Short Circuit Protection System At The Amplifier Output
This is the circuit diagram of 5V DC regulated power supply which featured with short circuit protection system. There are 2 kind of output that are regulated 5V DC with short circuit protection and without circuit protection. The main circuit is protected from any damage due to short-circuit in the additional power supply circuit by cutting off the derived supply voltage. The derived supply voltage restores automatically when shorting is removed. An indicator LED is utilized to show whether short-circuit exists or not.
Power Supply Block:
This circuit works just like the ordinary DC power supply adapter, in the main power supply circuit, 230V AC from main home electric source is stepped down by transformer X1 (230V AC primary to 0-9V, 300mA secondary), then rectified by a fullwave rectifier comprising diodes D1 through D4 with bridge arrangement (you may use a single bridge diode), filtered by capacitor C1 and regulated by IC 7805 to give regulated 5V DC output (O/P1).
Short-Circuit Protection for Low-Voltage DC Distribution Systems Based on Solid-State Circuit Breakers Sharthak Munasib. Munasib, Sharthak, 'Short-Circuit Protection for Low-Voltage DC Distribution Systems Based on Solid-State Circuit Breakers' (2017).Theses and Dissertations.
Short-Circuit Protection Block:
Transistors SK100 and BC547 are used to derive the secondary output of around 5V (O/P2) from the main 5V supply (O/P1).
The working of this circuit is quite simple. When the 5V DC output from regulator IC 7805 is available, transistor BC547 conducts through resistors R1 and R3 and LED1. As a result, transistor SK100 conducts and short-circuit protected 5V DC output appears across O/P2 terminals. The green LED (LED2)? lows to indicate the same, while the red LED (LED1) remains off due to the presence of the same voltage at both of its ends.
When O/P2 terminals short, BC547 cuts off due to grounding of its base. As a result, SK100 is also cut-off. Thus during short-circuit, the green LED (LED2) turns off and the red LED (LED1) glows. Capacitors C2 and C3 across the main 5V output (O/P1) absorb the voltage fluctuations occurring due to short-circuit in O/P2, ensuring disturbance-free O/P1. The design of the circuit is refer to the relationship given below:
RB = (HFE ? VS) / (1.3 ? IL)
where,
RB = Base resistances of transistors of SK100 and BC547
HFE = 200 for SK100 and 350 for BC547
Switching Voltage VS = 5V
1.3 = Safety factor
IL = Collector-emitter current of transistors
Build the circuit on a general purpose PCB and mount in a general circuit box. Connect O/P1 and O/P2 terminals on the front panel of the cabinet. Also connect the mains power cord to feed 230V AC to the transformer. Connect LED1 and LED2 for visual indication.
This circuit has been tested by the author, the circuit that already built is shown on the following image:
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The circuit is really easy to understand.
A resistor of low value (the resistor value will be explained later) is connected in series with the output of the power supply. As current starts to flow through it, a small voltage drop will appear on it and we will use this voltage drop to determine whether the power supply out put is overloaded or short circuited.
The 'heart' of this circuit is an operational amplifier (op amp) configured as a comparator (stage 2).
The way it works is really simple, you just need to follow this rule:
If the voltage on the non-inverting output is higher than the inverting output, then the output is set to 'high' level.
If the voltage on the non-inverting output is lower than the inverting output, then the output is set to 'low' level.
I put quote marks on 'high' and 'low' for a easier understanding of the op amp operation. It has nothing to do with logical micro controllers 5 volts levels. When the op amp is in 'high level', its output will be very approximate of its positive supply voltage therefore, if you supply it +12V, the 'high output level' voltage will approximate to +12V. When the op amp is in 'low level', its output will be very approximate of its negative supply voltage therefore, if you connect its negative supply pin to ground, the 'low output level' will be very near to 0v.
When we use op amps as comparators, we usually have an input signal and a reference voltage to compare this input signal.
So, we have a resistor with a variable voltage that is determined according with the current that flows through it and a reference voltage. Does this ring any bell on your mind? We're almost finished with the theory be brave and follow me.
As the voltage drop on the resistor in series with the power supply is too small, we need to amplify it a little bit because some op amps are not too accurate when comparing low voltages like 0,5 volts or lower. And that's why the first stage (stage 1) of this circuit is an amplifier using another op amp. A 3 to 4 times amplifications is more than enough in this case.
The op amp gain(av) is determined by the formula: av = (RF/R1)+1
In this case we've got 3.7 times of gain: av = (2700/1000)+1 = 3.7
The third stage of the circuit is the protection itself. Its a relay that you can connect directly directly with the output of your power supply if you are dealing with low current (2A) or you can connect it to a bigger relay if you are dealing with bigger current or even shut down a previous stage of you power supply forcing the output to shut down. This will vary with the power supply you've got. For example, if your power supply is based on a LM317, you can simply use the relay to physically disconnect the LM317 output pin from the power supply, as we are using the relay normally closed pin (I've uploaded a picture to better describe this example).
Short Circuit Protection System At The Amplifier Output System
The PNP transistor on stage 3 just act like a seal to keep the relay turned on after the short circuit so you can press a button to disarm it. Why I didn't use the relay itself to do this? It's because the relay is too slow to do it.
Just think about it: At the moment the relay turns off the output of your power supply, the short circuit does not exist anymore and the comparator goes from high level to low level. As there is no more current flowing at the NPN transistor base, there is no more current flowing through the relay coil as well. When all these steps happens, the relay contacts did not had enough time to complete its course and connect to the other contacts to close the seal. The behavior of the circuit if I used the relay itself to close the seal would be the relay madly trying to turn off the output, but without success. I know I could have used a capacitor to supply enough current to the relay, but I would need a big capacitor and no one can grant that it would work 100% of the times the output of the power supply is shorted. Electrolytic capacitors fails over time, and failure is not a good option in this circuit.
To disarm the circuit a normally closed switch is connected in series with the base of the NPN transistor. By pressing this normally closed switch, it would open its contact and disconnect the base of the NPN transistor from the rest of the circuit breaking the seal and resetting the power supply output.
The 1uF capaciton on the NPN transistor base is just a threshold so a little peak consumption don't trigger the protection.
You can feed this circuit 9V to 15V. Just be careful to correctly choose your relay voltage and the capacitors voltage. And just to be clear, do not connect this circuit supply pins directly with you power supply output or it will be useless. Just imagine, if your output is shorted, there won't be enough voltage to supply the protection circuit. You will need to connect it on a stage before the output, maybe a dedicated voltage regulator just for it. A LM7812 will be more than enough.