Did you look at post #69? I'm not sure where the negative numbers you guys keep referring to are coming from, but no negative power supplies are needed.

A 19V power supply is bit of a high starting voltage for a 555 or CMOS 555. It can be done, but it is not necessarily the best way. Using a 555 started out as a learning too.

Fact is, there are lots of chips that can do this more or less off the shelf. You could also use comparators I suspect.

Quick side note about MOSFETs in general. They take a large surge of current (duration should be as short as possible) to switch, but after the surge almost no current (less than nanoamps) to maintain. Imagine the gate is a capacitor measuring several nanofarads between gate and source. The quicker they switch, the less heat generated during the switching duration. This is why you see beefy switching circuits feeding the gate, even though the power used is extremely small. You are trying to force the gate capacitance to change voltage as quick as possible.

I can show you a hypothetical SMPS drawing if you want. We now have voltage specs, what are the current specs?

 

The only '12' in the datasheet is ID = 12A, the maximum rated drain current.

The maximum Vgs before the gate ruptures is +/-20V.

So let's start again with this...
Your PMOS has:
Vs = 19V
Vd = about 19V when PMOS on, about 0V when PMOS off
Vg = your driver.

Vgs = Vg - Vs
This will be your driver minus 19V.

For the PMOS to be off, you want Vgs = 0.
This makes Vg = 19V.
For the PMOS to be fully on, you want Vgs= -10V.
This makes Vg = -10 + 19 = 9V.

So, restating that, your gate driver needs to output 19V and 9V as high/low.

Your 555 timer will output 14V and 0V. When it outputs 14V the PMOS will be turned on with Vgs = -5V. When it outputs 0V the PMOS will be turned on with Vgs = -19V. It can never turn off the PMOS and remember that you permanently damage the transistor with Vgs = +/-20V.

You want Vgs = -10 or somewhere around there. In the datasheet you can see "Static Drain-Source On Resistance", under test conditions it says "Vgs = -10V, ID = -3.5A"
That means they tested this value with those conditions, so to get that spec you need to have similar conditions.

Gate threshold is given as between -2V and -4V. This is given under the condition that Vds = Vgs, or equivalent Vd = Vg. You don't want to be here, since threshold means 'barely on'. It says the drain current was only -250 uA in this situation. You need much higher Vgs, which is why we keep saying -10V.

The circuit I posted earlier allows you to properly switch the FET on and off using a 0V and >3V output logic chip, which fits a 555 perfectly.

And yes, it's the PMOS that's inverting the duty cycle relationship. When your gate driver is 'high' the PMOS is off and vice versa.

@Bill:
The negative voltage keeps coming up because of referencing gate and source. The datasheets stick to using Vgs which is indeed negative for PMOS, while in his schematic he has the pulse voltage source connected with the plus to gate, which requires a negative pulse for a negative Vgs, or equivalently a positive Vsg.

 

OK. Are you planning on using a 6.2V or a 6.8V Zener diode on the negative power supply for the 555? This would fulfill the conditions for the drive voltages, 19V for turning the MOSFET off and 6 to 7V on the gate for turning the MOSFET on. When I refer to voltages I tend to stick with ground. It is less confusing.

Such a scheme would require adjusting the 555 input to compensate. Not hard, but needed.

There are some advanced techniques out there using charge pumps to drive nMOSFETs. I don't think the OP is ready for them, but they are out there. They are mostly used for H Bridges as far as I know.

 

hi

"For the PMOS to be off, you want Vgs = 0.
This makes Vg = 19V.
For the PMOS to be fully on, you want Vgs= -10V.
This makes Vg = -10 + 19 = 9V.

So, restating that, your gate driver needs to output 19V and 9V as high/low.

