I need some help to complete my project . This is a simple 12 volt motor that needs to be protected when it goes to overload.
The data is shown and can help to design it.
The overload protection circuit should have minimum components because of not enough space to add it .
The input voltage is variable from 11 volt to 13 volt because of wiring length but the cut off overload should be happen when the V1 - V2 = > 0.7 volt .
Pls look at the attached overload diagram that should cut off if the amps increase more than 0.7 Amp. What is your idea about this diagram . Is it a complicated circuit or need to be added some components?
Referring to the above drawn 12v motor current control schematics, the concept appears to be correct, however the circuit implementation especially in the second diagram looks incorrect.
Let's analyze the diagrams one by one:
The first diagram explains the basic current control stage calculations using an opamp and a few passive components, and it looks great.
As indicated in the diagram as long as V1 - V2 is less than 0.7V, the output of the opamp is supposed to be zero, and the moment it reaches above the 0.7V, the output is supposed to go high, although this would work with a PNP transistor at the output, not with an NPN, ....anyway let's move ahead.
Here the 0.7V is with reference to the diode attached to one of the inputs of the opamp, and the idea is simply to ensure that the voltage on this pin exceeds the 0.7V limit so that this pinout potential crosses the other complementing input pin of the opamp resulting in a switch OFF trigger to be generated for the attached motor driver transistor (an NPN transistor as preferred in the design)
However in the second diagram, this condition will not get executed, in fact the circuit will not respond at all, let's see why.
In the second diagram when power is switched ON, both the input pins connected across the 0.1 ohm resistor will be subjected to almost an equal amount of voltage, but since the non-inverting pin has a dropping diode it will receive a potential that may be 0.7V lower than the inverting pin2 of the IC.
This will result in the (+) input getting a shade lower voltage than the (-) pin of the IC, which in turn will produce a zero potential at pin6 of the IC right at the onset. With a zero volts at the output the connected NPN will just refuse to initiate and the motor will remain switched OFF.
With the motor shut off there won't be any current drawn by the circuit and no potential difference generated across the sensing resistor. Therefore the circuit will stay dormant with nothing happening.
There's another error in the second diagram, the motor in question will need to be connected across the collector and the positive of the transistor for making the circuit effective, here a relay has no role, and is therefore not required.
The faulty situation discussed above can be reverted by simply swapping the input pinouts of the IC across the indicated points, that is across the sensing resistors, as shown in the corrected third diagram.
Referring to the third diagram as soon as power is switched ON, pin2 will be subjected to a 0.7V less potential than pin3 of the IC, forcing the output to go high at the onset.
With the output going high will cause the motor to start and gain momentum, and in case the motor tries to draw a current more the the specified value, an equivalent amount potential difference will be generated across the 0.1 ohm resistor, now as this potential begins rising pin3 will start experiencing a falling potential, and when it falls below the pin2 potential, the output will quickly revert to zero disrupting the base drive for the transistor and switching off the motor instantaneously.
With the motor switched OFF during that instant, the potential across the pins will tend to get normalized and will restore back to the original state, which in turn will switch ON the motor and the situation will keep self-adjusting through a rapid ON/OFF of the driver transistor, maintaining a correct current control over the motor.
The sensing resistor may be calculated as follows:
R = 0.7/current
Here as specified for a 0.7amp current limit for the motor the value of the current sensor resistor R should be
R = 0.7/0.7 = 1 ohm