how to monitor surge protection? - Page 1
Forum section: electronics / projects

author: calzap
author_rank: Frequent Contributor
timestamp: September 21, 2013, 04:21:38 am
content: I have seen power strips with surge protection that have an LED that indicates the protection is OK when the LED is lit. The mains breaker panel in my house has a two-pole breaker that provides surge protection and has similar indicators. The main (often the only) surge protection devices in these types of equipment are MOV's. So, what is the circuitry for the indicator LED's? MOV's have a high (multi-megohm) resistance to normal, non-surge voltages if they are functional and even higher resistance after they have failed. During the failure process, the resistance goes lower, perhaps very low, then the MOV causes a fuse to blow or a breaker to trip or blows like a fuse itself. So I assume a monitoring circuit is comparing the resistance of the MOV with some standard (like an ordinary resistor) and when it goes below a certain level (or to an extremely high level), power to the LED is removed (or the LED is burned-out).What would be a simple and cheap way to implement such a circuit? I've seen $US 20 power strips with this feature. I opened one, but the circuit board had been "snotted".One cheap, crude way would be to wire the MOV and the LED plus its current limiter in parallel and then have them both in series with a fuse. Of course, in this circuit, the MOV isn't really monitored, but if it fails, it might blow the fuse and the LED would go out. Problem is that the MOV might fail without blowing the fuse, but the LED tells the user all continues to be OK.Mike in California

author: SeanB
author_rank: Super Contributor
timestamp: September 21, 2013, 09:10:04 am
content: The last method you mention is how it is done. It assumes a surge big enough to blow the fuse will have damaged the MOV or that it went short circuit from repeated lower surges and blew the fuse.the only real way to test them is to apply a current limited voltage equal to the breakover voltage and measure that it both draws current and limits the voltage to the clamp voltage.

author: fpliuzzi
author_rank: Regular Contributor
timestamp: September 21, 2013, 03:20:26 pm
content: I haven't personally used these yet, but I came across the three legged LITTELFUSE "iTMOV" a while back. These thermally protected varistors allow their condition to be monitored. See page 5 of the datasheet below for some examples. Maybe this type of MOV can be of use to you.Regards,Frank

author: calzap
author_rank: Frequent Contributor
timestamp: September 22, 2013, 02:08:41 pm
content: I haven't personally used these yet, but I came across the three legged LITTELFUSE "iTMOV" a while back. These thermally protected varistors allow their condition to be monitored. See page 5 of the datasheet below for some examples. Maybe this type of MOV can be of use to you.Regards,FrankLittelfuse_Varistors_iTMOV_Datasheet.pdf---------------------------------------------------Thanks Frank. I wasn't aware of these devices; very neat. MOV and fuse in one package with extra lead connecting to the junction of the two. LittleFuse's Application Example 1 is exactly the circuit about which I speculated. However, I do wonder why they use a current-limiting resistor with mains voltage. At 120-240 V, the resistor will consume a lot more power than the usual types of LED's which are 20-30 mA. At 240 VAC RMS, the peak voltage is 339 V. For a 30 mA LED with a 2 V drop, the resistor would need to be (339 - 2)/0.03 = 11K ohms. Assuming a 50% duty cycle based on the circuit in their example, the LED would consume 0.03 watts and the resistor 2.5 watts! A transformer plus resistor or a capacitor or a mini switch-mode PS would be a better choice for current-limiting at higher voltages.Their example also shows a diode in series with the LED, which I don't understand. Unless that diode has a really whopping voltage drop (like 220 V for a 240 V mains voltage!), it's not going to significantly lower resistor power consumption nor protect the LED from high reverse voltage during the negative 1/2 of the AC cycle. Typically in an AC application, there is a diode in parallel, but opposite polarity, with the LED to protect it during the "reverse" half of the cycle. This configuration will give the resistor a 100% duty cycle though, so in the example above, the LED would consume 0.03 W and the resistor 5 W! That's a lot of power to throw away, and even a 10 W rated resistor is going to get more than a little warm at 5 W.If anyone knows of a diode that can be placed in series with an LED and resistor on mains voltage that will do the job of protecting the LED from high reverse voltage and allow low power consumption by the resistor, I'd like to know about it.Mike

author: SeanB
author_rank: Super Contributor
timestamp: September 22, 2013, 02:19:10 pm
content: Correct, but these typically use a high efficiency LED and drive it at around 1mA, which still gives a lot of light. Thus the resistor can be a 0.5W unit easily. The series diode half wave rectifies the voltage so the light is only on roughly half the time, but they will still be bright. Another alternative is to use a capacitive dropper, about 100n works well, though you do need a series resistor to limit the current spikes through the LED. This though is less reliable, as the capacitor does degrade with time. With this you either use a bridge rectifier to feed the LED or use another diode in reverse across it.

author: calzap
author_rank: Frequent Contributor
timestamp: September 22, 2013, 02:39:50 pm
content: OK. If you're going for 1 mA with a 2 V drop and 50% duty cycle, that's 1 mW. The dropping resistor for 339 V (peak for 240 V RMS) would be 340K ohms. Power consumption by the resistor would be 84 mW at 50% duty cycle. Yeah, the absolute power wastage is a lot less than running the LED at 30 mA, but the relative power wastage is still appalling. I don't understand the comment about the series diode. The self rectification within the LED will cause it to be on one-half the time with or without the series diode.Mike

author: SeanB
author_rank: Super Contributor
timestamp: September 22, 2013, 04:20:04 pm
content: Reverse breakdown of most LED chips is around 5V, you will need something a little better to have it withstand 330V plus reverse voltage.

author: NiHaoMike
author_rank: Super Contributor
timestamp: September 23, 2013, 12:16:57 am
content: Another bit of advice I have heard is to add some series inductance and plenty of parallel capacitance (motor run caps work well for hot to neutral). That combination would catch the smaller spikes before the MOV and load even see them.

author: fpliuzzi
author_rank: Regular Contributor
timestamp: September 23, 2013, 03:10:59 am
content: Even though I haven't used an iTMOV yet, I do have some of them in my parts bin (mouser #576-TMOV20R115M). I like the fact that the iTMOV includes a thermal fuse. I had a standard MOV in one of my surge protectors in my shop start on fire and scorch the pcb it was mounted on (luckily the unit was in a steel enclosure).I had planned on using a tiny neon lamp in place of the LED indicator circuit shown on the iTMOV datasheet. The neon lamps that I have only require about 1mA to operate and should be plenty bright.Noting the advice in this thread; if I do use a LED for the iTMOV indicator I'll be sure to use a high efficiency type and run it at about 1mA or so.Frank