I want to know if i can use this circuit to Drive an fet with a 555 and use a different power source not the same one that powers the fet.

 

The answer to your question is yes, but your circuit looks flaky.
What is the resistance of the coil? If it is low, the voltage at"60" will go to ~0V when the FET turns on. This will remove the supply voltage from your 555.
What do you hope to accomplish with the resistive voltage divider?
You will have pulsating DC at the output of the bridge rectifier unless you add a filter capacitor. If you do, you will have about 168V DC at the bridge output.
The LED will never turn on, because MOSFET gates have essentially infinite resistance.

What are you trying to do? I think you need help.

 

It will be a transformer with a air coil and the secondary will be a coil that can be moved. Like in a rechargeable tooth brush.

The resistive voltage divider divides the voltage to what i need to power the fet and the 555
12v for the 555 and 60v for the fet and coil.

Ok i will add a filter cap.
change the resisters for the 168v DC that the diodes will make. (How did you come up with 168?)
I will remove the led

How would you change it?

 

The resistors are not the only impedances the load will see. Your coil is also part of the load and must be accounted for.

Also note that 1.414 x 120 is not 120.

Also note that you have no filter.

If you can provide additional details (what is the load, what is the circuit supposed to do) we can provide better help.

 

Just did while you was typing

what is the 1.414? and duh 1.4 x 120 is not 120?

1.414 is the diode bridge.

 

Peak = 1.414 x RMS. If RMS =120 then peak = about 170 Volts. But, again, you have no filter!

 

I just put the filter in to the pic (need to reupload) but im working on the resister values.

But what fill the filter do? I know it will pass AC but block DC, but what will it really do?
How big should the cap be?

 

The filter will drastically reduce the ripple voltage.

I'm posting from my sister-in-law's, and cannot give a good capacitor formula right now. (One of these days I need to scan ALL of my books into my PDA.) If you can wait until Monday, I can proved the formula. IIRC, it is most influenced by tolerable percentage ripple, pulse frequency, and by output current.

 

how long will you be on? if you will be on for around 5 more mins I can reupload a new circuit.

 

Just enough time to go to the 'fridge.

 

Ok here it is.

I expect the led to die because of the high voltage that will go trough it.

On a smaller version with a 9v (7.6 actually) i can get about .6v across the led. this will be 6 times stronger.

ALSO!! will this have a shock hazard with the 170v dc? I got hit with 120 ac in the morning while fixing my power supply.
(i was working with one hand so im not dead(second time). But I also got hit with 120 on both arms(first time.(one week ago) now i only work with one hand with ac!). It was really weird. I thought someone came up behind me and grabbed my arms extreamly!! tight.)

 

ok i g2g sleep cant wait for a reply!! ill have this page on my psp so i can check for replays.

 

I'd go with a single capacitor at the bridge output.

Also, you still need to deal with the impedance of your transformer primary.

The junction of R10 and R2 won't be 60v unless the impedance of the transformer primary is on the order of several megohms. That xformer primary is in parallel with R2 through R6 and R11. The rules for impedances in parallel will apply. http://www.allaboutcircuits.com/vol_1/chpt_7/index.html

If your current draw will be half an Amp or less, may I suggest an LM5007 or equivalent?

 

Sorry to have missed you. The cherry pie in the fridge was just too tempting.

 

Are you planning to do this without a power transformer? If so, do you know how dangerous this is? You should be using a power transformer to get your voltage down close to the highest DC voltage you want. For example, if you want 60VDC, use a 45V transformer (or whatever you can find that is close). Without a transformer, if you touch the hot side of the line, and your other hand is touching something that is grounded to earth (soldering iron, oscilloscope probe ground or frame, etc.), you may wake up dead.

 

Using resistors in voltage divider configuration is not a good idea when designing power circuits (refering to the part of your circuit that converts power. Why not using a 12V relay?

Also, like said in the previous posts, using transformers would be a good idea. Don't forget that your load current varies (it is a 555, connected to a coil right?), and then the coil is not energized (time of less current biasing the resistors and circuit), the 555 may fry with overvoltage. Just the fact of having a transformer will much improve the regulation of your feeding circuit.

Also expect harmonics (at least, if not spikes and some degree of hazard) caused by that capacitor connected directly to the supply. I'm talking by my experience since I have an hair dryer that was assembled from factory with a cap directly connected in paralel with the mains. The story is that if I don't have the care of unplugging it before swiching it off (in order to discharge the cap) and if I have the misfortune of touching the plug, I will get a severe shock. Expect your circuit to behave the same way (the shock hazard could be avoided with a SPDT switch before the cap).

 

OK ill add a transformer.
120p/60s
then the 4 diodes?
Ill take the cap off of the primary. Ill leave one on the secondary.

But what should i do with the voltage divider? Can i leave it? i should change it. but with what?

 

There is really no point to having a cap on the secondary. The only place a cap will do any good is on the bridge output.

The problem with your voltage divider network is that it is in parallel with your load and you have still not accounted for this! Voltage regulators would serve you far, far better!

 

The problem with your voltage divider network is that it is in parallel with your load and you have still not accounted for this!

Why is that a problem?

 

Indeed, on the brigde output to filter current. Electrolytic capacitor are excellent for that. Why don't you use a 12V transformer instead? You just need to use a 12V coil/relay, which is quite common.

The formula to calculate the capacitor's capacitance:

C = Il / (2 x Vdc x ripple_factor x Fm)

Il - Load current;
Vdc - DC voltage;
ripple_factor - Factor of ripple you want for your project (0.1 or 10% is enought for amplifiers, 0.05 or 5% should be considered for your project);
Fm - Mains frequency (50Hz in Europe, 60Hz in the US).

Why is that a problem?

You will certainly overvoltage the 555 during the low current intervals. Better to use a transformer and totally exclude the voltage divider.