How short circuit protection is designed and works. Short circuit protection circuit Power supply with short circuit protection circuit
To protect the power supply when designing various circuits, it is recommended to add an overcurrent protection unit to the power supply output. A simple device circuit is built using a thyristor as a voltage protection control element.
As long as the supply voltage at the input is within normal limits, the zener diode and thyristor are closed, and current flows into the load. When the supply voltage exceeds 15.2V, the zener diode opens, followed by the thyristor, since there is a potential difference between its cathode and the control electrode sufficient to unlock it. Thyristor VS1 connected in parallel to the output of the power supply, when overloaded, breaks the fuse within a few microseconds if the output voltage is above the permissible value. The threshold for opening the thyristor, namely, triggering the protection, depends on the technical data of the zener diode. If the fuse blows, a piezo sound emitter with a built-in generator will turn on, which will signal an external fault, which also indicates a possible short circuit in the load. The alarm will sound until the utility power or load device is disconnected.
Video of the power supply protection circuit in action
Many homemade units have the disadvantage of lacking protection against power reverse polarity. Even an experienced person can inadvertently confuse the polarity of the power supply. And there is a high probability that after this the charger will become unusable.
This article will discuss 3 options for reverse polarity protection, which work flawlessly and do not require any adjustment.
Option 1
This protection is the simplest and differs from similar ones in that it does not use any transistors or microcircuits. Relays, diode isolation - that’s all its components.
The scheme works as follows. The minus in the circuit is common, so the positive circuit will be considered.
If there is no battery connected to the input, the relay is in the open state. When the battery is connected, the plus is supplied through the diode VD2 to the relay winding, as a result of which the relay contact closes and the main charging current flows to the battery.
At the same time, the green LED indicator lights up, indicating that the connection is correct.
And if you now remove the battery, then there will be voltage at the output of the circuit, since the current from the charger will continue to flow through the VD2 diode to the relay winding.
If the connection polarity is reversed, the VD2 diode will be locked and no power will be supplied to the relay winding. The relay will not work.
In this case, the red LED will light up, which is intentionally connected incorrectly. It will indicate that the polarity of the battery connection is incorrect.
Diode VD1 protects the circuit from self-induction that occurs when the relay is turned off.
If such protection is introduced into , it’s worth taking a 12 V relay. The permissible current of the relay depends only on the power . On average, it is worth using a 15-20 A relay.
This scheme still has no analogues in many respects. It simultaneously protects against power reversal and short circuit.
The operating principle of this scheme is as follows. During normal operation, the plus from the power source through the LED and resistor R9 opens the field-effect transistor, and the minus through the open junction of the “field switch” goes to the output of the circuit to the battery.
When a polarity reversal or short circuit occurs, the current in the circuit increases sharply, resulting in a voltage drop across the “field switch” and across the shunt. This voltage drop is enough to trigger the low-power transistor VT2. Opening, the latter closes the field-effect transistor, closing the gate to ground. At the same time, the LED lights up, since power for it is provided by the open junction of transistor VT2.
Due to its high response speed, this circuit is guaranteed to protect for any problem at the output.
The circuit is very reliable in operation and can remain in a protected state indefinitely.
This is a particularly simple circuit, which can hardly even be called a circuit, since it uses only 2 components. This is a powerful diode and fuse. This option is quite viable and is even used on an industrial scale.
Power from the charger is supplied to the battery through the fuse. The fuse is selected based on the maximum charging current. For example, if the current is 10 A, then a 12-15 A fuse is needed.
The diode is connected in parallel and is closed during normal operation. But if the polarity is reversed, the diode will open and a short circuit will occur.
And the fuse is the weak link in this circuit, which will burn out at the same moment. After this you will have to change it.
The diode should be selected according to the datasheet based on the fact that its maximum short-term current was several times greater than the fuse combustion current.
This scheme does not provide 100% protection, since there have been cases when the charger burned out faster than the fuse.
Bottom line
From an efficiency point of view, the first scheme is better than the others. But from the point of view of versatility and speed of response, the best option is scheme 2. Well, the third option is often used on an industrial scale. This type of protection can be seen, for example, on any car radio.
All circuits, except the last one, have a self-healing function, that is, operation will be restored as soon as the short circuit is removed or the polarity of the battery connection is changed.
