I tried to make a longer title, but I kept getting the blank screen while trying to post my first post

A more appropriate title would be "Help with 555 idle current"

Can a moderator change?

Hello,
I need help to figure out how to keep the idle current draw at a minimum for this circuit. I built the attached 555 Schmitt trigger circuit on a breadboard.


http://www.allaboutcircuits.com/vol_6/chpt_8/2.html

However I substituted 2 relays for the 2 leds and then attached a motor so when the 555 sinks or sources the current it will energize the coil(s) and turn the motor. I also added a photo resistor next to the potentiometer to create a voltage divider going into the 555 so when it is dark the motor will turn one way and when it is light the motor will turn the other way.

In practice the motor will hit a limit switch to de-energize the coils - for now I just do that manually on the breadboard.


Ok so here is my question....

How can I have very little or no idle current draw when in "standby" mode?

The idle current draw of the circuit is around 2ma when using a semi depleted 9 volt battery. I plan on using 4 alkaline batteries 6V. So because of the current draw the batteries won't last long enough because of the 555 chip. I also tried the CMOS version, but it is about the same.

How do I get around this? Can I use another IC somehow, but it has to be able to energize the relay coils - I can't remember but probably aroundish 100ma.


Many thanks for any help you can provide,
Matt

 

You can use the cmos version, but you will need to add a transistor to the output to drive your relay. Then make the pot 100k instead of 10k.

 

Thanks ronv.

I tried the cmos version and no difference. It had enough current to energize the coils of the relays.

I also physically removed the 2 relays and the current was still around 2ma.

 

What about changing power supply like a lipo battery or walwart? ???

 

I thought about that but really want to stay with consumer batteries without any A/C.

I also don't want the potential, but unlikely, fire hazard with the lipos. I use them all the time in R/C planes, but don't want them in this project.

 

You could design a sleep mode controller out of CMOS amplifiers. First, you have to figure out what condition you want to wake up the sleep circuit. I don't think you have stated that condition here, yet.

 

A CMOS 555 will draw very little current, in any condition. This suggests your output, such as the LEDs. You aren't still powering up LEDs with it, are you?

BTW, I wrote the article in question.

 

The condition would be whether it was day or night - changing the resistance of the photo resistor. I guess either day or night could wake up the circuit.

Are you talking about op amps? I would like the output to swing positive and negative to ultimately drive a single dpdt latching relay instead of two separate relays. I think that would be the simplest.

 

Thanks Bill for writing it - something I can more or less understand!

No, the CMOS still draws almost the exact same current as the regular 555. I substituted relays for the LEDs. Limit switches would cut current off the coils so no current would be used there.

Also in trying to figure out the 2ma current drain in the beginning (I didn't know it was the 555) I physically removed both relays from the breadboard and the 2ma still persisted.

 

The max supply current for the LMC555 with a 5V supply is specced as 250uA and for a 12V supply is 400uA. So something's wrong if yours draws 2mA.
To reduce circuit current the input pot resistance could be much higher, say 100k or more.
A simple very-low-current Schmitt is a CD40106 (or CD4093).

 

I bought it at Radio Shack and just switched out the 555 with the low power one and was disappointed that 2ma was still there. I will double check again.

The circuit is identical to the one here by Bill, but added a photo resistor to the input and replaced the LED's with relays.

 

Without a schematic, it's impossible to say for sure, but your voltage divider may be where the current is going.

 

Here's a good question, is it stopped from oscillating and still using 2 ma?

I guess that's what tracecom just said.

 

Here's an alternative circuit for a 6V supply, with an idle current of ~50uA, for switching a bistable 3V single-coil latching relay.

 

tracecom and #12
I'm not sure of the answer to your question. I attached the schematic.

This circuit is driving me crazy. After turning "on" I disconnect from the coil and the circuit draws 1.85ma.

I remove one relay and the circuit still draws 1.85ma
I also remove the other relay and it still draws 1.85ma
I then remove the 555 and still 1.85ma
I then remove the resistors and it is now 0ma

But if I connect it all back together and remove the resistors - still the 1.85ma
Then remove the 555 and 0ma

I guess I just don't understand the circuit enough.

 

Wow. Thanks Alec_t!

I'll order the components and give that one a try.

What does the 100m mean on the relay?
Since it is 6v, couldn't I use a 6v or 4.5v or 3v relay if I wanted?

What is the function of the capacitors on the relay?

 

100 milliwatts would fit the image.

If you use a lower voltage relay, it will need more current to operate. You know, same amount of power at half the voltage requires twice as much current.
The capacitors allow current to flow for just a little while, long enough to set the latch of the relay. Then they stop the current from being wasted by flowing long after you don't need it any more.

Brilliant use of a cmos Schmidt trigger to minimize parts count.

 

Thanks #12

I think I understand about the capacitors - after the relay is energized no more current flows because of the capacitor (reached its full capacitance), but yet when current is reversed the relay is switched in the other position and then no current until the next switching.

So the coil is rated at 100 milliwatts and 3 volts, but he is supplying 6 volts and thus the coil only needs ~50 milliwatts to energize - the same amount of power.

I didn't realize it could actually work that way. I just assumed the relay would draw and need 100 milliwatts no matter what the voltage.

So in theory I could supply 6v and still use a 6v 100mw relay, but of course that would use twice the power and the capacitors would have to be adjusted accordingly?

 

You got lost on, "milliwatts". A milliwatt is a power rating. The power (milliwatts) required to latch the relay remains the same, no matter what voltage you use (as long as you have ENOUGH voltage). Only the voltage and current change. Voltage more = milliAMPS less (and the other way around).

That circuit in post #14 slaps 100 milliwatts through the relay coil for a tenth of a second and the mechanical latch takes over from there.

The math on this one is in milliwatt-seconds. Watts = volts times amps. milliwatt-seconds = volts times milliamps times seconds. In theory, 6 volts applied to a 6 volt relay wouldn't slap it hard enough to get the dose of milliwatts into it in a tenth of a second. If it did work, you would need half as much capacitance for double the voltage because twice the voltage only has to supply half the current to get the same milliwatt-seconds done in the same amount of time.

It's one of those brilliant moments when the obvious method goes right out the window and everything converts to the raw physics of the beast.

 

Thanks. Yes I know power in V and Amps. I must have had 100ma on the brain for the coil instead of mw even though I wrote mw. Digikey only lists coil current on their latching relays.

OK so
watts is hours
and milliwatts is seconds and 100 milliseconds is one tenth of a second

100mw = 3V times 33ma

But this circuit drives the coil at 45ma and that is ok because it is only 1/10 of a second?

This dpdt relay is 4.5v and 22.2ma so it should work perfectly in this same circuit?
http://www.digikey.com/product-detail/en/IM42DGR/PB1176TR-ND/1828448

Did I get that all correct? I'm not used to thinking in milliseconds.