I think the constant drone about using microcontrollers is crippling to kids coming up in the hobby or industry. There is a time and place for microcontrollers and a time and place for discretes.

Thanks RK. Much appreciated and welcome aboard.

60V nice!

Are you using the recommended MOSFET? A MOSFET is nearly zero resistance when fully off and many Megohms resistance when fully on. It is a near perfect switch. But when not fully on or off it's resistance is anywhere from tens of ohms to Kilohms. This is almost always what heats or overheats FETs and MOSFETs. You must be using a MOSFET that doesn't have triggering as low as the recommended one. Or try a lower value pull down resistor. Another possibility is that the MOSFET has a large gate capacitance and the 555 can not hit it with enough current to turn the gate on and off quickly enough. This leaves it in heating element mode to long each cycle. Good luck. This stuff is hella fun when it works.

When the fet closes the input voltage is essentially shorted to ground. The fet will have max current going through it, or as much as the Vin can supply. You could put a low value resistor in series with the fet to limit the current to whatever the fet is rated for

Exactly my question too!

There ya go! I want to hear about the results

Do you know an alternative for a nce85h21tc mosfet?

Exactly

+TheJavaSync thanks

Aluminum electrolytic caps will tolerate some over-voltage for a short time, but you should never design for continuous over-voltage. You could use a 35 or 50 volt cap for 29 volts.

Which pin are you talking about? Pin 3? That's going to gate of the MOSFET. Maybe I need to rewatch the video, but it looked fine to me. Edit: OK, I'm looking at 3:45, and now I see what you're talking about. There's a wire there though, not a resistor, and that's why I was confused. My best guess is that it's a leftover from when Paul was testing the timer, needed to sink the signal on the OUTPUT pin, but forgot to remove it before tying it to the gate of the MOSFET.

You are correct, the 10k resistor is superfluous since the 555 will pull the output both high and low.

Resistor is there to pull the gate to 0v. Mosfet will still be on even after removing gate voltage,so you need to tie gate to ground through resistor to turn it off. In your case,without the resistor connecting gate to the ground,the MOSFET is always on and no pusle is generated.

@@naingtheawperc3817 Yes, but in this case the 555 timer square wave will pull it both high and low, so really it is not necessary.

In years gone by switchers were made with bipolar junction transistors. However, very fast switching is important for good efficiency and though you have to swing quite a bit of gate current into and out of a power MOSFET gate to get it to switch if fast, they do switch very fast and can handle some things that kill BJTs. BJTs often need quite elaborate "snubber" circuits to protect them (a boost converter is more or lesss "self snubbing"). The MJ1300x types were fairly popular for switchers operating at moderate frequency (20 to 40 kHz, typically).

Use a voltage Divider at output followed by an Opamp( As a Comparator) which is connected to an Microcontroller or Any IC . The Microcontroller will increase or Decrease the PWM signal Duty cycle Based on the Requirements of load for a stable Output.

dwpetty3723 if you want to change the output voltage you would need to change either the inductor value or the switching frequency. the resistor at the end of the circuit just stabilizes the output. thanks for the compliment.

what is the value of the resistor?

use a high voltage capacitor

@@big_o1952 ok

It is a fundamental property of an inductor that you cannot instantaneously change the current through it. When the switch is closed you store energy in the inductor as a magnetic field. If the current in the inductor were zero at the time the switch closed the current would start to rise linearly with respect to time. The magnitude of current if would reach is directly proportional to the time and to the voltage. Let's say that it reached 1 ampere. At the instant the switch opens, that current will still be 1 ampere. The voltage across the inductor will essentially instantaneously rise to whatever it needs to be to make that current flow. It doesn't matter if the voltage across the inductor at that instant can only be 10 millivolts or 200 volts, it will make 1 ampere flow. That's how you get the high output voltage. A switchmode power supply is a _power_ converter. If you want 10 watts out, you must put 10 watts in, assuming 100% efficiency. With a boost converter the output voltage is higher than the input voltage. If you had a 10 volt input supply you'd take 1 ampere from it for the 10 watts. You could take the output at 100 volts and 0.1 amps or at 20 volts at 0.5 amps. How you actually do this in a real circuit is by controlling how much time the inductor is charging versus how much time it is discharging. Usually this is done by controlling the duty cycle with fixed frequency. In a boost converter the discharge is much faster than the charge because the voltage across the inductor is larger. di/dt = V/L the rate of change of current in an inductor is equal to the voltage across the inductor divided by the inductance i is in amperes, t is in seconds, V is in volts and L is in henries

Right on!

