What wattage resistors for 555 timer circuit?

Thread Starter

smokemachine

Joined Apr 15, 2016
6
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
 

TheButtonThief

Joined Feb 26, 2011
237
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.
 

jjw

Joined Dec 24, 2013
540
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
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
 

Thread Starter

smokemachine

Joined Apr 15, 2016
6
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:

Dodgydave

Joined Jun 22, 2012
9,196
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:

dannyf

Joined Sep 13, 2015
2,197
555 timers have a max output current of 200ma so P=VA, P=0.2x12 = 2.4w ?
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.
 

dannyf

Joined Sep 13, 2015
2,197
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.
 

GopherT

Joined Nov 23, 2012
8,012
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:

DickCappels

Joined Aug 21, 2008
6,451
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
 
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