Hi,

I need 3.7 volts for this circuit (see attached schematic). I’d like to use non-rechargeable coin cell batteries, due to space issues. Can I use 3, LR44 batteries in series, with a resistor (for the extra .8 volts) If not, what are my options for coin cells, or other small battery options?

Components in this circuit:
3x 5mm LEDs (3V)
1x 50V, 10uF capacitor
on/off switch
555 IC
1k & 22k resistors

 

Which timer are you using? Bipolar or CMOS? The bipolar version isn't guaranteed to operate below 4.5V.

I redrew your sad "schematic". I corrected the wiring problem for the timing components (because I couldn't make myself draw it the way you indicated), pin 5 (which has to be 8 or the timer wouldn't have power), but not with the LEDs. Some leave the control pin floating, but the company who did the original design recommended bypassing it.

With a Li-ion battery, they're probably not going to be very happy. With 3 button batteries, they might survive.

If not, what are my options for coin cells, or other small battery options?

What is the volume constraint on batteries?

 

2” x 2” x .25” is the volume constraint.

 

ne555 hlf 21a8 is the print on the chip

 

2” x 2” x .25” is the volume constraint.

I'd just go with LR44 and not expect much runtime.

ne555 hlf 21a8 is the print on the chip

A bipolar timer is not guaranteed to operate below 4.5V.

I don't think the batteries will be strong enough to damage LED1 or LED2 (from my schematic), but I'd skip LED3. If the switch is closed, either LED1 or LED2 will be on, so you don't need a power indicator. LED1 might be stressed a bit until the battery voltage drops.

High output voltage is 1.7V from the supply voltage (typical) and guaranteed to be 2.25V from the supply. So LED2 won't be overstressed much.

 

Can I use 3, LR44 batteries in series to power the chip, then just put a resistor before LED 1 (your schematic) to drop the extra 1.5 volts?

 

LR44 button batteries are so small that the old NE555 will shut down in less than one hour.

 

Can I use 3, LR44 batteries in series to power the chip, then just put a resistor before LED 1 (your schematic) to drop the extra 1.5 volts?

If you only use 1 resistor, I'd put it on the anode of LED1. If you use 2, put the second on the anode of LED2. Assuming you drop LED3.

Are your LEDs ultra-brights?

The voltage when the batteries are fresh will be above 1.5V, but that won't be long.

From Renata with a load of 190uA:

 

My LEDs are not UBs, but, I’m going to need to rethink my supply power options…. Thanks