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In my device 6.5V is generated from 5V USB. This can be done by boost converter, but if something goes wrong (external resistor voltage divider break up), due to overvoltage output condition (for example 16V) all parts connected to boost converter output will be destroyed. Output should never go over 7V.

I am looking for bullet-proof solution (low cost and minimum number of parts, of course). Required current is about 100mA, and efficiency is not so important, because this part of device is used very rare, and it is active during short period of time (1 second).

Closest one that I found is tps61220, but per datasheet, maximum output voltage is 6.0V, while overvoltage protection threshold (internaly) is in 6.0-7.5V range. I am afraid that due to output voltage close to 6.5V, overvoltage condition will be fired non-stop.

I am not interested in boost converters where overvoltage protection treshold is defined by external resistors, because again there things can go wrong.

I can use (100 mA) voltage doubler (charge pump) from 5.0V (USB) to 10V, and then regulate it down by LDO to 6.5V. Or regulate 5.0V (USB) to 3.3V by LDO and than double it by charge pump to 6.6V.

Edit: Let say that due some reason one of 0603 on board resistor fail. Board will not work anymore, but failing of this resistor will not cause any other damage. After replacing this resistor board will continue to work normally. There is LDO (5V from USB to 3.3V or lower voltage selectable by user) on board that can fail, but 5V passed from USB directly will not destroy the board. With boost converter this is not a case, and failing (because of any reason, with overvoltage on output) will destroy board (almost) completely.

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    \$\begingroup\$ "external resistor voltage divider break up" : parts falling off your board isn't something you usually design protection against, unless the thing to be protected is damn expensive. \$\endgroup\$ – peufeu Feb 11 '18 at 23:24
  • \$\begingroup\$ Checkout This answer \$\endgroup\$ – Trevor_G Feb 12 '18 at 0:21
  • \$\begingroup\$ Bulletproof, simple and for free. Put a slave with voltmeter and a switch. \$\endgroup\$ – Gregory Kornblum Oct 5 '18 at 11:00
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First, keep in mind that nothing is completely safe. All you can do is decrease the probability of a failure causing additional damage. You can protect against failure of specific components, but that introduces other components. The failure of those can't usually be guaranteed to be harmless.

You should also consider the cost of failure times the probability of failure, versus the fixed cost in every unit of protecting against the failure. For example, if the failure blows up $10 of parts and might occur in 1 out of every 100,000 units during their lifetime, then adding $1 of parts to protect against the failure is bad economics. If, on the other hand, a failure means a $500,000 house catches fire, with the possibility to kill people, then rather more than $1 is justified.

You haven't given us any guidance where in the spectrum of failure cost your product fits. Proper protection against the boost converter over-voltage is therefore impossible to judge.

However, given no other information, and taking on face value that over-voltage is worth a dollar or two to protect against, I suggest a shunt circuit. This circuit draws all available current when the voltage gets above some threshold. That's basically what a Zener diode does, but those can't usually dissipate the required power. There are such things as power Zeners. Look around.

A easier way to make a high-current shunt is usually with a Zener turning on a transistor. The Zener does the voltage detection, and the transistor does the heavy lifting of handling most of the current:

You have to look carefully at power dissipation. Find what maximum current the boost regulator can deliver at 7 V output, and then make sure the Zener/transistor combination can dissipate that power. For example, let's say the maximum is 200 mA at 7 V. That comes out to 1.4 W. That's enough that a single transistor in TO-220 package probably needs at least some heat sink. That will cost space and money. Protection isn't free, which is why the tradeoffs should be carefully analyzed.

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You could use a zener diode on the output to limit an over-voltage situation to the breakdown voltage of the zener. To protect the zener from burning you should add a 150 mA fuse between booster output and zener.

The 1N5921 is a 6.8 volt 3 watt zener diode but its tolerance (5%) may not be tight enough for this application. There are 2% zeners available but you could also use a crowbar circuit.

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