Attached files:
How to make a simple Power Bank with your own hands: diagram of a homemade power bank
When setting up various electrical and radio equipment, sometimes everything does not go as we would like and a short circuit (short circuit) occurs. A short circuit is dangerous both for the device and for the person installing it. To protect the equipment, you can use a device, the diagram of which is presented below.
Principle of operation
Relay P1 acts as a monitoring element against a short circuit; it is connected in parallel with the load. When voltage is applied to the input of the device, current flows through the relay winding, the relay connects the load, and the lamp does not light up. During a short circuit, the voltage at the relay will drop sharply, and it will turn off the load, while the lamp will light up and signal a short circuit. Resistor R1 is used to adjust the current threshold; its value is calculated using the formula
R1=U network /I additional
U mains – mains voltage, I additional – maximum permissible current.
For example, the network voltage is 220V, the current at which the relay will operate is 10A. We consider 220 V/10 A = 22 Ohm.
Relay power is calculated using the formula 0.2 * I add.
Resistor R1 should be taken with a power of 20 W or more.
That's all. If you have comments or suggestions regarding this article, please write to the site administrator.
List of used literature: V.G. Bastanov Moscow worker. "300 Practical Tips"
Modern power switching transistors have very low drain-source resistances when on, which ensures low voltage drop when large currents pass through this structure. This circumstance allows the use of such transistors in electronic fuses.
For example, the IRL2505 transistor has a drain-source resistance, with a source-gate voltage of 10V, only 0.008 Ohms. At a current of 10A, the power P=I² R will be released on the crystal of such a transistor; P = 10 10 0.008 = 0.8 W. This suggests that at a given current the transistor can be installed without using a radiator. Although I always try to install at least small heat sinks. In many cases, this allows you to protect the transistor from thermal breakdown in emergency situations. This transistor is used in the protection circuit described in the article “”. If necessary, you can use surface-mounted radioelements and make the device in the form of a small module. The device diagram is shown in Figure 1. It was calculated for a current of up to 4A.
Electronic fuse diagram
In this circuit, a field-effect transistor with a p channel IRF4905 is used as a key, having an open resistance of 0.02 Ohm, with a gate voltage = 10V.
In principle, this value also limits the minimum supply voltage of this circuit. With a drain current of 10A, it will generate a power of 2 W, which will entail the need to install a small heat sink. The maximum gate-source voltage of this transistor is 20V, therefore, to prevent breakdown of the gate-source structure, a zener diode VD1 is introduced into the circuit, which can be used as any zener diode with a stabilization voltage of 12 volts. If the voltage at the input of the circuit is less than 20V, then the zener diode can be removed from the circuit. If you install a zener diode, you may need to adjust the value of resistor R8. R8 = (Upit - Ust)/Ist; Where Upit is the voltage at the circuit input, Ust is the stabilization voltage of the zener diode, Ist is the zener diode current. For example, Upit = 35V, Ust = 12V, Ist = 0.005A. R8 = (35-12)/0.005 = 4600 Ohm.
Current-voltage converter
Resistor R2 is used as a current sensor in the circuit, in order to reduce the power released by this resistor; its value is chosen to be only one hundredth of an Ohm. When using SMD elements, it can be composed of 10 resistors of 0.1 Ohm, size 1206, with a power of 0.25 W. The use of a current sensor with such a low resistance entailed the use of a signal amplifier from this sensor. The DA1.1 op amp of the LM358N microcircuit is used as an amplifier.
The gain of this amplifier is equal to (R3 + R4)/R1 = 100. Thus, with a current sensor having a resistance of 0.01 Ohm, the conversion coefficient of this current-voltage converter is equal to unity, i.e. One ampere of load current is equal to a voltage of 1V at output 7 DA1.1. You can adjust the Kus with resistor R3. With the indicated values of resistors R5 and R6, the maximum protection current can be set within.... Now let's count. R5 + R6 = 1 + 10 = 11kOhm. Let's find the current flowing through this divider: I = U/R = 5A/11000Ohm = 0.00045A. Hence, the maximum voltage that can be set at pin 2 of DA1 will be equal to U = I x R = 0.00045A x 10000 Ohm = 4.5 V. Thus, the maximum protection current will be approximately 4.5A.
