Drive an fet with a 555 and a different power source/not the same one that powers fet

bloguetronica

Joined Apr 27, 2007
1,541
Except for that missing connection, the project is quite good. Keep in mind that the LM555 doesn't need to be fed using a regulator, although the regulator will improve the 555's precision.
 

Thread Starter

arthur92710

Joined Jun 25, 2007
307
No, the 555 does need the regulator. If it does not use the regulator it will burn during the time when there is a low current in the coil. Or something like that, someone mentioned that earlier.

So as long as i change the fet this will work as is?
 

Ron H

Joined Apr 14, 2005
7,063
The gate voltage of the FET must be kept below Vgss(max), which is specified on the FET's datasheet. The 555's output swing is very near the value of VCC applied to it.
The 555 must be run with Vcc less than 18V. For some FETs, this is more than Vgss(max).
 

SgtWookie

Joined Jul 17, 2007
22,230
Ron H - look again!

1. His full-bridge rectifier isn't wired correctly - I had to look at the thing three times. :rolleyes:
2. Yes, the 555 does need a ground. I suggest that he use ground symbols instead of drawing a ground trace everywhere.
3. Right - but instead of swapping out the regulator, he could put a cap in series (to block DC levels) at the output of the 555, and then wire in a pot (say 10k-50k Ohms) across the FET with the wiper tied to the FET input, to establish a reliable trigger point.
4. Current-limiting resistor for the FET is a good idea. Just use Ohm's Law, assuming that the coil will show as a dead short, to limit the JFET current below it's rating. Put it between the JFET and the coil, or above the coil.
 

thingmaker3

Joined May 16, 2005
5,083
The IRF540 Vgs is 20 Volts, and the E45NK80ZD Vgs is a whopping 30 Volts. But Vgs for the SUB85N02 is only 12 Volts. The devil is in the details, and the details are in the datasheets.
 

Thread Starter

arthur92710

Joined Jun 25, 2007
307
aw i thought we were done with the circuit. :(

SgtWookie can you show how the rectifier should be wired? I had it wrong the ron fixed it but now you say that its wrong.

About the ground: is the "ground" the - or "negative" form the diode bridge Or is ground somewhere else?
if its the - from the diode bridge should i just put a ground symbol on every thing that should be negative(or ground)?

OH Now i see how the diode bridge has to be done.
 

JoeJester

Joined Apr 26, 2005
4,390
1. His full-bridge rectifier isn't wired correctly - I had to look at the thing three times.
I fail to see a problem with the diode bridge.

Overall, as long as he doesn't draw appreciable current, the supply should work. Trying to supply an amp will cause a few problems.
 

JoeJester

Joined Apr 26, 2005
4,390
Now your drawing is incorrect.

In the first example, you shown in the previous posting, the bridge is rotated 90 degrees counter-clockwise from your original drawing. Pay attention to the details of the anode and cathodes.

The same applies to your second drawing illustrated in that same posting.

Look at the attached file.
 

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Ron H

Joined Apr 14, 2005
7,063
The pot in series with the FET gate won't work. There is no path to GND to establish a DC level on the gate. Even if you connected it as Sgtwookie intended (with the other end of the pot grounded), it won't work. As others have said, the gate drive impedance needs to be low. It won't be low with a big pot in series. Also, the AC coupling is a bad idea. The gate does not need to swing below GND to turn it off. I would go back to what you had, with the series gate resistor of 10 ohms and no capacitor. You don't need a pot as long as you keep the regulator, and make sure you use a FET that can handle 12 volts of gate drive.
 

nomurphy

Joined Aug 8, 2005
567
Arthur,

You are correct that the diodes in a bridge all point the same direction.

Where the cathodes join is the (+), where the anodes join is the (-), and the other two points are for the AC. It doesn't matter how the bridge is rotated in the schematic, as long as you make these proper connections.

But, you still have a few R1 resistors.
 

bloguetronica

Joined Apr 27, 2007
1,541
No, the 555 does need the regulator. If it does not use the regulator it will burn during the time when there is a low current in the coil. Or something like that, someone mentioned that earlier.

So as long as i change the fet this will work as is?
It was I who mentioned that earlier, but that case does not apply when you have a 12V DC power supply already. I was refering to the case where no transformer was used, or a 60VAC transformer was used instead.
 

nomurphy

Joined Aug 8, 2005
567
I suggest you make the following changes (see attached):

U1 --> LM317 Adjustable
add R6 = 124 ohm
add R7 = 909 ohm (or 866 ohm)
This will give you ~10V and reduces the supply voltage for less drive current required from the 555 (by using an LM317 you can adjust this voltage if needed by changing R7).

Add R3 ~= 10 ohm, 1W
change C2 = 220uF
add C7 as a ~0.1uf - 0.22uF near U1 input for noise
C1/R3/C2 creates a pi filter that will help with noise from coil switching and helps reduce power dissipation from the regulator. The regulator will need to be a TO-220 with heatsink.

Delete the 470 ohm, this is now done by R6/R7.
Note: Vout / (R6+R7) ~= 10mA

And, you don't need that humongous 1N1202C diode, a B2100 or B290 or similar should work okay (also see attached).
 

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Thread Starter

arthur92710

Joined Jun 25, 2007
307
lol, That was from like june! man iv been doing this for so long :(


You are correct that the diodes in a bridge all point the same direction.

Where the cathodes join is the (+), where the anodes join is the (-), and the other two points are for the AC.
about the diode bridge the diodes do not all point in the same direction. the bottem is like a 'V' and the top is like a '^'

If the catodes join they do not point in the same direction
So I was wrong.
 

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