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I'm building a wireless device that have to last with batteries as long as posible. The device has an rf module with a microncontroller that manages the control, that part is bought and works fine. When that module recieves an especific rf signal, has to activate a motor, using a motor driver (LB1930MC).

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I'm not a expert in that field, but i achieved this with literaly 5 components (module, 2 capacitors, the motor driver and the connection header for the motor and the JTAG for the module), and that makes me think i'm doing something wrong. If i want to sell it i need a device that is ready for produccion, and i need help in that field. Is something i'm missing?, is something needed in the power part of the circuit or for safety norms that i have to have in mind for production?

And by this I do not mean certification as the FCC for the radio module or anything like that. I mean what is necessary for my product to be a final product, safe and ready to go.

Second question:

In the power part of the circuit, the microcontroller can work from 3.8 to 1.8 volts, so i'm using 2 AA batteries. If i want to replace it with a lithium battery, i can't because when is fully charged it's voltage is 4.2 volts. The simplest-cheapest conclusion was to put a diode in series with the lithium battery, so now the maximun voltage of the battery drops 0.7 volts (3,5 v), and the minimun voltage of the battery (2.7-3 V for lithium battery) least 0.7 volts is still above the 1.8 minimun voltage of the microcontroller. My question is: a low dropout linear regulator or a DC-DC converter is a better choice for achiving more battery life? (both alternatives are more expensive, and the product needs to be very cheap)

enter image description here

One last question is:

The device needs to run 24/7 countinously forever, so if i use lithium batteries i have to charge it at the same time the device is running.

I read that i can't charge the battery and have a load at the same time because in the constant current part of the charge the current is low and if the device uses more power, the battery keeps discharging and never charges. Achiving this needs an extra circuit that disconnects the battery from the load and powers the load from another source.

The power consumption of my device will be less that 1 mA normaly, with some peaks of current when the rf module is transmiting (the motor is powered from other source).

If i have such low current, that means i can charge the battery and still power the device with no extra components? or is that a risk?

EDIT:

  • As i said the average current of the device is less than 1 mA, with peaks of 50-60 mA for the Rf module.
  • I have not the option of pluging it into the wall.
  • The number of times the device will wake up and activate the motor are from 0 to 10-15 times in a day, with an average of 6-7 times/day.
  • I need at least 4 months of battery life (time between recharges).
  • I was thinking in using a 500-1000 mA lithium battery.
  • The motor used is a little dc motor. It's not dangerous, but the old system that is already used for the motor should continue to function without any problem, so my system must interfere with it as little as possible (in my example only interferes getting in parallel with the motor).
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    \$\begingroup\$ Why would you run this off a battery if you have a motor power source close by? \$\endgroup\$ – stefandz Aug 14 '15 at 16:27
  • \$\begingroup\$ My device is an addon for another battery powered device. I can't use their battery because the other device is crucial and i don't want to reduce his battery life. Also the voltage is higher and if i use it i have to go for a regualtor or a dc-dc converter. \$\endgroup\$ – juan jesus pinuaga Aug 14 '15 at 17:59
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    \$\begingroup\$ "The device needs to run 24/7 countinously forever," why not plug it into the wall? \$\endgroup\$ – MathieuL Aug 14 '15 at 18:13
  • \$\begingroup\$ You asked a lot of questions. I will weigh in on the interesting ones (to me). Use an LDO to drop the lithium battery voltage. The diode might not drop enough voltage to protect your uC, and at 1mA, a DC-DC converter will have a hard time competing with an LDO in efficiency. If you want to keep the Lithium battery charging, look for a charger with "power path" feature. Basically, this type of charger has three power connections. One for USB (or whatever) one for the battery, and one for system power. It manages all current flows intelligently just the way you want it to. \$\endgroup\$ – mkeith Aug 15 '15 at 3:29
  • \$\begingroup\$ Oh, if the average load is only 1mA, then you might be OK without the power path, or you could use a dual diode to power your LDO so that it will draw power from the DC power source when it is available (assuming it is greater than 4.2V), or switch over to battery automatically when the DC power source goes down. \$\endgroup\$ – mkeith Aug 15 '15 at 3:33
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OK, here goes...

In terms of safety, you will want to have some sort of protection against overcurrent faults at the power input stage. This could be a fuse, PTC resettable fuse, dedicated power management IC or a custom analog solution - but for this I would look at a PTC resettable fuse vs a traditional wire fuse. PTC should be cheaper (assuming you fit a holder for the wire fuse) but the wire fuse will likely have a lower resistance, increasing efficiency.

