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Li-Ion power supply for CC111x

Alexey Petrov
Prodigy 250 points

Hello,

I'm going to use Li-Ion cell for powering my device with chip CC111x. In datasheet I see that proper maximum VCC is 3.6V (absolute maximum is 3.9V), but Li-Ion can produce up to 4.2V. So it's impossible to connect CC111x to the battery directly. Of course I can use LDO linear regulator for this but it increases device's cost - it is critical in my case. So I have an idea - connect CC111x to the battery through silicon diode (in forward direction). It "eats" about 0.4V or more and VCC will be in permissible range. I'm sure the idea will work when CC111x is in active mode but what it will be in sleep when CC111x is comletely off? I mean if VCC is not more than 3.9V in this case?

Thanks in advance,

Alexey

  • 0
    Mastermind 8170 points

    Li-Ion batteries can be either 4.3 or 4.2Vdc depending on the technology. If they are charged in your circuit you can see over 4.6 Vdc due to IR losses in the wires and battery terminals.

    A couple of thoughts.

    1.) With a passive approach the power used, and therefore the battery life, is 4.3V x current x time. Whether you drop .4V in a diode and the rest in the radio chip or 1V in a diode and the rest in the radio the numbers do not change. So I recommend you target the 2.9  - 3.3V range for the radio. Operating the radio near it absolute rating will lead to failures.

    2.) You are right in worrying about the diode drop when the radio is in a low power sleep mode. When the radio current is very low the diode drop due to leakage will be nil and the radio will see the battery voltage. If it was just gates that start to leak in the radio their increased current would cause sufficient voltage drop to keep the voltage in the proper range. Unfortunately there are other failure modes that will not self-limit until sudden breakdown damage occurs so you need more than just a diode.

    3.) Instead of just a diode consider a transistor / diode pair with a resistor. Due to the gain of the transistor the resistor can be much larger and have negligible additional impact of battery life.  It is much cheaper than a LDO and better leakage performance than just a diode.  Other variations include 2 transistors used alone.

    4.) Unless weight matters consider other battery technology, Li-Ion are expensive and have some self discharge

     

  • 0 in reply to H Stewart
    Expert 1835 points

     There are some very inexpensive LDOs out there. I would highly suggest one. TI makes a few very good ones with very low noise for less than a buck in single piece quantities. Hamilton is right about operating the IC near it's ab max for any length of time. I have blue smoked a couple at 3.9 volts, and changing them out is tricky at best.

    Ron

  • 0 in reply to ronnym
    Expert 1670 points

    I agree with the others - a single diode won't be enough protection. The LP2950-33 is a good 3.3v linear regular for around $0.10 - $0.15 in volumes over 10k

  • 0
    Prodigy 250 points

    Thanks guys for your very useful notes!

  • 0 in reply to H Stewart
    Prodigy 250 points

    Hamilton, hi,

    Could you please give me example scheme how to use transistor / diode pair with a resistor? I have no idea :-(

  • 0 in reply to Alexey Petrov
    Expert 1020 points

    The easiest circuit is going to be zener stabiliser circuit.  You will need to chose the correct components accordingly.  The voltage output is going to be based on the zener diodes clamped voltage, plus the voltage across the base and emitter, typically 0.6V.  So in this case 5.6-0.6 will give you an output voltage of 5.0V, as long as the input is greater then 6 volts.  The output current is going to be based on R1 and what gain you are looking for.  Hope this helps.