I remember good book i think its . Power electronics Hart

I get you. You can build this circuit without the 555.. And without the mofet if you can push a switch a few hundred times a second. The basics are that a switch (MOSFET) is fast and this turns on and off, to get the build up in the coil. There are many ways to make the switch, a 555 is easy and has two ((asynchronous switching)), which isn't needed, there are MOSFET drivers, bjt's And if you don't have a IC you can make a charge switch using resistors and capacitors to the gate on the FET. That works by building a charge in a capacitor which has to pass a resistor when it builds and then reaches the gate and is discharged, this charge and discharge will occur in ms, and youll will have to add this to drive the switch which will change the voltage. Make one resistor variable and you can change the voltage. You also need to add a fly back which would be the other half.

Thank you very much.i understand.

VibeX MaTxeMa you can (it's what i am doing) just be carefull to not fry your arduino :)

Optoisolator, and yes, you can also look up a pot to the analog input and feedback to another analog input through a voltage divider if you know you're not going to exceed the ratings of your Arduino.

Optoisolator isn't really necessary

No. A boost converter is simply a switch mode power supply. So it follows those conventions.

Was wondering if current loss needed to be factored in when using a buck converter. Thanks

Look at the videos demonstrating how to configure a 555 in astable mode. Then notice that pin 3, the output pin, is hooked to the gate of the MOSFET. As the square wave generated goes high and low, the MOSFET opens and closes.

Use the same setup but with a 48v capacitor.

There is no feedback in this circuit ,hence the output current and Voltage are not constant. Bigger Load Will drop the output current and Voltage. Knowing about the Power,you can determine Volts and Amperes.

12V at 1 amp is 12 watts. That's 0.25 amps at 48V.

Huh

Irf2907l

Voltage divided by resistance

Yes

sova713 i.stack.imgur.com/7uveJ.png

Yep..I think so too....it was about 5 mili Henry

about an amp

Well say Ian supplying it with 12v and 150ma, will it still give me an amp and atleast 12v?

It depends on wattage. It can't produce extra wattage out from no where. You said 12v and 150mA, but how much exactly this 12v supply can deliver of amperage. If we say that this is 12v 2A supply, so the input wattage is 24W (12*2). In this case the boost voltage can't give power more than 24W (if we considered efficiency is 100%). So with 24W input and you boost your voltage to for example 48v, it will only capable of delivering 0.5A (48*0.5=24W) if not lesser.

Get a capacitor that big...

You need to check for the inductance of the inductor used more the inductance higher the voltage can get and frequency of the oscillator is also proportional to the voltage output also use an appropriate capacitor at the end for smooth output voltage . By the way this is not the circuit for high voltage output this one is nice if you want to drive 20volts load from 5volts battery but if you want something to play with having 500volts output build something with appropriate transformer not with a inductor circuit like this because the high voltage peaks can damage your battery but in case of transformer based circuit your battery is safe .... : ) ignore typing mistakes if you find some

Ne

Deepak khatri yes, of course

Hicham Bougaa 👀

he said may god bless you

Kshitij Patil ne555

learnelectronics is it necessary to connect timmer? I saw in some other video they didn't connect the timmer? Sorry for disturbing you but i want to make this

Have you tried to recreate this circuit to test your theory?

When he puts the resistor load, he puts it in series, which means he has the resistance of the resistor summed up with the resistance of the multimeter which is really high, so his voltage is not really affected by this load. If you would put the load to the ground, and be that a small resistor 100Ohm, this circuit will most likely drop in voltage significantly. Don't try to reinvent the wheel, just buy it from ebay, they are much more stable than anything you can build at home.

@@MitkoNikov dude you are right that's why he got low output current 👍

Not much

Yes

Buy them....

Don't worry, everybody here will sooner or later connect a real load to their converters. And then they will lol also:)

Yes, we do our best work in the dark.

555 Very fast switching trigger for the Mosfet - 555 is a timer IC