Additionally, since the motor power source is likely to be an even more powerful (and hence potentially dangerous power source) you will want similar current protection here - likely a traditional fuse as close to the power source as possible. This should be rated to protect the cable from exceeding its maximum allowable current.

Since you are controlling a motor, you need to consider the impact of the system failing with the motor on. Is that dangerous? If so, you will need to implement radio packet checksumming on the microcontroller to ensure that you don't respond to accidental interference from other sources. You should also really have a "keep-alive" signal coming from the control radio which you monitor using a hardware watchdog - if you don't receive a "keep-alive" every X seconds, the hardware watchdog kicks in and cuts out the motor.

Additionally, you need to consider the environment that the product will sit in and ensure that you encase it appropriately to protect from humidity, water ingress, dust ingress etc.

You have already hinted at EMC certification - you will have to go through this for both your unit and the transmitter that we can't see in your diagram. You will also have to go through the relevant product certification for your area regarding the type of device this is classed as. For both of these you will be needing consultancy from a firm specialising in compliance engineering, as this is clearly not your area of expertise.

As far as the battery voltage regulation goes, a switched-mode DC-DC converter will bring higher efficiency (if properly designed) than a linear regulator for all but the tiniest of power draws. For low cost, you should be looking at an inductorless step-down charge pump for this.

Regarding battery charging - if you are able to charge the battery, surely it is better to use the available power to run your unit and forget the battery altogether? Unless you are designing this for an area with unreliable or intermittent power.

The schematic below allows for powering from a Lithium polymer cell with minimal loss (no diode drop) along with automatic disconnect should a charging source become available and charging of the cell itself. The charge pump follows after, and unlike some comments have described, can indeed reduce voltage! Although it will be about as efficient as a linear regulator for the 1mA periods of operation, it will be more efficient when larger bursts of current are required.

schematic

simulate this circuit – Schematic created using CircuitLab

To be honest, if you want to turn this into a product you are going to want to find an electronic engineer with experience of getting stuff to market to take on the design for manufacture and handle certification. But my answer should at least give you the gist of what you are looking for.

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Here is a picture of the diode "power path." I am assuming you have some power source available since you indicated that you want to charge the battery while the system is on.

diode power path

Just use an LDO. Let's say you run your LDO at 2.7V. The average battery voltage is around 3.7V over the full range. So the average efficiency of the LDO is around 2.7/3.7V. That is 73%. This assumes a low quiescent current LDO (make sure to check this). You will have a very difficult time improving upon that efficiency with a buck converter (at 1mA).

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  • \$\begingroup\$ The charge pump comment here is completely wrong. Charge pumps absolutely can and are used to reduce voltages - see ti.com/lit/ds/slvs391b/slvs391b.pdf for an example. Although at the 1mA level a charge pump will be about the same efficiency as an LDO, during the high current bursts, the charge pump will be far more efficient. Please stick to known facts as opposed to conjecture. \$\endgroup\$ – stefandz Aug 15 '15 at 7:55
  • \$\begingroup\$ Also, the dual diode-path circuit shown here is inefficient as you lose a diode drop for both power sources. The circuit I linked in my answer includes a diode drop for the mains power source, but only a saturated MOSFET (very low resistance) for battery operation, keeping you much more efficient when you need it most. \$\endgroup\$ – stefandz Aug 15 '15 at 7:58
  • \$\begingroup\$ That is a pretty cool part. Have not come across that before, and honestly I did not know that type of charge pump existed. The switched capacitor architecture of that part is not something I would have described or recognized as a charge pump. I will edit my answer. \$\endgroup\$ – mkeith Aug 15 '15 at 9:43
  • \$\begingroup\$ You know, I have started to look at this in more detail just now (for another project I am doing) and I'm starting to think that synchronous bucks can be even more efficient - check out cds.linear.com/docs/en/datasheet/338813fa.pdf for 90% efficiency (claimed) at 3.3V in, 1.8V out and 1mA draw! \$\endgroup\$ – stefandz Aug 15 '15 at 9:48
  • \$\begingroup\$ If you are using something with higher efficiency than an LDO, then the diode drop will hurt you. If you are using an LDO, the diode drop doesn't really matter because the processor can run all the way down to 1.8V. Please keep in mind that the OP is trying to keep cost under control. \$\endgroup\$ – mkeith Aug 15 '15 at 9:49

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