Voltage comparator
A voltage comparator is assembled on the second op-amp, which is part of this MS. The inverting input of this comparator is supplied with a reference voltage regulated by resistor R6 from stabilizer DA2. Non-inverting input 3 of DA1.2 is supplied with amplified voltage from the current sensor. The load of the comparator is a series circuit, an optocoupler LED and a damping adjustment resistor R7. Resistor R7 sets the current passing through this circuit, about 15 mA.
Circuit operation
The scheme works as follows. For example, with a load current of 3A, a voltage of 0.01 x 3 = 0.03V will be released at the current sensor. The output of amplifier DA1.1 will have a voltage equal to 0.03V x 100 = 3V. If in this case, at input 2 of DA1.2 there is a reference voltage set by resistor R6, less than three volts, then at the output of comparator 1 a voltage will appear close to the supply voltage of the op-amp, i.e. five volts. As a result, the optocoupler LED will light up. The optocoupler thyristor will open and bridge the gate of the field-effect transistor with its source. The transistor will turn off and turn off the load. You can return the circuit to its original state with the SB1 button or by turning the power supply off and on again.
A protection design for any type of power supply is presented. This protection circuit can work together with any power supply - mains, switching and DC batteries. The schematic decoupling of such a protection unit is relatively simple and consists of several components.
Power supply protection circuit
The power part - a powerful field-effect transistor - does not overheat during operation, therefore it does not need a heat sink either. The circuit is at the same time a protection against power overload, overload and short circuit at the output, the protection operation current can be selected by selecting the resistance of the shunt resistor, in my case the current is 8 Amperes, 6 resistors of 5 watts 0.1 Ohm connected in parallel were used. The shunt can also be made from resistors with a power of 1-3 watts.
The protection can be more accurately adjusted by selecting the resistance of the trimming resistor. Power supply protection circuit, current limit regulator Power supply protection circuit, current limit regulator
~~~In the event of a short circuit and overload of the unit output, the protection will instantly operate, turning off the power source. An LED indicator will indicate that the protection has been triggered. Even if the output short-circuits for a couple of tens of seconds, the field-effect transistor remains cold
~~~The field-effect transistor is not critical; any switches with a current of 15-20 Amps or higher and an operating voltage of 20-60 Volts will do. Keys from the IRFZ24, IRFZ40, IRFZ44, IRFZ46, IRFZ48 line or more powerful ones - IRF3205, IRL3705, IRL2505 and the like are ideal.
~~~This circuit is also great for protecting a charger for car batteries; if the connection polarity is suddenly reversed, then nothing bad will happen to the charger; the protection will save the device in such situations.
~~~Thanks to the fast operation of the protection, it can be successfully used for pulsed circuits; in the event of a short circuit, the protection will operate faster than the power switches of the switching power supply have time to burn out. The circuit is also suitable for pulse inverters, as current protection. If there is an overload or short circuit in the secondary circuit of the inverter, the inverter's power transistors instantly fly out, and such protection will prevent this from happening.
Comments
Short circuit protection, polarity reversal and overload are assembled on a separate board. The power transistor was used in the IRFZ44 series, but if desired, it can be replaced with a more powerful IRF3205 or with any other power switch that has similar parameters. You can use keys from the IRFZ24, IRFZ40, IRFZ46, IRFZ48 line and other keys with a current of more than 20 Amperes. During operation, the field-effect transistor remains icy. therefore it does not need a heat sink.
The second transistor is also not critical; in my case, a high-voltage bipolar transistor of the MJE13003 series was used, but there is a large choice. The protection current is selected based on the shunt resistance - in my case, 6 0.1 Ohm resistors in parallel, the protection is triggered at a load of 6-7 Amps. You can set it more precisely by rotating the variable resistor, so I set the operating current to around 5 Amps.
The power of the power supply is quite decent, the output current reaches 6-7 Amps, which is quite enough to charge a car battery.
I chose shunt resistors with a power of 5 watts, but 2-3 watts is also possible.
If everything is done correctly, the unit starts working immediately, close the output, the protection LED should light up, which will light up as long as the output wires are in short-circuit mode.
If everything works as it should, then we proceed further. Assembling the indicator circuit.
The circuit is copied from a battery screwdriver charger. The red indicator indicates that there is output voltage at the power supply output, the green indicator shows the charging process. With this arrangement of components, the green indicator will gradually go out and finally go out when the voltage on the battery is 12.2-12.4 Volts; when the battery is disconnected, the indicator will not light up. 