Need help with using transistors as a switch with 555 timer

Status
Not open for further replies.
A

amzjl

Member level 1
Joined
Apr 9, 2014
Messages
34
Helped
1
Reputation
2
Reaction score
1
Trophy points
8
Activity points
247
I am trying to create a circuit that will delay coming on and then turn on 6 LED's. As far as I can tell, my diagram is correct (please correct me if something is wrong.), but I do not know the values of the resistors connected to the npn and pnp transistors I am using or how to calculate them.
 

Attachments

  • unnamed.jpg
    27.5 KB · Views: 358

You saved and posted your schematic and text on it too small, with no contrast and as a fuzzy JPG file type instead of larger and as a very clear high contrast PNG file type. Therefore I cannot read most of your text.

What is the current in each LED?
What is the battery part number?

You wired the circuit so the IC and transistors did not connect to the negative terminal of the battery so I fixed it.
You used an NPN transistor (that is not needed) as an emitter-follower so the PNP transistor would NEVER turn off. I eliminated the NPN transistor because the datasheet for an ICM7555 (the same as a TLC555 or LMC555) shows that it has plenty of output low current to drive the PNP transistor.

I cannot read your calculations so I do not know the resistor values.
 

Attachments

  • Delayed LEDs.png
    79 KB · Views: 268
Reactions: Eshal

    E

    Eshal

    Points: 2
    Helpful Answer Positive Rating
Sorry about the image. I was drawn on graph paper and scanned into my computer. The bank of LED's (group of 6) has a voltage of 3.4 and the one by itself is 2.6. The batteries are 2 CR123 ran in series (6V). To restate what I am trying to create, when the unit is turned on, it delays it for a certain time BEFORE sending power to the lights. I do not want it to just come on for certain time and go off, I need it to delay the set amount of time, then come on. I was using the npn to do that. I was told it would work. The resistor value going into the 3.4V LED's is 140 ohms. I want to delay it for 3 seconds. The chip I want to use is a TLP555IP made by Texas Instruments. The calculations at the bottom are where I worked out the timing portion and the resistor values I needed.
 
The Cmos 555 output pin 3 goes high when it is timing then goes low at the end of timing. Then my PNP is turned off during timing and turns on at the end of timing which is what you want. But the LEDs light the moment the power is applied until the button is pressed.

140 ohms is not a standard resistor value but when the battery is new then each white 3.4V LED will have a current of (6V - 3.4V)/140 ohms= 18.6mA and the total for six is 111mA plus current for the red LED which is fine for your battery. The base current for the transistor needs to be 11mA so the value of R7 will be (6V - 0.7V - 0.5V)/11mA= 437 ohms which is not a standard value, use 430 ohms. The 0.7V is Vbe of the transistor and the 0.5V is the saturation voltage drop of the Cmos 555 when it goes low.
 
Reactions: SR19

    S

    SR19

    Points: 2
    Helpful Answer Positive Rating
Do you mean the LED's come on as soon as batteries are inserted or when the switch is depressed? Does it then go off for the timing interval and then comes back on? I need it off all the time until the the switch is closed to start timing. Only then can it come on. Is that resistor value you gave good for all them around the transistors? I think there are 3. Is there a better way to create what I'm wanting? I only want it as small as possible and would rather not use something I have to program.
 

1) When the circuit is powered then the output of the Cmos 555 is low which turns on the PNP transistor which lights the LEDs.
2) When the button is pressed then the timer starts timing and its output goes high which turns off the PNP transistor which turns off the LEDs.
3) At the end of the timer period the output of the Cmos 555 goes low which turns on the PNP transistor which lights the LEDs.

That is how a 555 works. It is made to make something turn on for a timer period.
 

How is it instantly powered on when the circuit is open due to the switch no being pressed? The capacitor nearest pin 2 is supposed to cause the trigger to go low which in turn makes pin 3 go high. Since it is hooked up to the NPN, that transistor allow power to flow to the base of the PNP, keeping it off for the timing interval. Once pin 3 goes low, the NPN is clamped off so power can flow through the PNP. I have been told this circuit works as far as the 555 and transistors are concerned. This wasn't even my circuit. I got it from the guy that said it worked, I just adapted it with the values I needed for my application. What am I missing? If this doesn't work, is there another way?
 


