POWER BUTTON SWITCH

.

The switches to momentary action button at low current, as types “tactile” for PCB mounting, They are inexpensive and available in many different styles. The types of hooking, on the other hand, They are often larger, more expensive and available only in a relatively limited range of styles. This can be a problem if you need a small and inexpensive switch to lock the power supply to a load. The solution is to convert the momentary action of a button in a function block.

The circuit described below, It requires only two transistors and a handful of passive components in order to obtain this result.

The circuit of Figure 1 (a) It is configured to block the positive supply to a load (reported on the ground). It works in mode “toggle”; that means, the first switch closure applies energy to the load, the second removes the power supply and so on.

Figure 1

The circuit converts a pressure switch Momentary action in a snap ignition switch.

To understand how the circuit works, suppose that the DC power supply, +Vs, has just been applied, that the capacitor C1 is initially not loaded and Q1 is turned off. The P-channel MOSFET, Q2, It is kept turned off by R1 and R3, who work in series to bring the gate up to + Vs, so that VGS is zero. The circuit is now in the state “not locked”, where the load voltage at the terminal OUT (+) it is zero.

If the normally open push button switch is closed momentarily, C1 – being devoid of charge – It brings the gate of Q2 to 0 V, thus turning the MOSFET. The load voltage of OUT (+) now immediately increases towards + Vs, and Q1 receives the base bias through R4 and ignites. In these conditions, Q1 saturates and lowers the gate of Q2 through R3, thus keeping the MOSFET turned on when the switch is open. The circuit is now in the state “stuck”, in which both transistors are active, the load is energized and C1 charges up to + Vs through R2.

When the switch is closed momentarily for the second time, the voltage on C1 (now approximately equal to + Vs) It is transferred to the gate of Q2. Since the gate-source voltage of Q2 is now approximately equal to zero, the MOSFET is turned off and the load voltage drops to zero. Even the voltage of the base emitter of Q1 drops to zero and the transistor is turned off. Therefore, when the switch is released, there is nothing that can hold active Q2 and the circuit returns to its state 'not engaged', where both transistors are off, the load is de-energized and C1 is discharged through R2.

The resistor R5 through the output terminals is an optional component that serves as a pull-down. When the switch is released, C1 discharges through R2 in the load. If the load impedance is very high (that is similar in magnitude to R2), or it contains active devices such as LEDs, the load voltage at Q2 is turned off can be large enough to solicit Q1 on R4, thereby preventing the circuit turns off properly. The presence of R5 carries the terminal OUT (+) a 0 V when Q2 turns off, thus ensuring that Q1 turns off rapidly and allowing the circuit to return to the state not locked correctly.

If the transistors are chosen properly, the circuit will operate over a wide voltage range and is suitable for driving loads such as relays, solenoid, LED and so on. However, attention that some DC motors and fans keep running when you remove their power. This rotation may generate an EMF large enough to solicit Q1, thus preventing the shut down circuit. You can eliminate this problem by inserting a blocking diode in series with the output, as shown in Figure 1 (b). You must also include R5 to ensure that Q1 OFF properly.

The complementary circuit proposed in Figure 2 It is intended to loads “high-side” connected to the positive supply line of the relay as shown in this example.

Figure 2 complementary circuit is intended to loads with common positive.

Note that Q1 has been replaced with a PNP transistor Q2 and is now an N-channel MOSFET. The circuit operates in a manner similar to that described above. who, R5 acts as a pull-up resistor which pulls the OUT terminal (-) until + VS when Q2 turns off, thus ensuring that Q1 turns off quickly. As in the previous circuit, R5 is optional and is only necessary for the load types mentioned above.

Note that in both circuits, produced by the C1-R2 time constant involves the separation of the switch contacts in button. Usually, a value between 0,25 e 0,5 seconds should be adequate. Constants of smaller time can lead to irregular behavior, while a longer time constant increases the waiting time between the closures of switches necessary to ensure that C1 is loads and discharges correctly. With C1 and R2 = 330nF = 1MW as shown, the time constant is nominally 0,33 s. This is usually sufficient to eliminate the bouncing of the contacts and to enable the load power to be activated after a couple of seconds.

Both circuits are designed to lock and unlock in response to short and momentary closures of switches. However, They have been designed to ensure the proper functioning even if the push-button switch is kept closed for a certain period of time. Consider the circuit shown in Figure 2 when Q2 is active. When the switch is pressed to unlock the circuit, the gate is lowered to 0V (since C1 is not loaded) and the MOSFET turns off, allowing the climb towards + Vs junction of R1-R2 through R5 and the load impedance. At the same time, Q1 also turns off, in such a way that the gate of Q2 is brought to 0 V by the combination of R3 and R4 series. If the switch is released immediately, C1 will charge just to Vs + via R2. However, if the switch is kept closed, the gate voltage of Q2 will be defined by the potential divider formed mainly by R2 and R3 + R4. If we assume that the terminal OUT (-) It is approximately equal to Vs + when the circuit is unlocked, the gate-source voltage of Q2 is given by:

VGS = (+Vs) × (R3 + R4) / (R2 + R3 + R4) = 0,02 (+Vs).

Although he Vs + 30 V, the resulting gate-source voltage of approximately 0,6 V will be too low to turn the MOSFET. Consequently, both transistors remain disabled until the opening of the switch contacts.

The circuit in Figure 2 It is blocked momentarily closing the switch when C1 is charged up to + Vs which causes OUT (-) fall in 0 V when Q2 turns on immediately, quickly followed by Q1. A temporary closure of the switch would allow C1 to discharge via R2 to zero after the opening of the contacts. However, if the switch is kept closed, the gate voltage of Q2 will be defined by the potential divider formed by R2 and R3. Since Q1 is saturated, the junction of R3-R4 on the collector of Q1 will be pulled up to + Vs, and the R1-R2 junction will be reduced to 0 V tramite Q2.

Therefore, with the switch kept closed, the gate-source voltage of Q2 is given by:

VGS = (+ VS) × R 2 / (R2 + R3) = 0,99 (+Vs).

Consequently, provided that the supply voltage is at least equal to the gate-source threshold voltage of Q2, both Q2 and Q1 will remain active until the opening of the switch contacts.

Both circuits provide an inexpensive way to derive a lock function by a momentary switch and, just as a mechanical lock switch, the quiescent power dissipation (without block) is zero, therefore, also make it suitable for battery-operated circuits.

VOTE
1 reply

Leave a Reply

Want to join the discussion?
Feel free to contribute!

Leave a Reply