Hello, I am a total hobbyist and know very little about electronics, but I try. I made a LED on a switch that stays on for six hours when I press a button then resets waiting for another press. I used a 555, capacitor and resistors. The problem is the 9 V battery dies in three days. How could I make it last much longer? I was going to use a smaller LED and try to make it flash? Could I somehow use the 555 to make it flash also? Thanks all

 

The simplest solutions are:
1. Use a higher value resistor in series with the LED so that it draws less current.
2. Use a larger battery.
That is all I can suggest without seeing a circuit diagram.

 

A schematic would help….
A high output LED could run at lower currents, extending the battery life, but a 9v battery does not have much capacity to start with.

 

Welcome to AAC!

The problem is the 9 V battery dies in three days. How could I make it last much longer? I was going to use a smaller LED and try to make it flash? Could I somehow use the 555 to make it flash also?

A schematic would be helpful.

If you're not using the 555 to make the LED flash, then you don't need the timer.

LM3909 will flash LEDs with just a capacitor and the LED with very low drain on the battery. But it wants to operate from 1-5V.

 

S.C.H.E.M.A.T.I.C . . .

And - what is the exact part number of the 555 you are using. There are two basic types, bipolar and CMOS, and there is a large difference in how much energy the chip uses internally to function. If you are using the bipolar 555, changing to a CMOS version will increase battery life.

Separate from that, if you configure the 555 as a medium-frequency oscillator, such as 1 kHz or more, the apparent brightness of the LED will decrease very little while the actual power dissipated in the LED will decrease by 50%. The problem with this is that if you stick to the standard circuits, one 555 can be either a timer or an oscillator, but not both.

To have both a 6-hour timer plus the power-reducing oscillator requires three more external parts, 1 - R, 1 - C, and 1 - diode.

If you show us your current schematic, I'll post one with the combo circuit.

Note - Schematics are important. Paraphrasing Rear Admiral Joshua Painter,

"Engineers don't take a dump, son, without a schematic."

ak

 

I used this. I changed the cap to 1000uF and the resistor to 470,000? This gave me a time on of 8 hours after pressing the button. I did not know 9v batteries did not have good capacity. Thank you. So using this could I make a smaller led flash for the time on? How does changing the value of the cap and resistor change the time on? If you could dumb it down

 

I used this. I changed the cap to 1000uF and the resistor to 470,000?

How hard is it to show your actual schematic?

This gave me a time on of 8 hours after pressing the button.

How accurate does the time need to be? 555 timers don't do long delays well.

I did not know 9v batteries did not have good capacity.

Google is your friend.

So using this could I make a smaller led flash for the time on?

You'll need another timer, or oscillator.

How does changing the value of the cap and resistor change the time on?

On time is determined by the time it takes the capacitor to charge to 2/3 VCC.

\( \large V_f = V_i(1-e^{-\frac{t}{RC}}) \)
\( \large t=-RCln(1-\frac{V_f}{V_i}) = -(470k\Omega)(1000uF)ln(1- 2/3) = 516s = 8.6\ minutes\)

Capacitor leakage current will affect timing.

This nomograph from National Semiconductor makes it easy to select component values for times up to 100s:

They made a mistake. There's only one resistor.

 

First pass at a schematic that has both an on-timer and a modulated LED to increase battery life.

R1 and C1 set the operating time. R1 is very large to reduce the current used by the oscillator.

Pressing SW1 charges C1 up to 9 V, and the oscillator starts. When the button is released, R1 starts discharging C1.

When R1 discharges C1 down to around 0.5 V, the 555 is reset and the output goes low. With a CMOS 555, this is a very low power state.

When the 555 is oscillating, R2 and C2 set the oscillator frequency. The LMC555 datasheet has the output frequency equation. Adjust R3 to change the brightness.

ak

 

Along with using a high-brightness type at a lower current, you could use a CMOS type 555 (e.g. LMC555) which draws a very low current.
Beyond that, you could use a high-frequency switching circuit that adds an inductor in series with the LED instead of a resistor to reduce the 9V to the LED voltage.

Are you really getting 6 hours of time from a 555 circuit?

 

I used this. I changed the cap to 1000uF and the resistor to 470,000? This gave me a time on of 8 hours after pressing the button.

Those numbers do not agree with the standard equations. 6 hours is 21,600 seconds. With a 1 M timing resistor, you would need around 20,000 uF. So how are you getting 6 hours of timing out of a 1,000 uF capacitor?

BTW, the timing equation for your circuit is: (delay time) t = 1.05 x R x C

A possible cause is that your large-value electrolytic capacitor is leaky. Basically its like having a resistor in parallel with the capacitor. The resistor provides a bypass path for some of the charging current, so the cap charges up much more slowly than a "perfect" capacitor. Convenient for you, but not something you should rely on.

Other culprits are the input current required by the 555 Threshold input and leakage current through the Discharge transistor. Both of these steal current away from the timing capacitor. They would be much lower with a CMOS part, such as the LMC555.

ak

 

LM3909 will flash LEDs with just a capacitor and the LED with very low drain on the battery. But it wants to operate from 1-5V.

LM3909s haven't been made in 20+ years and are well and thoroughly obsolete.

 

One of the reasons a 3909 is (was) so efficient is that it dumped the energy accumulated in the timing capacitor into the LED for illumination. A typical 555 circuit does not do this. However, it can with some rearranging of the components. Couple this approach with a CMOS 555 and you have a very efficient flasher. I thought about this for post #8, but decided to go with a very high impedance timing network instead. In that circuit, the timing capacitor peak current is only 6 uA.

A similar thing can be done with unijunction transistors, silicon bilateral switches, SCRs, and a common 2-transistor avalanche-style circuit. And don't forget the Dick Cappels (The Master) 1-transistor oscillator.

ak

 

When R1 discharges C1 down to around 0.5 V, the 555 is reset and the output goes low.

My calculation shows a discharge time of appx 47 minutes.

 

The problem is the 9 V battery dies in three days.

Just changing the chip to a LMC555 would decrease the current demand appx 43 times.
NE555 at 9 volts, appx supply current 5400 microamps
LMC555 at 9 volts, appx supply current 125 microamps.

 

The NE555 or LM555 were designed 53 years ago when all ICs drew a lot of current. The newer LMC555 or TLC555 use much less current.

There are 3 different 9V non-rechargeable batteries. The Super Heavy Duty (carbon-zinc) junk from thrift stores, the newer Name Brand Alkaline and the newest Energizer Advanced Lithium that produces power for a long time.

 

My calculation shows a discharge time of appx 47 minutes.

Same as I calculate, 48.167 minutes (2890 seconds)
Could change R1 to 7.5Meg

 

What's the slowest you would you would want the LED to flash to save power?
I have some thoughts on a circuit to do that for six hours.

 

My calculation shows a discharge time of appx 47 minutes.

Same as I calculate, 48.167 minutes (2890 seconds)

Interesting that your numbers are a little over 2% different. If you are talking about the schematic in post #8, I get a different number.

ak

 

I'm not the TS, but I would guess only long enough to see that its timing. Maybe a short flash every 15 or 30 seconds (just cause I wouldn't want to stare at it very long).

 

Interesting that your numbers are a little over 2% different. If you are talking about the schematic in post #8, I get a different number.

ak

Yes...i'm referring to post #8

Vc = Vs*e^(-t/(R*C))

t=time elapsed
Vs= supply
Vc= voltage across cap

Edit: I've confirmed the calc with LTspice. Matches almost exactly.