Your 555 timer will output 14V and 0V. When it outputs 14V the PMOS will be turned on with Vgs = -5V. When it outputs 0V the PMOS will be turned on with Vgs = -19V. It can never turn off the PMOS and remember that you permanently damage the transistor with Vgs = +/-20V."

so my 555timer must satisfy this conditions right,

Vg > 19V (OFF),
Vg < 9V (on)?

hmmm, and the circuit for your pmos how does it differ from mine... when u use a pulse of 5V it works? Mine doesnt tho

 

another thing is that how can i ammend the 555 circuit so i can produce that? and also one question about buck and mosfet..... from the simulations.... i realized that when its "working" my inductor shows a ripple like saw tooth... doesnt that shows that the mosfet is switching on and off? because if it is always on it will be a constant current?


to bill:
How does the zener help in controling this voltage to 19V and 7V? Cheers. cna u explain it in detail? and which way the anode and cathode shoiuld be facing? where shouldi place it on my model? isit the empty drawing space?

 

This will keep all the devices in spec. The 7555 will be powered by 12.8V, and the MOSFET will receive the voltages it needs to turn off and on. Just use pin 1 as equivalent to ground if you are trying to use any 555 circuits (same with op amps). The 6.2V point is a pseudo ground.

 

hi bill, on my previous post #85... my schematic... im not sure where to place to zener diode? can u show me the basic calculations from there how the zener will help them?

Ive tried simulating while placing the zener on the blank space however i get a spike and a DC voltage thats all...

 

Same place I did, between ground and pin 1 (which is ground the for the 555 chip).

Your schematic has a serious flaw though, the battery is feeding an open wire. It needs to be connected between Pin 8 buss (there are other pins, I'm using Pin 8 as a reference) and ground. You also need 19V, not 5V.

Something similar to this. As a convention, you put the beginning of the signal on the left, the end on the right.

 

right now i need some idea... since the PUlse of the output of 555timer is always reference to 0V..... i need it to reference to 9V... meaning on 19V off 9V... how can i achieve this ..

oppps just saw ur post bill gonna read what u replied.

 

wow.. ive just tried... then it works.. haha but im using a slightly different zener diode model with 10V....... do u think thats aceptable? and also whats the "logic" behind putting a zener there? can i say that i reference it to the voltage drop of zener instead of the 0V?

 

You need a bit less voltage on that zener, 9V or less. This is to provide the gate of the MOSFET with a full 10V range.

 

or so let me recap ya... less than 9V to on the MOSFET and more than 19V to off mosfet? am i right?

 

The 19V is a fixed voltage for the MOSFET. You set the number, but having set it it is locked in. You do not want to increase the voltage, just make it equal to the Drain (which is easy).

If you look at the other circuit on Post #69 you will notice the other type of MOSFET still has a fixed voltage, that of ground, and it is still the Drain.

Dropping the gate to 9V or less provides the 10V differential needs, so yes, you got that right.

 

alright thanks... i will try working on it and get back to u guys again... really relaly really appreciate that u guys take so much time and effort to help me.... thanks alot!

 

hi guys, so ive build my 555timer only on the breadboard, i found that the voltage max output will only reach 6V and didnt have time to try other stuff... do u think if i use an op amp to boost the voltage will it work? thanks..

 

hi guys, so ive build my 555timer only on the breadboard, i found that the voltage max output will only reach 6V and didnt have time to try other stuff... do u think if i use an op amp to boost the voltage will it work? thanks..

short answer? yes.

 

hi guys, so ive build my 555timer only on the breadboard, i found that the voltage max output will only reach 6V.

The output voltage of an ordinary 555 reaches as high as 1.3V less than its supply voltage. Then your supply voltage for your 555 is 7.3V.

 

yeah the Vcc for the 555timer even though ive changed it to 20V it still 6V.... so later im going to try to put an op amp to see whether i can amplify it to 16V... thanks

 

Hey guys, what are the important parameters for op-amp? I simulated a LM324.... it gave me some output of only 6mV or something? Whereas the theoretical values i used i should get a gain of 4... input voltage was 5V

 

Since the Vcc of your 555 is +20V then its output high should be +18.7V.
Its max allowed supply voltage is 16V so its "ground' should be +4V or higher.

Your LM324 opamp doesn't work because it does not have a power supply.
If you want the gain to be 4.3 and the input is 5V then its output should be 21.5V and its power supply should be +23.0V and ground. Its load should be 2k ohms or higher.