This article shows you how to make a turn ON and turn OFF push button switch.
This circuit can be done with two switches. You press ON one switch and the light bulb turns ON. You press ON another switch and the light bulb turns OFF. However, this Instructable shows you how to make a similar device with buttons, instead of switches. The relay acts like a latch, activated by the first button. The second button turns this latch OFF.
Components: relay (low power), power source (batteries/power supply), 100 uF capacitor, 10-ohm resistor (high power) - 2, general-purpose diode - 1, light bulb / bright LED, light bulb harness, push buttons - 2, cardboard, sticky tape (masking/clear).
Button ON charges the capacitor Crelay and turns the relay ON. Button OFF discharges the capacitor Crelay and turns the relay OFF.
I used a 9 V battery for a 12 V relay. This is a risky method. Unfortunately most relays, are 12 V relays. Thus 9 V might not be enough to turn the relay ON. However, there are 6V relays.
I used a 12 V light bulb for my circuit because this is what I already had in stock. A bright LED needs to be biased at 2 V. It will burn at voltages above 2 V.
Calculate the resistor value that needs to be connected in series with a bright LED:
Rled = (Vs - Vled) / IledMax = (12 V - 2 V) / 10 mA = 1,000 ohms or 1 kohms
Calculate the maximum supply current when the two buttons are ON at the same time:
IsMax = (Vs - Vd) / R1
Vs = 9 V: IsMax = (9 V - 0.7 V) / 10 ohms = 8.3 V / 10 ohms = 0.83 A = 830 mA
Vs = 12 V: IsMax = (12 V - 0.7 V) / 10 ohms = 11.3 V / 10 ohms = 1.13 A = 1130 mA
Calculate the maximum current across the Button OFF during switching OFF:
Ib2Max = Vs/R1 + Vrelay/R2
We can approximate this calculations by assuming the Vrelay = Vs:
Ib2Max = Vs/R1 + Vs/R2
Vs = 9 V: Ib2Max = 9 V / 10 ohms + 9 V / 10 ohms = 1.8 A
Vs = 12 V: Ib2Max = 12 V / 10 ohms + 12 V / 10 ohms = 2.4 A
The Relay diode ensures that the Vrelay voltage does not become negative when the battery is disconnected. The relay input is a coil. Discharging coil can be modeled as a decaying current source that causes current to flow the same direction as when the relay is ON for a few seconds or milliseconds even after power is disconnected. This can damage your electrolytic Crelay 100 uF capacitor if you capacitor is not bipolar.
Simulations show that after the Button ON is released (at 0.5 seconds), there is a small drop in output voltage. Both the relay and load voltages fall by 0.1 V. However, the relay remains ON.
The voltage 0.1 V drop occurs because the relay conduction resistance is not 0 ohms. Solid state relays that use semiconductors instead of mechanical switches can have those characteristics. This will not be the case for mechanical relays that use mechanical switches.
At 1 second point in time, Button OFF is pressed and the relay turns OFF. At 2 seconds point in time, a new cycle starts.
In the constructed circuit, R2 is a short circuit (I did not use that resistor), and Relay is not included. I also used a diode in series with Button OFF to discharge the relay because my diode had a resistance of 10 ohms (not all diodes have this resistance). Later I modified this Instructable, to replace the diode with R2 resistor (I also had to change the connection of R2 - not connected in series with Button OFF). Relay was added later.
You can see the color codes on the three yellow resistors.
The colors are:
Yellow - 4
Purple - 7
Black - 0 (number of zeros after 47)
This means the resistor value is 47 ohms. The gold band is tolerance in resistors, means 5%. That means the resistor value could be anywhere from 47 * 0.95 = 44.65 ohms to 47 * 1.05 = 49.35 ohms.
I used three 47 ohm resistors and the ceramic resistor is 56 ohms.
R1 = 1 / (1 / 47 ohms + 1 / 47 ohms + 1 / 47 ohms + 1 / 56 ohms) = 12.2418604651 ohms
This is approximately 10 ohms.
The two buttons are from an old VCR (Video Cassette Recorder).
