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We are currently designing a system which will use a relay, a MCU and a sensor (powered in 3.3V with 100mA max current consumption) + a few LED (2.2V type). it will be powered by a 9V battery

After inspecting a similar product we find out they use MSP 430 from TI which use 3.3V, and a relay HK4100F-DC5V which supposedly work with 5V DC (datasheet http://img.ozdisan.com/ETicaret_Dosya/445413_4369639.pdf/) . I read in this datasheet that coil voltage can also be 3V.

I need to choose the PMU which is going to be used to power this system.

1./ Do i really need to use 3 LDO/ DC-Dc to power 3.3V,, 5V and 2.2V or can I group all of those under a 3.3V power rail ? I was thinking to use TPS563200

2./ How do I determinate the current required by the coil of the relay ? Do I just take the voltage applied to it and divide by its resistance?

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    \$\begingroup\$ Don't try powering your LEDs directly from 2.2 volts - some will work fine, some won't even light and some will burn. \$\endgroup\$
    – Andy aka
    Commented Jul 21, 2016 at 7:21
  • \$\begingroup\$ When you say 9 V battery, do you mean a PP3-style 9 V battery? I'm asking since they don't deliver 100 mA. If you buck down 9 V to 3.3 V, you may get away with it. And to answer your question, can you find a 3.3 V relay instead? It would make things less complicated. \$\endgroup\$
    – winny
    Commented Jul 21, 2016 at 7:28
  • \$\begingroup\$ @winny, yes it is a PP3-style. How do I manage then to supply more than 100mA from this kind of battery ? \$\endgroup\$
    – chris
    Commented Jul 21, 2016 at 7:32
  • \$\begingroup\$ @Andyaka, what do you recommend then ? \$\endgroup\$
    – chris
    Commented Jul 21, 2016 at 7:32
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    \$\begingroup\$ @chris Then you need to be aware of its limitations. Only good ones/rechargeable/lithium can even support 100 mA and only for a short period. Choose a relay with minimum power consumption and buck down to 3.3 V. If a 9 V relay is more energy efficient, go for that and MOSFET to drive it. \$\endgroup\$
    – winny
    Commented Jul 21, 2016 at 10:18

2 Answers 2

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  1. use a LDO or DC-DC converter to give you 3.3V for the MSP430 and the sensor. The LEDs will also be powered from this, using a series resistors to drop 1.1V. Chose the value of resistor to give you the LED current and brightness that you require.
  2. Rather than use a 5V relay, why not use a 9V relay from the same manufacturer? Then you can power the relay directly from the battery. The 5V relay cannot be guaranteed to operate correctly from 3.3V since it has a Max Operate Voltage of 3.75V. If you must use a 5V relay then you will need a separate LDO or DC-DC to give you 5V.

The relay current is, as you say, Voltage / Coil resistance.

There are 3 versions of this relay, specified as the power consumption of the coil, ie 0.15W, 0.2W, and 0.36W. If you use the most sensitive (0.15W), and use the 9V version, then the current required is 16mA. A PP3 will be able to supply this current, but not for very long.

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    \$\begingroup\$ Keep in mind that a 9V battery near the end of it's life is going to be at around the 7V mark. For the relay he specified though it shouldn't be a problem \$\endgroup\$
    – user103993
    Commented Jul 21, 2016 at 7:36
  • \$\begingroup\$ @Hayman indeed. For the 9V version the max operate voltage is 6.75V. \$\endgroup\$
    – Steve G
    Commented Jul 21, 2016 at 7:38
  • \$\begingroup\$ I have spotted this TPS563200 to go from 9V to 3.3V . I have used Webench from Ti website to spot this device considering the 6V to 10V range of battery discharge @SteveG considering that uncontrolled drop of the battery voltage, would not be better to use a 3.3V or 3V relay on the output of the DC DC to make sure it can switch properl y? \$\endgroup\$
    – chris
    Commented Jul 21, 2016 at 7:40
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    \$\begingroup\$ The 9V relay is guaranteed to work down to 6.75V. Once your battery has fallen to that level it has little charge left in it, so it will soon fall to 6V. Using a 3V relay from the 3.3V DC-DC could give you a little more life, but you will need to do the maths to see if it really does, because the DC-DC will be around 88% efficient, so you loose some power, ie some battery life, by using the DC-DC. Also, the 3V relay consumes 55mA @ 3.3V which adds to the load on the DC-DC. Personally I'd use the 9V relay but you are the designer! \$\endgroup\$
    – Steve G
    Commented Jul 21, 2016 at 8:08
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I'll try to answer each of your questions separately.

  1. No, you don't need multiple LDO's. You could get away with using 3.3V for the entire circuit as long as your MCU is 3.3V compatible and not 5V only. In the datasheet where you read the voltage could also be 3V, that's a different relay from the same family. The 5V relay has a maximum actuating voltage of 3.75V, this means that it is possible that 3.3V will actuate the relay, but sourcing a 3V relay from the same family would be a much more reliable alternative (HK4100F-DC3V) as you can't guarantee the 5V relays will work every time, you might order some which work fine and then order the same again but they don't work.
  2. Yes, the current that will be drawn by the relay will be the supply voltage divided by the coil resistance

For your 2.2V LEDs, remember that this is the forward voltage of the LED. Try not to think of it as an operating voltage. The LEDs will drop 2.2V across them but will draw as much current as they feel like until you end up with a dead LED. You need to put a resistor in series with the LED that will drop the remaining 1.1V (If we use 3.3V). We then select an appropriate resistance as to limit the current flowing through the LED, this can be done using the equation \$R = (V_{cc}-V_{f})/I\$. Where Vcc is your 3.3V Vf is your 2.2V LED forward voltage and I is your required current which you can find from the datasheet.

The other thing to keep in mind is that dropping from 9V down to 3.3V at the current you require is going to be about 1W power dissipation, if you do use an LDO you'll need a heatsink for that more than likely. The other alternative would be to use a switching regulator instead (TSR 1-2433 is the sort of thing I'm talking about)

Looking at the current consumption of this device, if you have the 100mA drawn by the sensor, 20mA from the LEDs and 20mA from the relay coil then your 9V battery will need to source ~150mA. 9V batteries are around 550mAh, this means your circuit is, at a worst case scenario only going to run for about 4 hours

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  • \$\begingroup\$ I was thinking using TPS563200 for droppîng from 9V to 3.3v \$\endgroup\$
    – chris
    Commented Jul 21, 2016 at 7:39
  • \$\begingroup\$ the device will be mostly on low power mode, It will not run continuously. \$\endgroup\$
    – chris
    Commented Jul 21, 2016 at 7:43
  • \$\begingroup\$ You could use the TPS563200 chip but you'll need a couple extra external components like the LC filter and voltage dividing resistors \$\endgroup\$
    – user103993
    Commented Jul 21, 2016 at 7:45
  • \$\begingroup\$ yes sure there are some passive components around. I am just concerned about the Relay actually. \$\endgroup\$
    – chris
    Commented Jul 21, 2016 at 8:06
  • \$\begingroup\$ I'd try to get a 3V relay. Don't try looking for a 3.3V relay as it's not really a standard value so you won't find one. G5V-1 3DC is a good relay. it's a 3V DC relay that can operate between 2.4V to 6V \$\endgroup\$
    – user103993
    Commented Jul 21, 2016 at 8:10

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