1 Battery 3.7V but Circuit Requires 5V?

[arrow]

Hi

I would like to use one battery which has a V=3.7V. However my circuit (sensor and uC) requires 5V to operate (stable 5V).
Can someone please tell me how to do this?

One option is to double the voltage of the battery to 7.4V using a Dc-Dc converter (e.g.MAX1683). Then to use a Voltage regulator to ensure that I have a STABLE 5V. (I notice that the Dc-Dc converters V is not stable and decreases with Vin- is this right)?
Can someone please comment on this approach?
Are there better solutions?

Thank you in advance for your help.
a.

 

[philba]

That's an inefficient approach as you will waste 2.4V*load current. Why not build a boost SMPS tha outputs 5V directly? National, Linear, Maxim and others have plenty of designs to accomplish this. National and Linear have tools that will basically give you a design if you specify input V, output V and output current.

For example, national's webbench recommends 15 different possible chips to use (Vin 3 to 4.2, Vout of 5V, Iout 100 mA) which have complete design schematics. Just buy the components and build them.

 

[arrow]

Hi Philba

Thank you for your reply.

I have never done this before. What worries me is how clean will the output V be doing what you suggest?
My sensor requires a clean 5V regulated voltage.

Also can you please recommend a part number that I can look up?

Thank you
a.

 

[arrow]

Hi Philba

I understand that I shall waste P=I*V=output current*2.4
However I have never understood the following: my battery is rated at say 1000mAhours. Can you please explain to me how will that power dissipation drain my battery any quicker (more current will not be drawn by my application- right?).

Thank you
a.

 

[philba]

It's not a question of quicker but rather how much longer a more efficient solution would allow.

so lets say you double the 3.7 to 7.4 and then use an LDO 5V VREG. Further, assume your circuit draws 100 mA. Ignore efficieny for a minute. Your doubler will need to draw 200 mA from the battery - giving you a battery life of 50 hrs. during this 50 hrs, you are wasting 2.4*.1 (.24 Watts) as heat. that gives you a 67% (5/7.4) efficiency. However, your doubler has some loss as well so in practice it will be lower than 67%. All this, however, presumes your battery gives out a constant 3.7V - it doesn't so the analysis is more complex but sufficient for now.

In the case where you use a boost regulator you should be able to get to around 80% or better efficiency. The efficiency difference will translate into a longer run time.

Now, if I were you, I'd try to figure out how to run off of the 3.7V battery directly. You would need to deal with the changing voltage but it would give the absolutely longest run time. worth the effort.

 

[arrow]

Hi Philba

Thank you for your good explanation- a lot of thigns have been cleared up in my mind!

I have another question: Say I were to use the SMPS with a 5V regulated output as you suggest, I have looked up the output from these devices (MAXIM parts), and they seem to have:
(a) ripple output of 10mV and above
(b) the Vout is dependent on Vin; (My Vin will fall since its a battery powered application.)
If my sensor needs a clean 5.0V to operate (it is strain gauge bridge), how would you approach this with a SMPS?

Once again I appreciate your help.
a.

 

[philba]

no, a boost smps will have a range of input voltage over which it will produce the designed output voltage. Ripple is a fact of life in SMPSs.

I'm not sure how ripple will effect your sensor and ADC. You might need to filter results or compensate in your microcontroller.

Does it absolutely need 5V? There may be a way to compensate for falling voltage (as your battery drops) - If your ADCs reference and the sensor Vin are the same, won't the sampled results be constant/consistent?

 

[arrow]

Hi Philba

Thank you- you have a point- the Vref on the A/D might be effected in the same way. I will look into it.

I appreciate your help.
Best Regards
a.