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Help with a flasher circuit with 555 Timer IC

Bike Lights Hi+Low+Flasher.png

Hi Guys :)
The Purpose of this circuit :
1- Creating Low and Hi levels of lights with the three positions switch " SW1"
the 1st position will light up the upper group of LEDs " the two LEDs ", and 2nd position will light up all the exciting LEDs.

2- Creating a Flasher with the five LEDs together with " SW2 "

The problem is finding a way to connect leg 3 of the 555 IC with the five LEDs without affecting the first circuit.
 
My old brain is not too good so I had to find a book to remind me of what pin3 is. I deduce that it is the output.
The book also says that 4.5V is the minimum supply voltage for the 555.

The voltage required by the leds is not specified. It would be better to use a higher voltage and place a current control resistor is series with each led.

Diodes can be used to stop voltage going from one load to another but this involves another 0.6V drop.
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
I read your description and looked at your circuit. I don't get it?

Chris
 

Harald Kapp

Moderator
Moderator
There are a few issues with this circuit:

  1. LEDs in paralle are not a good idea. The one with the lower pass voltage will drain most of the current, the other one will be only dimly lit, if lit at alll.
  2. LEDs without current limiting series resistor are not a good idea. Without current limiting, the LEDs will draw too much current and will nurn. See our ressource section for more information on that issue and how to deal with it.
  3. You are likely to burn off the B-E junction of your transistor.
    Add a resistor in series to the base to limit the currrent.
    Apart from that, Q1 has no function at all. When sw1 is in the low position, it will drive a current through the be-junction of Q1 through LEDs D2, D3 and D4 (see above for that issue) and the LEDs wil be simply on, just as when sw1 is in the top position and LEDs D1 and D5 will be on. You can simply leave out the transistor, but don't forget the current limiting resistors, one per LED.
  4. When sw1 is in the on position, the 555 is supplied with power, but nothing apparent will happen as the output pin 3 is unconnected.

What is it that you want to achieve with this circuit? Your explanation is not all too clear to me. Perhaps a table showing switch positions and LED function could be useful.

Combining logic functions of two (or more) switches can be achieved by logic gates or by some simple diode-transistor logic. This way you can combine the flashing output of the 555 with a constant control signal from the switches to let the LEds either flash, be on or off.
 
Is this the sort of thing you want?
3ModeLEDs.gif
I've included a push-switch to prevent the battery being drained by the 555's standby current.
 
I'll revise the suggested circuit if you can identify the issues, Chris. (My interpretation of the OP's intentions may be wrong.)
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
Alec, when I viewed your schematic earlier I could have sworn the Trig pin was floating. I don't see that now. What's the Diodes for?

Chris

Edit: Another look and I see what D7, D8 (isolation) are doing but what's D8 used for?
 
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Hi Guys :)
The Purpose of this circuit :
1- Creating Low and Hi levels of lights with the three positions switch " SW1"
the 1st position will light up the upper group of LEDs " the two LEDs ", and 2nd position will light up all the exciting LEDs.

2- Creating a Flasher with the five LEDs together with " SW2 "

The problem is finding a way to connect leg 3 of the 555 IC with the five LEDs without affecting the first circuit.

Why dont you just use the 555 in unstable mood and then just use the switch to connect the pairs of led's to the output (pin 3) with one resistor on the cathode of each led group ? that would make your leds flash when pin 3 goes high or low if you reverse polarit and use the resistor on the anode.
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
Alaa, just a bit curious. .. Are you designing a rear bike light with brake light capability?

Chris
 
I see what D7, D8 (isolation) are doing but what's D8 used for?
D8 provides a voltage drop so that the LED brightness will be the same whether the LEDs are turned on via the manual switches or the 555's discharge transistor on pin 7.
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
Just my opinion but it's a needless waste of power. Power is going to be dissipated with it or without it. I'd rather dissipate it in the LEDs. I cycle an average of 11,000 miles a year. My wife has been side swiped, knocked off her bike and injured. I've (by the grace of God) escaped these lunatics more times than I can count. In short,.. if there's battery power to burn I'd rather burn it in my LEDs.

Chris
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
One other thought. I see you designed this with a 6V source. It would be more power efficient to series up some of those LEDs. Rear bike lights are Red LEDs.

Chris
 
The 6V source was chosen for proof of concept because we don't know the LED type or spec, and the 3.7V source in post # 1 doesn't leave much headroom for current control if white LEDs are required.
Agreed, if we're talking red bike lights (the OP hasn't specified use or LED colour/spec) and efficiency then D8 isn't needed and series LEDs would work. There are several ways to skin this particular cat.
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
Hi Alec, just so you know... I believe I'm a bit guilty of the "Peck The New Guy Syndrome". :D However, you took the pecking like a pro! ;) I didn't detect so much as a little rattle in your replies. :p

Welcome!
Chris
 
Hi Chris, Clearly you don't check you're fly zapper enough. Some smoke and others don't.
I prefer smoke!! Sometimes a banggg too.

Martin
 
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