How to get 5 volts from 6.4 volts

alec_t · Sep 5, 2012

In that case why not just use a low drop-out 3V regulator?

Boncuk · Sep 5, 2012

Richardcavell said:
Boncuk,

I have several rechargeable Nickel-metal hydride batteries. My workplace encourages us to use them as a gesture for the environment. They produce 1.2 volts when fully charged. If you want to allow the end user to use rechargeables, you couldn't use the voltage as an indicator of charge status. Really, the cell is depleted when it produces 0 volts. It is useless when the combined output is less than the requirements of your components (in this case, 1.8 volts total). Assuming the microcontroller and all other components still work acceptably at 1.8 volts, then that's all you need the cells to produce.

Richard

Oops!

Completely different story!

Please refer to post #1.

Richardcavell · Sep 5, 2012

Well, look, I guess engineering is all about compromise. Ultimately, you can accept a widely varying input voltage if your components can tolerate it, and you might get identical function from your components or tolerably different behaviour given different input voltages. Or you could try to fix the voltage and accept loss of power as heat. A low drop out regulator limits your margin for error so that you may not be able to use rechargeables at all. One way or another, you're taking the worst of it somewhere.

WTP Pepper · Sep 5, 2012

View attachment 66958 View attachment 66959

A well designed oscillator will be supply voltage tolerant within reason.

The second attachment has both outputs overlapping exactly.

The circuit shown is a simple Wein RC oscillator and simulation is done with LTSpice.

Hope this helps!

electronicsbasecom · Sep 5, 2012

Ok you want 5V, no current details but I can tell that 250mA for uC is enough and less than 1A. It is best to use 4 rechargable batteries so input is 4.8V max, so you need charge pump to get 5V. For example, use MAX862 https://www.electro-tech-online.com/custompdfs/2012/09/74840.pdf. For this chip you can also use 3 batteries instead of 4.

Boncuk · Sep 6, 2012

Richardcavell said:
Well, look, I guess engineering is all about compromise. Ultimately, you can accept a widely varying input voltage if your components can tolerate it, and you might get identical function from your components or tolerably different behaviour given different input voltages. Or you could try to fix the voltage and accept loss of power as heat. A low drop out regulator limits your margin for error so that you may not be able to use rechargeables at all. One way or another, you're taking the worst of it somewhere.

Very niceley said! In other words: Well, look, it's your baby!

Boncuk

grizzly1 · Sep 6, 2012

Check out an LT1129, I have had good luck with these in many designs. The dropout voltage is pretty low around .3 volts at 300 mA and increases to about .6 at 700 ma which is the max for this. -- You don't mention anything about current (or I missed it) but usually the dropout voltage is a function of the current and most suppliers have a graph in their data sheet. If the microcontroller wil run 1.8 to 5.5 -- use the adjustable regulator and set the microcontroller bias voltage to 4 v then either of your battery choices will do the trick

WTP Pepper · Sep 8, 2012

Why not try and design the circuit properly instead of trying to design a botch regulator to make the oscillator stable for a fixed voltage?

Then we have temperature variations that can and probably negate all the effort.

How to get 5 volts from 6.4 volts

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