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

omnispace

Joined Jul 25, 2007
27
I'm not entirely sure about this, but usually when coils are being switched on and off, you need some kind of back-emf protection. I don't know if that's a problem for your specific setup, maybe the capacitors acting as low-pass filters take care of that. Can someone with more experience in this area tell us if this might be a problem?
 

bloguetronica

Joined Apr 27, 2007
1,423
Although the resistor R4 has an E24 value, keep in mind that may not be available in most stocks. Retailers prever to use E12 values when asking for 5% tolerance resitors (E12 was used for 10% tolerance resistors).
Also, your 9.1pF capacitor might be hard to find.

I normally use E12 for 5% tol. resistors, E48 for 1% tol. resistors, and E6 for 10% tol. capacitors. But then again I live in Portugal, also known as the "Bananas Republic", and those are the values that you mostly get here.
 

nomurphy

Joined Aug 8, 2005
567
The formula for an astable 555 timer is:

Fo = 1.49 / [Ct*(Ra + 2Rb)]

Which, with the values given, puts you at 7MHz or above (using 10pF, not 9.1pF):

Fo = 1.49 / 10pF (1K + 2*10K)

= 1.49 / (10pF * 21K)

= 1.49 / 0.21E-6

= 7.1MHz

Is this really what you intend? Please verify before you go any further with this.
 

Thread Starter

arthur92710

Joined Jun 25, 2007
307
right the coil produces back emf which can destroy components. a diode in parallel association to the coil will fix that.

The capacitor is not a 9.1pF its a 91pF sorry that my mistake.
 

omnispace

Joined Jul 25, 2007
27
The diode (facing backwards against the normal current flow) is a good place to start, and what is usually done. I don't think it's optimal though. Better to redirect that back-emf and store it up in one of your capacitors. But, that's another topic, for an advanced circuit.
 

nomurphy

Joined Aug 8, 2005
567
The 91pF still puts you at >700KHz, so what is your definition of "basically works." And, why not use 100pF which is a much more common value?

I think your pushing the limits of the 555, it may not work or will overheat at that rate, especially trying to drive the gate capacitance of your FET with a Ciss of 6600pF. I would suggest using an IRFR024 or IRF540, or something like that.

You should have an RC snubber across the Drain/Source of the FET, and an RCD clamp between the Drain and the supply -- these are not simple issues.

You really need to determine the inductance and series resistance of the inductor/transformer. If this is is not an OTS component with a datasheet, then you need to measure it. If you have access to an LCZ meter (perhaps at school), you should measure the component for the values of L and Ls.
 

Thread Starter

arthur92710

Joined Jun 25, 2007
307
Yes the 555 will (should output around 680kHz) work. It worked fine with a smaller setup.

What is Ciss and how is it 6.6nF?

I have an LCR meter I can mesuer L but whats Ls?

I have 3 coils i might use.
1. Short wrapped evenly around 7 Screws. .002mH
2. the green coil from the 3 pack at radio shack 1.2mH
3. a long coil wrapped around a big ceramic(i think) cup .25mH

I think the best coil is the one with 1.2mH but the short one also worked. so did the cup.
 

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nomurphy

Joined Aug 8, 2005
567
The reactance (Xl) of your coils at 700KHz is 8.8 ohms, 5.28K, and 1.1K respectively (Xl = 2piFL).

What is the secondary rating of the power supply transformer T1? You need to know this to determine the diode and other ratings, because this will be the voltage across that reactance and the resulting current the FET has to pass, and the wattage the coil needs to dissipate. However, these are basics and without knowledge of any secondary on the coil, loading, or leakage, etc.

"Ciss" is the input capacitance of the FET, you will find it on the datasheet(s). Also, place a 10 ohm 1/4W series resistor between the 555 and the FET gate (and near the gate pin).

"Ls" is the series resistance of the inductor, you can measure it with an ohm meter (DMM) as well.

U1 could simply be an LM7812 without the xstr and resistors, or an LM317 without the xstr but with adjusted resistor values. Regardless, I would also increase C3 to 10uF for stability, and suggest placing a 470 ohm resistor in parallel with C3 to help maintain regulation.
 

Thread Starter

arthur92710

Joined Jun 25, 2007
307
But what does Ciss do? do i have to input that capacitance to the gate or what?

Ok I added the extra resisters and increased the cap to 10uf.

I dont have any LM7812's so i put a LM7805. I will change it to 7812 to make it simpler later
 

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Ron H

Joined Apr 14, 2005
7,014
But what does Ciss do? do i have to input that capacitance to the gate or what?

Ok I added the extra resisters and increased the cap to 10uf.

I dont have any LM7812's so i put a LM7805. I will change it to 7812 to make it simpler later
R5 should be 10 ohms, not 10k ohms. You missed a connection. See below.
 

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nomurphy

Joined Aug 8, 2005
567
Well, you see, this is where engineering and math really come into play for understanding circuit performance.

i = C * dv/dt

So, you have 6600pF of input capacitance being driven by an ~12V signal from a 555 timer with a rise/fall time of ~200ns (see datasheet).

So let's plug those in:
i = 6600pF * 12V/200ns
i = 0.396A (396mA)

That's ~400mA at 12V (4.8 Watts) every 715ns (700KHz*2) from the 555 timer in order for it to overcome that input capacitance whenever the waveform rises or falls.

Let's try it with an IRF540:

i = 1960pF * 12V/200ns
i = 0.118A (118mA)

1.4W = 12V * 0.12A

Much better, but I suspect it's still a problem.

This heat is only generated during the rise and fall times of the waveform (sourcing and sinking), and will be averaged out by the package and the period of the waveform. Can you determine the waveform averaging?
 

thingmaker3

Joined May 16, 2005
5,084
Yes. All FETs have an input capacitance which is characteristic to the FET type. We charge that capacitance to turn the FET on and we discharge it to turn the FET off. Ciss for SUB85N02 is 6600pf. Ciss for E45NK80ZD is 2600. This information is found in the datasheets.

Your circuit will work better if you either add a "driver" to more effectively turn your FET on and off - or if you switch to a different FET as suggested by Nomurphy.
 

Thread Starter

arthur92710

Joined Jun 25, 2007
307
Ok thank you for explaining that. i will try to get that other fet is there any other thing i should change?

so a lower capacitance makes it work faster?
 

thingmaker3

Joined May 16, 2005
5,084
It makes it turn on or off faster. It also means less current is needed to turn it on or off. You are driving the FET directly from the 555, so the 555 must source and sink the drive current.

FETs are often referred to as being operated by voltage rather than by current, but wherever we have one we find the other.
 
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