This article shows you how to make a turn ON and turn OFF push button switch.
This circuit can be done with two switches. You press ON one switch and the light bulb turns ON. You press ON another switch and the light bulb turns OFF. However, this Instructable shows you how to make a similar device with buttons, instead of switches. The relay acts like a latch, activated by the first button. The second button turns this latch OFF.
Components: relay (low power), power source (batteries/power supply), 100 uF capacitor, 10-ohm resistor (high power) - 2, general-purpose diode - 1, light bulb / bright LED, light bulb harness, push buttons - 2, cardboard, sticky tape (masking/clear).
Button ON charges the capacitor Crelay and turns the relay ON. Button OFF discharges the capacitor Crelay and turns the relay OFF.
I used a 9 V battery for a 12 V relay. This is a risky method. Unfortunately most relays, are 12 V relays. Thus 9 V might not be enough to turn the relay ON. However, there are 6V relays.
I used a 12 V light bulb for my circuit because this is what I already had in stock. A bright LED needs to be biased at 2 V. It will burn at voltages above 2 V.
Calculate the resistor value that needs to be connected in series with a bright LED:
Rled = (Vs - Vled) / IledMax = (12 V - 2 V) / 10 mA = 1,000 ohms or 1 kohms
Calculate the maximum supply current when the two buttons are ON at the same time:
IsMax = (Vs - Vd) / R1
Vs = 9 V: IsMax = (9 V - 0.7 V) / 10 ohms = 8.3 V / 10 ohms = 0.83 A = 830 mA
Vs = 12 V: IsMax = (12 V - 0.7 V) / 10 ohms = 11.3 V / 10 ohms = 1.13 A = 1130 mA
Calculate the maximum current across the Button OFF during switching OFF:
Ib2Max = Vs/R1 + Vrelay/R2
We can approximate this calculations by assuming the Vrelay = Vs:
Ib2Max = Vs/R1 + Vs/R2
Vs = 9 V: Ib2Max = 9 V / 10 ohms + 9 V / 10 ohms = 1.8 A
Vs = 12 V: Ib2Max = 12 V / 10 ohms + 12 V / 10 ohms = 2.4 A
The Relay diode ensures that the Vrelay voltage does not become negative when the battery is disconnected. The relay input is a coil. Discharging coil can be modeled as a decaying current source that causes current to flow the same direction as when the relay is ON for a few seconds or milliseconds even after power is disconnected. This can damage your electrolytic Crelay 100 uF capacitor if you capacitor is not bipolar.
Simulations show that after the Button ON is released (at 0.5 seconds), there is a small drop in output voltage. Both the relay and load voltages fall by 0.1 V. However, the relay remains ON.
The voltage 0.1 V drop occurs because the relay conduction resistance is not 0 ohms. Solid state relays that use semiconductors instead of mechanical switches can have those characteristics. This will not be the case for mechanical relays that use mechanical switches.
At 1 second point in time, Button OFF is pressed and the relay turns OFF. At 2 seconds point in time, a new cycle starts.
In the constructed circuit, R2 is a short circuit (I did not use that resistor), and Relay is not included. I also used a diode in series with Button OFF to discharge the relay because my diode had a resistance of 10 ohms (not all diodes have this resistance). Later I modified this Instructable, to replace the diode with R2 resistor (I also had to change the connection of R2 - not connected in series with Button OFF). Relay was added later.
You can see the color codes on the three yellow resistors.
The colors are:
Yellow - 4
Purple - 7
Black - 0 (number of zeros after 47)
This means the resistor value is 47 ohms. The gold band is tolerance in resistors, means 5%. That means the resistor value could be anywhere from 47 * 0.95 = 44.65 ohms to 47 * 1.05 = 49.35 ohms.
I used three 47 ohm resistors and the ceramic resistor is 56 ohms.
R1 = 1 / (1 / 47 ohms + 1 / 47 ohms + 1 / 47 ohms + 1 / 56 ohms) = 12.2418604651 ohms
This is approximately 10 ohms.
The two buttons are from an old VCR (Video Cassette Recorder).
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