I want to protect my circuit against a wrong conneted power supply (changed polarity). To do this i want to use a diode. If I use the S1A for fairchild semicondutor. The maximum voltage at 1A is 1.1V. The maximum power dissipation is 1.4W. In my appilication the current that is needed is 0,4A. So the maximum voltage with this current would be:
in figure 2 of the article:
When the power supply is reversed connected. the TVS diode is forward biased, creating a short. The current is the power supply voltage divided by the on-resistance of the TVS diode. Because this resistance i low the current can be very high. Therefore a PPTC device is needed. when the current is increasing above the Trip point of the PPTC the resistance of the PPTC is increasing causing a lower current. The PPTC and the TVS diode are connected parallel. Do i also need a normal diode in series with the PPTC or not?
If i look at the resistance change of the PPTC it's a bit low. it changes form for example 0.6ohm to 2.6ohm. So the current at 12V supply would be 12V/2.6=4.61A???? This calculation is probably wrong so how can i calculate the current in reversed polarity.
rudie,
Your analysis assumes that the forward voltage is directly proportional to current, i.e, the diode is a linear device. However, this is not the case. The diode voltage is a logarithmic function of current. You are better off using the "Forward Voltage Characteristic" curve from the datasheet to determine the diode voltage. The data sheet at **broken link removed**
has such a curve.
Regards,
Kral
If i look at the resistance change of the PPTC it's a bit low. it changes form for example 0.6ohm to 2.6ohm. So the current at 12V supply would be 12V/2.6=4.61A????
Against the reverse polarity protection I think you can use a Schottky diode like the MBRS130LT3 (30V / 1A), just check the maximum reversing voltage that you need for your application.