Since Q1 will be 0.7 volts, the current through R2 is 0.7/470k or 1.5uA. So the current through that needs to be accounted for. Resistor R2 keeps Q1’s base low when it is in its off state. The minimum current into the base of Q1 will thus be 800uA/beta or 800uA/180 = 4.5uA. The needed collector current for Q1 saturation depends on the value of R3 (since the collector is close to zero volts and current will no longer flow through the base of Q2). Using a 15k resistor gives us 800uA, which satisfies that need. In order to push it completely into saturation, we’ll want to increase that by a factor of 2-10. This is the minimum current at Q2’s base that will cause 33mA of collector current to flow. For the maximum current draw case, you’ll want to divide the 33mA by Q2’s beta of about 180, which gives you 183uA. This translates to an output impedance range of approximately 365 – 100,000 ohms. This circuit was tested with a load resistor of as high as 100k, so the current sink range is from 120uA to 33mA. If using the relay, the current will be 12V/365 ohms or 33mA. The expected current through load LED1 is (Vcc-VLed)/R1 or (12-2)/680 or 15mA. Each one operates in either saturation or cutoff. Transistors Q1 and Q2 are configured as inverting switches. Once this happens, Q2 is turned off, and the cycle continues. The 0.7 volts at capacitor C1 is pushed into Q1’s base – switching it on. Since Q1’s collector is 0.7 volts, capacitor C1 charges to this value via R1. The circuit has now latched into its “on” state. The voltage at Q2’s collector now drops to approximately zero volts, which causes current to flow through the load LED1. This action forward biases that junction, and turns Q2 “on”. When Q1 is turned “off” the voltage at its collector rises from approximately zero volts, to 0.7 volts, which is the maximum value that the base to emitter junction of Q2 will allow. When the pushbutton is pressed, the low voltage value on C1 is “pushed” into the base of Q1. Capacitor C1 stores this low voltage value via R1. Since Q1 is “on”, the voltage at its collector will be close to zero volts. This voltage causes enough current to flow through R4 to turn transistor Q1 “on” (saturation state). As such, the voltage at the collector of Q2 is (Vcc minus the LED voltage drop), or around 10 volts. Referring the basic circuit diagram, the circuit operates as follows:Īssume that transistor Q2 is “off” (cutoff state). It is the transistor equivalent of a two-inverter latching circuit where the states of Q1 and Q2 invert with respect to each other upon each button press. The operation of this circuit is straight-forward. Transistor Push-on, Push-off Latching Switch with Relay Schematic Diagram Two-transistor Push-on, Push-off Switch Circuit Two-transistor Push-on, Push-off Switch Circuit with RelayĬlick here for a our recommended programmable bench power supply. Transistor Push-on, Push-off Latching Switch (1) 0.47uF 25v (or greater) electrolytic capacitorīelow are pictures of both the basic circuit, and the one used to control a relay, along with the schematic references.(2) 2n3904 general purpose NPN transistors.(1) IN4148 switching diode (if using relay).(1) General purpose LED (or 12v SPST relay – 350 ohms or greater – RT334012). Minimum breadboarding equipment (see here).Parts Needed (if you’d like to build this circuit) With this relay, you could drive a substantial load if desired. One that turns on an LED, and another that energizes a relay. The circuit is depicted in with two options. This electronic equivalent of a latching push-button is accomplished by a two-transistors and a few supporting components. So when you press it once, it is on and press it again, it is off. The circuit described in this article is the equivalent of a latching push-button switch. Build a Transistor Toggle (push-on, push-off) Switch on a Breadboard What is a Push-on, Push-off Switch?
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