + ( ωRC)
2
(sin(ωt − φ) + sin(φ) e
− ωt / tan φ
)
The DIAC will breakdown at ωt = α when hen the the vol volta tage ge acro across ss the the cap capac acit itor or is equa equall to VFBO . In other words,
sin( α − φ) + sin( φ) e − α / tan φ =
VFBO VM
1 + (ωRC) 2
(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-k Ω 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 k Ω k Ω, 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-k Ω 4.5-k Ω setting of the potentiometer. Hence, R = 4.5 k Ω, C = 1 µF, VFBO = 60 V, and VM = 120 2 = 169.7 V. tan φ = ωRC = 120π × 4500 × 1 × 10 −6 = 1.696
φ = tan −1 (tan φ) = 1.038 rad (or 59.48 o ) Equation (5) can now be written as sin(α − 1.038) + sin(1.038) e −α / 1.696 =
60 169.7
1 + 1.696 2
We can solve this equation numerically or graphically. The numerical solution yields
α = 1.351 rad
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(or 77.41o )
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Light-dimming Circuit
To solve the nonlinear equation graphically, the voltage across the capacitor as
VM
v C ( ωt ) =
1 + 1 . 696
2
[sin( ω t − 1 . 038 ) + sin( 1 . 038 ) e − ω t / 1 .696 ]
After couple of adjustments for the angle and the magnitude, the plot of voltage as a function of ωt (degrees) is shown below.
) V ( e g a t l o v r o t i c a p a C
200 180 160 140 120 100 80 60 40 20 0 60
Angle at which TRIAC turns on
62.5
65
67.5
70
72.5
75
77.5
80
82.5
85
87.5
90
Angle in Degrees
From the graph, it is evident that the capacitor voltage reaches a value of 60 V when
ωt = 77.5o . This is in agreement with the numerically obtained result. We can now write an expression for the output voltage across the lamp as
VM sin( ωt ) V, α ≤ ωt ≤ π
v o (t) =
VM sin( ωt ) V, π + α ≤ ωt ≤ 2π
The input and the output waveforms are sketched below.
s i x a e g a t l o V
Input and OUtput Voltages
200 160 120 80 40 0 40 80 120 160 200 0
30
60
90
120
150
180
210
240
270
300
330
360
Angle in Degrees
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Light-dimming Circuit
