Modern powerful key FETs have a very low the resistance of the channel in the open state, often even below resistance closed contacts of the electromagnetic relay or a mechanical switch, because resistance of mechanical contacts is influenced corrosion, pollution, scorching. These weaknesses in the key field-effect transistor no. Besides low resistance of the open channel, even with large current and a large load power, making power dissipation on the transistor itself minimum. So, often, for switching kilowatt load key field effect transistor does not require even a simple radiator.
Here's a schematic of the electronic switch two loads voltage power from 5 to 20 V At a current up to 20 A. the Basis of the scheme are two key field transistor APM2556NU in which the resistance of the open channel does not exceed 0,006 Ohms. This means that at a voltage of 20 V and a load current of 20 A (i.e., when power load 400 watts) power on the open channel of the transistor will not exceed 2...4 watts.
The switch is controlled by two quasi-touch (non-latching) buttons short presses which can be switched loads. At the same time the load must not be activated, even while pressing both buttons both load off. There's an emergency entrance lock, when applying for which voltage from supply voltage up to 50 off In both loads. This input can be used in various protective circuits, when you need to urgently disable any of the included loads, and the possibility of their inclusion buttons to block.
The load is connected between positive power supply and the respective output of the circuit. The state of the switch is indicated by two LEDs.
Schematic diagram shown in the figure.
Device management is the RS-flip-flop on the chip D1 pins 2 and 12 are used to switch steady States of the trigger. These findings through resistors R1 and R3 tightened to zero. The resistance of the resistor is taken is relatively small (usually, in such circuits use resistors of tens to hundreds of the shortfall). In the first embodiment were resistors 56 Ohm, but later it turned out that at the time of turning on a power the load pulse occurs is a hindrance which resets the trigger and puts the scheme in self-oscillating mode. To avoid this, the resistance of the trigger inputs had to be lowered by lowering the resistance of the pull-up resistors, as well in addition to installing the capacitors C2 and C3, enhancing the stability of the trigger under impulse noise.
The S2 button leads to the emergence of logical units on the output 13. Transistor VT2 is opened and includes the load 2. The pin 1 is zero, therefore, VT1 turned off and the load 1, respectively off. When S1 is pressed unit appears at pin 1 D1 and the load transistor VT1 1 is enabled, and the pin 13 appears zero, so the load 2 is turned off. Resistors R6 and R7 are needed to reduce the influence of the gate capacitance of the field transistor to the circuit output. The gate capacitance is quite high, so when a sudden change in voltage on it is pretty large charging current that capacity. The resistors limit the current to a safe for the chip level. Diodes VD3 and VD4 help to discharge the gate capacitance at the closing of the transistor.
United together conclusions 3 and 11 are used to generate the point of locking. These the findings of the resistor R2 pulled to zero, so that as the pin lock no voltage (or the voltage a little) they do not affect the operation of the trigger. But the voltage is applied to them the level of logical units is forced the translation of the two elements of D1.1 and D1.2 to a logic zero at the output. Then to eat when on that point logical unit both load off independently from the previous state.
The voltage at the input of the lock may come from any scheme or system lock. The magnitude of this voltage, preferably, should not be more voltage to power the circuit. However, the presence of the Zener diode VD1 and the resistor R4 allows you to use to lock the voltage to 50 kV (can be more, but there is a danger of damage to the Zener diode, and after him, and chip).
The supply voltage of the load can be from 5 to 20 V. the voltage the chip should not exceed 15 V. To reduce the maximum voltage power D1 installed circuit R5-VD2. This circuit is powered from a source In more than 15 works like a parametric stabilizer and will not tolerate over-voltage on a chip. When the supply voltage is below 15 V, the scheme is not working as the stabilizer, since the Zener diode is closed, and only in conjunction with C1 as locking RC-chain on the food chain.
To reduce the voltage below 5 V, as the voltage on the gate an open transistor is insufficient for its full opening. Channel transistor will not fully open, i.e. will have a higher resistance, and this will lead to the fact that the power dissipated on it dramatically increases, which can damage the transistor.
At installation it is necessary to ensure a sufficient width of the path that goes to the drain and the source transistors from the load and from the minus power. Circuit conductors should too to be quite thick. The conductors of the control circuit of D1 can be thin, is there any reasonable thickness, so that the current small.
Transistors APM2556NU can be replaced by others with similar characteristics. If transistors with such a low resistance of the open channel to purchase not possible, but there are transistors with twice as much resistance - you can instead one transistor to use two in parallel. Or work on a lower maximum current, or use the radiator to drain the excess heat.
Zener BZV55C15 can be replaced by 1N4744A, X, X, DD. In principle, you can use any of the Zener voltage is not below 10 In and not more than 15 V.
Chip CLI you can replace the analogue CD4002 or chip CLE (similar CD4025). Chip CLE is characterized in that it has three three-vchodovych items OR NOT. Two are used in this scheme, and one extra remains free. So it is not damaged by static electricity inputs free item must be connected to 7 or 14 pins. All elements chips are physically interconnected, so even unnecessary damage item may have a negative impact on other elements of the chip. Else to use chip CLP, it has two three-vchodovych nor gate and one one-shadowy the inverter, it remains free (his entry is connected with pin 7 or 14).
The 1N4148 diodes you can substitute almost any low-power pulsed diodes, for example KD522.
Varistor FNR05K220 can be replaced by a varistor voltage of about 20 V.
LEDs - any indicator.
Assembled without error, the device, when the health of all parts of the building not requires.
Author: R. Lyzhin