Can anyone confirm what I posted in my last entry? If anyone has a better idea I'm all ears.
 

Get rid of the NPN transistor because it prevents the PNP and LED from turning off.
Since the Cmos 555 is triggered when pin 2 is low then maybe it is triggered when power is applied because then the capacitor is discharged. Maybe sometimes it will begin timing and other times it will not. Try it to see.
 

I don't want it to turn off. I want to push a button, then for 3 seconds nothing happens. Once the time interval is over, the lights come on and stay on until I turn it off and then back on. I thought 555 set up for monostable would do that.
 

amzjl said:
I don't want it to turn off. I want to push a button, then for 3 seconds nothing happens. Once the time interval is over, the lights come on and stay on until I turn it off and then back on. I thought 555 set up for monostable would do that.
Click to expand...
No. For 3 seconds after you push the button you want the PNP transistor and the LED to be turned off. your NPN transistor prevented the PNP and the LED from turning off.

Normally a 555 timer uses a logic low signal on its trigger pin 2 which is used to turn on something for a time period then after it times-out the thing is turned off.
 

I am at a total loss then. I've been looking at this stuff for the better part of 3 months, been on multiple forums, talked with people in person and this is the first place that has said the part I was sure is now wrong. OK, then how do I do it?
 

I like that it is simpler than using a 555. Can you walk me through how it does what it does? I understand the parts but since most of my stuff has been wrong, I don't want to assume. Do I need the zener since that is used primarily for protection from the relay? Is the transistor in the center missing a connection or is it just not used? Is there any specific power requirements in addition to what the LED's need? Where is the best place to put a switch? I know how to adjust the timing. Will this simply delay it from coming on for a set time when the switch is pressed and then stay on until the switch is pressed again?
 

The 1N400x diode across the relay is not needed - it is there to short out any back EMF from the relay.

The transistor in the center that seems to be missing a connection - is correct. Connected in this way makes the transistor perform like a zener diode.
So if you think of it as a zener, the operation is quite simple. When it is powered up, the capacitor is slowly charged through the 47K resistor - the voltage rises - when the voltage exceeds the "zener" voltage of the transistor it turns on the Darlington arranged transistors, and so the load is powered.

The switch would be a SPST and can go between the 47K resistor and the supply voltage.
 

Would I use the same part numbers listed in the first link? Are those the transistors I need to use? So all I would need to do is replace the relay portion with my LED bank and add a switch? One other thing I don't understand is what happens with the timing portion once it activates whatever it powers. How does it "know" to not start over or repeat? In the first link, what is the 10K resistor on the left used for?
 

The transistors are not critical - use similar transistors.

Yes, just add a SPST switch , and your LEDs replace the relay and 1N400x diode.

Once the capacitor is charged there is nothing to "un-charge" it. It will remain charged [and so the load is activated] until the voltage is removed [ie switch is flipped].

The 10K resistor helps remove the charge on the capacitor quickly once the power is removed.
 

If I am using a 6 volt power supply and set the resistor and capacitor for the time delay I need, I do I find out which transistor to use as the zener? I'm assuming I would have to pick one with the correct reverse biased load or it won't let the Darlington pair turn on. I guess I'm confused how it knows when to power to load. A 555 works off of the 1/3 and 2/3 voltage levels but there is no chip here to tell it when that happens. Is it the backwards transistor that does it? What calculation do I need to o in order to pick the right values for the resistor, capacitor, and transistor to create a 3 second delay?
 

It won't work with 6VDC as the transistor acting like a zener has a break-down voltage of between 6 and 9V. You can try it without that transistor, by using a much larger cap, and a slightly larger value for the resistor.
 

So if I just give it more voltage it will work? I know how to set the time but how does that effect the transistor at the right time?
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…