# What wattage resistors for 555 timer circuit?

Discussion in 'The Projects Forum' started by smokemachine, Apr 15, 2016.

1. ### smokemachine Thread Starter New Member

Apr 15, 2016
6
0
Hi,
I am somewhat inexperienced when it comes to electronics but I am wanting to build a fuel injection tester using a 555 timer chip. See link here..
http://hackaweek.com/hacks/?p=1156

Fuel injectors draw approx 1 amp and operate on 12vdc. The link is fairly detailed in terms of parts needed but I am wanting to know what size (wattage) resistors I need for the circuit and how do I work it out?

thanks

2. ### TheButtonThief Active Member

Feb 26, 2011
219
38
Power (in Watts) is Amps x Volts
P=A.V

...so your power consumption is 12W.

For the resistor, I'd add 20% to that so you're looking for a 14W or 15W resistance.

3. ### jjw Member

Dec 24, 2013
173
31
The resistors in series with the leds are too small ( depending on the leds used ), 470 ohm would be ok for 20mA leds.
All resistors can be 0.25W

4. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
That's the consumption for the whole circuit when it is on - the bulk of which goes to the injector.

Doesn't look like you need much power dissipation. i would use 1/4w resistors or whatever you have.

5. ### smokemachine Thread Starter New Member

Apr 15, 2016
6
0
Thanks, the ones he uses appear to be the standard 1/4w style. Anyone care to explain why 1/4w is suitable it in terms of a formula though so then I can understand it?

555 timers have a max output current of 200ma so P=VA, P=0.2x12 = 2.4w ?

Last edited: Apr 15, 2016
6. ### Dodgydave AAC Fanatic!

Jun 22, 2012
5,136
767
Wattage = Voltage x Current

Or.... Voltage squared/ Resistance

Or... Current squared x Resistance

So if you're using a 12v supply, i would use 470ohms for the led resistors R1,R9.

Last edited: Apr 15, 2016
7. ### TheButtonThief Active Member

Feb 26, 2011
219
38
The maximum current it can deliver is 200mA, though you're not drawing as much as that with something like an LED, the LED will consume only as much as it needs (or more accurately, as much as you limit it to by means of a resistor)

8. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
That's only the max.

Here, the 555 is driving a mosfet (=capacitor here) and not much is delivered in a sustained fashion.

The two largest power dissipating resistors will be the resistors in serial with the leds, particularly the always on one. Its power dissipation will be just shy of 12*12/100 = 1w, if the device is powered on for a long period of time. I would actually size up that resistor considerably, especially if you use a high brightness led.

9. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
The whole thing is a little bit convoluted. All you need is a pulse generator (the 2nd 555) and a button that powers it. Hold the button down for as long as you want, or 1 second or 5 second, ...

No point in having the first 555 in my view.

10. ### GopherT AAC Fanatic!

Nov 23, 2012
6,300
4,020
Look at the 555 on the right. The output pin is pin 3. It goes to a IRF510. This is essentially a relay. A voltage applied to the gate (g) pin 1, will connect the D and S pins (pins 2 and 3).

The main current flow for the injector is from the 12 V battery (+) to the injector, from injector to the D pin of the IRF510 and then to the S pin and down to ground (12 volt battery (-) pin).

Note that essentially no current flows into the gate of the IRF510 after the first few microseconds.

1/4 W was recommended for everything else because that is what most people keep in stock nd they fit perfectly into prototyping breadboards.

Last edited: Apr 15, 2016
11. ### smokemachine Thread Starter New Member

Apr 15, 2016
6
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Will 35vdc capacitors be ok or should I use a lower voltage like 25vdc?

12. ### Dodgydave AAC Fanatic!

Jun 22, 2012
5,136
767
Yes any voltage higher will do, 16v will be fine for 12v circuit.

13. ### DickCappels Moderator

Aug 21, 2008
2,752
665
The 100 ohm resistors in series with the LEDs will have to be 2 watt. It might well be that you don't need 100 ma or even want 100 ma through the LEDs. Consider using a higher series resistance.

Figure each LED will drop about 2 volts. That leaves 10 volts across the resistor.

Current = 10 volts/resistance
Power = 10V x current

14. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
Those leds are specified to be 12v / automotive leds.

DickCappels likes this.
15. ### DickCappels Moderator

Aug 21, 2008
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Thank you dannyf, you are very good at scoping out the details.