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This simple circuit is designed to cutoff power to the load when the supply voltage either exceeds a certain maximum input voltage (over-voltage condition) or is shy of a minimum input voltage (under-voltage condition). It cuts off power to the load if the supply voltage is either lower than 10V or higher than 20V, i.e., it connects the input to the load only if the input voltage is between 10V and 20V. This version is simpler than the previous version, but is less precise and shows a lot of temperature dependence.
Previous version: http://everycircuit.com/circuit/5407749949882368
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Problem:
We need to create a circuit that "trips" when the input voltage is outside the allowed voltage range. The allowed voltage range is 10V - 20V.
Solution:
Let us use a PMOS to connect/disconnected power to the load. Now this PMOS must be turned off when the input voltage is outside the allowed voltage range and turned on when the input is within the range.
By connecting a resistor from the gate of the PMOS to ground through a BJT and also to the source through a pull-up resistor, the PMOS can be turned on by turning on the BJT only when the input voltage exceeds 10V. This can be achieved by a voltage divider. But this only solves half the problem. We still need to shut the PMOS down when the input exceeds 20V. This can be achieved by shorting out the pull-up resistor with a parallel BJT, which is triggered by a voltage divider.
To reduce the number of parts, instead of using 2 seperate voltage dividers (4 resistors in total), the 2 circuits can be combined to get a voltage divider with 3 resistors with 2 outputs. The outputs are connected to the bases of the 2 BJTs. For convenience, let us replace the 3 resistors with 2 potentiometers, whose wipers connect to the BJTs as shown.
Calculations:
Let the resistances of the 2 potentiometers be 10K.
Let the resistance of the pull-up resistor be 1M.
Let the resistance of the pull-down resistor be 10K.
Note: The above values are arbitrarily (to an extent) chosen.
Let the resistance of the section of the upper potentiometer connected between the supply voltage and it's wiper be 'x'.
Let the resistance of the section of the lower potentiometer connected between ground and it's wiper be 'y'.
Let the Base-Emitter threshold voltage of the BJTs be 0.5V.
The NPN BJT must turn on when the input voltage exceeds 10V. Therefore, the drop across y should be 0.5V when the input is 10V. Applying voltage divider theorem,
0.5 = 10(y/20k) => y = 1K
The PNP BJT must turn on when the input voltage exceeds 20V. Therefore, the drop across x should be 0.5V when the input is 20V. Applying voltage divider theorem,
0.5 = 20(x/20k) => x = 0.5K
The values have been determined.
But there is a problem. If we use these values as it is, the circuit does not cutoff the supply at 20V, but it does so at 29.7V. The UVLO happens at 11.3V, which is not too big of an error (caused due to the fact that the actual value of Vbe(th) is different from the assumed value), but the OVLO is way off. Click on the link to see the problem: http://everycircuit.com/circuit/6209590067462144
Now set the input source to DC and manually sweep the input, while monitoring the 2 values of Vbe. You will observe that the PMOS turns on when the input voltage is 11.4V, for a Vbe of 570mV. But the PMOS turns off only when the input reaches 29.7V, for a Vbe of 748mV! The two types of BJTs, NPN & PNP, have identical I-V characteristics, so why is the PNP only turning on for a Vbe of 748mV and not around 570mV?!? Please let me know!
Note: A zener diode to protect the PMOS's gate from blowing up is intentionally omitted due to the fact that EC MOSFETs don't blow up and also because it doesn't contribute to the lockout function. Not having the "non-mandatory" components makes the circuit easier to understand.
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