Design Article
Comment
pcsalex
also the first diode -which shunts the input voltage for half period, could be ...
pcsalex
if the power line carries transients --which one dos not?-- that transient could ...
Switch circuit controls lights
C Castro-Miguens, University of Vigo, Spain, and JB Castro-Miguens, Cesinel, Madrid, Spain; Edited by Martin Rowe and Fran Granville
4/7/2011 1:00 PM EDT
Cities and towns worldwide are
considering and installing LED
streetlights to help save electric energy,
reduce costs, protect the environment,
and improve lighting for their citizens.
Despite this trend, the lamps’ turn-on/turn-off time control is receiving little
attention.
A suitable control
can achieve
an important energy
saving because
lights can operate
too late, too early,
or both, wasting
energy or providing
insufficient
light. Using a twilight
switch can significantly reduce
energy consumption
in all types of
lamps (Figure 1).
It offers a cost-effective,
compact,
and reliable way of
providing lighting
time control.
The circuit does not use a relay. Therefore, it has no moving parts, and it is not prone to contact oxidation. It uses a TRIAC (triode alternating current) that can switch hundreds of watts.
The circuit requires little power. It uses a charge pump to feed the circuit from the ac line, drawing less than 37 mW for a 220V-rms ac line. It uses just a few low-cost components.
You can adjust the circuit’s darkness and illuminance level that switches the light on and off using only onboard potentiometer R1. The circuit automatically turns on the lamps at nightfall and turns them off at daybreak. You can use it with incandescent lights, fluorescent lights, or LEDs.
The circuit uses an LDR (light-dependent
resistor) to measure the ambient-light
level (Figure 2). Be sure that the
LDR you use has a spectral response similar
to that of the human eye to achieve
good performance. It uses a hysteresis
comparator because a basic comparator
configuration oscillates or produces
a noisy output when the illumination
level is close to the edge between natural
light and darkness. Hysteresis creates
two switching thresholds in the circuit:
The upper threshold voltage is 8.47V
for the rising input-voltage change from
natural light to darkness, and the lower
threshold voltage is 7.75V for the falling
input-voltage change from darkness to
natural light. The relationship between
the 82-kΩ and the 4.7-kΩ resistors controls
the 0.72V hysteresis. This value is
adequate to avoid the false triggering
that light noise can cause.
When the ambient light falls below the level that R1 sets, the input voltage, VI, rises above the upper threshold voltage and the output of the comparator decreases, switching on the TRIAC. When the ambient light rises above the level that R1 sets, the input voltage decreases below the lower threshold voltage and the output of the comparator increases, switching off the TRIAC.
You must provide a mechanical isolation
between the LDR and the lamp
light to prevent the formation of a
feedback path to the LDR. Otherwise,
the lamp light will cause an oscillation
at the comparator’s output and
then in the lamp’s state. The BTA16-600SW, which is available from many
sources, is suitable for switch lamps
operating at more than 2000W.
The comparator drives a Vishay IL4216 or BRT12-F optocoupler with a TRIAC output (Figure 3). The optocoupler, in turn, drives the BTA16-600SW TRIAC that controls the lamp.
A suitable control
can achieve
an important energy
saving because
lights can operate
too late, too early,
or both, wasting
energy or providing
insufficient
light. Using a twilight
switch can significantly reduce
energy consumption
in all types of
lamps (Figure 1).
It offers a cost-effective,
compact,
and reliable way of
providing lighting
time control.The circuit does not use a relay. Therefore, it has no moving parts, and it is not prone to contact oxidation. It uses a TRIAC (triode alternating current) that can switch hundreds of watts.
The circuit requires little power. It uses a charge pump to feed the circuit from the ac line, drawing less than 37 mW for a 220V-rms ac line. It uses just a few low-cost components.
You can adjust the circuit’s darkness and illuminance level that switches the light on and off using only onboard potentiometer R1. The circuit automatically turns on the lamps at nightfall and turns them off at daybreak. You can use it with incandescent lights, fluorescent lights, or LEDs.
The circuit uses an LDR (light-dependent
resistor) to measure the ambient-light
level (Figure 2). Be sure that the
LDR you use has a spectral response similar
to that of the human eye to achieve
good performance. It uses a hysteresis
comparator because a basic comparator
configuration oscillates or produces
a noisy output when the illumination
level is close to the edge between natural
light and darkness. Hysteresis creates
two switching thresholds in the circuit:
The upper threshold voltage is 8.47V
for the rising input-voltage change from
natural light to darkness, and the lower
threshold voltage is 7.75V for the falling
input-voltage change from darkness to
natural light. The relationship between
the 82-kΩ and the 4.7-kΩ resistors controls
the 0.72V hysteresis. This value is
adequate to avoid the false triggering
that light noise can cause.When the ambient light falls below the level that R1 sets, the input voltage, VI, rises above the upper threshold voltage and the output of the comparator decreases, switching on the TRIAC. When the ambient light rises above the level that R1 sets, the input voltage decreases below the lower threshold voltage and the output of the comparator increases, switching off the TRIAC.
| Read More Design Ideas |
The comparator drives a Vishay IL4216 or BRT12-F optocoupler with a TRIAC output (Figure 3). The optocoupler, in turn, drives the BTA16-600SW TRIAC that controls the lamp.
|
|
|
|
|
Navigate to related information
Marian Stofka
4/7/2011 2:45 PM EDT
Any of the 1N4007 diodes in supply in Fig. 1 are potentionally threatened and could be destroyed at connecting the supply to ac-grid at a time, when the ac voltage is close to its positive or negative peak.
