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I have a battery powered device with an 8051-base MCU running off a 3v coin cell (with a DC/DC converter down to ~2.1v). I'm attempting to drive a 3v dual-coil latching relay using two GPIO pins of the MCU with the following schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

However, when running off the battery, the above does not work. It does work when powered off the debugger, which supplies 3.3v, without the DC/DC converter.

The parts in question are:

  • CC2541 MCU
  • DMG6968U N-Channel MOSFETs
  • RT314F03 Latching Relay:
  • 1N4148W Diode

Is there something inherently wrong with the design? Is there something I'm missing? Something else I can test?

Perhaps there's a better way to drive this (or another) relay from a battery powered device?

Thanks in advance!

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    \$\begingroup\$ What coin cell? CR2032 or something bigger? Possibly not enough current available for the relay coils. The datasheet implies 200mA (3V at 15 ohms). \$\endgroup\$ – Roger Rowland Feb 13 '15 at 15:39
  • \$\begingroup\$ In the case where it is not working (battery operated), when the output pin is high (gate voltage present), what is the voltage at the gate pin and the drain pin of the FETs? \$\endgroup\$ – tcrosley Feb 13 '15 at 15:46
  • \$\begingroup\$ Roger Rowland probably has the answer. If coin cell is CR2032, examine the data sheet page 2 : Voltage drops to 2.8 volts with 100 ohm load. Expect much lower battery voltage since your load is 15 ohms. data.energizer.com/PDFs/cr2032.pdf \$\endgroup\$ – Marla Feb 13 '15 at 16:04
  • \$\begingroup\$ It is indeed a CR2032... I don't have precise enough tools for the measurement. I might be able to have a friend help out on Sunday. \$\endgroup\$ – Assaf Inbal Feb 13 '15 at 18:45
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Several things:

  1. Do you really really need to make 2.1 V to run the micro from the 3 V coin cell? There are many micros that will run nicely from the full range of voltage a "3V" coin cell will produce over its lifetime. A converter seems wasteful and unnecessary.

  2. You really expect a "coin cell" to drive a relay!? Unless these are very special tiny extra small low power minuature relays, that's not gonna happen. As soon as you attempt to activate a relay, the coin cell voltage collapses, which probably resets the CPU.

Since you say your relays are latching, you only need a pulse of power for a short time. One possibility is to put a large capacitor across the coin cell. This would hold up the voltage long enough for a relay to latch in the opposite state, which is 10 ms according to the datasheet you provided. However, that will still cause significant current drain on the battery to recharge the cap after the relay is tripped. That will decrease the lifetime of the coin cell. They are intended for µA continuous drain, and a few mA pulsed at most.

A better option for the battery is to charge up a capacitor to a higher voltage ahead of time, then discharge it across the relay. You can then control the switching power suply that charges the cap to only drain power from the coin cell slowly.

However, no matter what you do, basic physics limits how many times a single coin cell can activate the relay. It takes a finite amount of energy to flip the relay, and there is a finite (and small) amount of energy in a coin cell. Even with theoretical 100% conversion efficiency, that coin cell won't live long.

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  • \$\begingroup\$ 1. According to TI's documentation, the MCU's boltage regulator isn't very efficient and do they suggest on of their own converters for a longer battery life. 2. I don't expect much as I'm only starting to step into the world of electronics and, pretty much, have no idea what I'm doing. Would a CR123A make more sense? BTW, a previos design where the relay was used incorrectly (the common leg of the relay was connected to GND and transistors to each end of the coil) had the GPIO trigger an PNP transistor which was connected to a NPN transistor which powered the same relay, which worked. \$\endgroup\$ – Assaf Inbal Feb 13 '15 at 19:00

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