Determining PPTC Resistance in Your Application
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Figure 1. Typical Operating Curve for a PPTC Device.
As a field application engineer, one of the most common questions I am asked is, “What is the resistance of a PPTC device in its tripped state?” The short answer is, “That depends.” PPTC (polymer positive temperature coefficient) devices are constant power devices in the tripped condition. In other words, in the “untripped” condition, the resistance is constant. But in the tripped condition, the resistance will vary as the voltage varies in order to maintain constant power dissipation, as shown along the curve in Figure 1. Each PPTC device has a power dissipation (Pd) specification. Keep in mind that this is a typical number, not a minimum or maximum.
Design engineers normally see variations in the device’s power dissipation, depending on things that may act as a heat sink, such as circuit board thickness, trace thickness, etc. There are so many variables in the field that we do not specify an absolute minimum or maximum.
In order to calculate the resistance in the tripped state, we simply need to look to our old friend, Ohm’s Law, Pd = V2/R, or R = V2/Pd, and I = Pd/V where Pd is power dissipation, R is resistance and V is voltage. It may seem counter-intuitive in some cases, but lower voltage means higher leakage current in the tripped condition. Also, many customers are surprised at how low the tripped state resistance is. But remember that in a low-voltage situation a little bit of resistance goes a long way. With our high-voltage PPTC devices, the tripped state resistance can be very high. With our low-voltage surface-mount parts, such as those used in USB applications, the tripped state resistance can be quite low.
Consider the example of a small surface mount PPTC device in a 5V USB port, where a short circuit causes that device to trip. The Pd for this device could be 0.8 watts. So, the leakage current would be: I = 0.8 watts/5 volts = 160 milliamps. The resistance in the tripped condition would be R = (5 volts)2/ 0.8 watts = 31.3 ohms.
At the opposite extreme, consider a line voltage through-hole PPTC device tripped at 240 volts AC rms. Since it is a larger device, it could have a power dissipation rating of 2.9 watts. The leakage current here would be: I = 2.9 watts/240 volts = 12.1 milliamps. The tripped resistance would be: R = (240 volts)2/2.9 watts = 19.9 kohms. This is a significant difference, but in both cases, the resistance is sufficient to limit the fault current and help prevent damage to sensitive components.
Calculating PPTC resistance in the tripped state is simple, as long as you have a good power dissipation spec and let Ohm’s Law take it from there. And remember that this power dissipation is normally a “typical” spec, so depending on circuit board heat-sinking conditions, there will be some variation.
Barry Brents
FAE Manager
TE Circuit Protection
Barry Brents is a Field Application Engineering Manager with TE Connectivity – Circuit Protection. A member of IEEE, and former US Navy nuclear submarine officer, he has been involved in the communications industry since 1991. Barry received his B.S.E.E. from Texas Tech University.