Although a non-repetitive 30 A, 8.3msec, half-sine-wave pulse is allowed for the 1N4007; a 100 nF series-capacitor; initially uncharged, can cause current peaks such as 100 A, decaying within 300 nsec.
A resistor of some 100 Ohm in series with the 100 nF capacitor could be a cure.
Sign in to Reply
anonymous user
4/8/2011 6:09 PM EDT
That is not a charge pump. it is simply a capacitor used as a resistor in ac.
Sign in to Reply
larryparts
4/8/2011 6:13 PM EDT
The series resistor suggested by Marian is absolutely necessary for a second reason: without the resistor, a line transient can easily charge the input capacitor to hundreds or even thousands of volts; this can destroy the capacitor. The addition of a series resistor creates a low-pass filter which can prevent transients from damaging the capacitor. Line transients of several thousand volts are very common.
Also note that the half-cycle turn-on current surge may damage the zener diode as well as the rectifier diodes.
Sign in to Reply
anonymous user
4/9/2011 9:16 AM EDT
The guys are right. The current surge that arises when a capacitor is connected to the low impedance mains may affect reliability of the whole circuit. A series resistor 20 to 100 Ohm 1W is a must!
Sign in to Reply
rickman1
4/9/2011 2:37 PM EDT
I tried simulating this in LTspice and it doesn't work for me. If I use a 1000 ohm load resistor the output voltage drops to about 3.7 volts and the current 3.7 mA. Certainly the circuit this is supposed to power is will draw at least this much. Did I miss something important?
Sign in to Reply
anonymous user
4/11/2011 3:28 AM EDT
Why using the optocoupler? The OPAMP is not isolated from the AC thus its output can be used to drive the triac. The power supply circuit should be modified depending on the triac driving quadrants.
Sign in to Reply
Bob Phare
4/11/2011 11:16 AM EDT
Some sort of time delay is needed to prevent short light transients (e.g. passing car lights, lightning) from toggling the street light unnecessarily.
Sign in to Reply
anonymous user
4/13/2011 4:24 PM EDT
The half-wave rectifier leaves the 470uF to discharge for half a cycle. Since optical isolation is provided by the trigger device, there's no need for line neutral to be ground in the control circuit. Might as well use a full-wave bridge. Filter cap then only supplies current while the AC line drops ~17v (counting the Vf of the bridge) from its peak value. The filter cap can then be reduced by about a factor of 8, yielding space and cost savings.
I would also add not only a suitably rated surge limiting resistor but also a Varistor downstream of it to clamp line spikes. This limits repetitive stress on the rectifier and input cap. When one considers the scale of deployment of these controls, the need for reliability is clear.
Why the 8.2v Zener? Ambient light threshold corresponds to a resistance ratio, so why not use a pair of resistors for Vref?
Sign in to Reply
DickN
4/13/2011 4:24 PM EDT
The half-wave rectifier leaves the 470uF to discharge for half a cycle. Since optical isolation is provided by the trigger device, there's no need for line neutral to be ground in the control circuit. Might as well use a full-wave bridge. Filter cap then only supplies current while the AC line drops ~17v (counting the Vf of the bridge) from its peak value. The filter cap can then be reduced by about a factor of 8, yielding space and cost savings.
I would also add not only a suitably rated surge limiting resistor but also a Varistor downstream of it to clamp line spikes. This limits repetitive stress on the rectifier and input cap. When one considers the scale of deployment of these controls, the need for reliability is clear.
Why the 8.2v Zener? Ambient light threshold corresponds to a resistance ratio, so why not use a pair of resistors for Vref?
Sign in to Reply
Marian Stofka
4/18/2011 1:52 PM EDT
The previous poster is really very inspirative with its suggestion to replace the half-wave rectifier by a Graetz one.
The mains voltage has, however, to drop by 2x(Vz + 2Vd) ~ 2x 17 V = 34 V from its peak value Vpeak, to start to charge the 470 uF capacitor again by decreasing part of this sine voltage. It can be derived, that a factor of lowering the requirement for value of the 470 uF capacitor is approximately:
(Pi=3.14)/arccos(1-2(Vz + 2Vd)/Vpeak)
As the author lives probably in New Hampshire, then for his ac mainns of 120 V the Vpeak = 169.7 V.
By inserting it into above formula we get factor of 4.877, which is still attractive.
Sign in to Reply
pcsalex
1/23/2013 11:02 AM EST
if the power line carries transients --which one dos not?-- that transient could kill the diode serial with the input capacitor, and also the Zener diode, a low value but high power resistor in siree with the input capacitor could limit the transient current and save the circuit.
Sign in to Reply
pcsalex
1/23/2013 11:11 AM EST
also the first diode -which shunts the input voltage for half period, could be replaced with the Zener diode -- which would work with forward voltage as a normal rectifier and shunt the half period, and in the next half period would limit the voltage to it's Zener voltage, the serial diode connected to the capacitors is still required, that modification will not just reduce the parts cost but the power dissipation on the Zener diode too
Sign in to Reply