Hi, this is my first post and I am a student/electrical hobbyist so I apology if something is incorrect and would like to learn. I learned about boost converters and understand their basic function. I have built a boost converter my self with some parts I had lying around. Right now, my circuit looks like the schematic above. I am using a 555 timer in astable mode to switch the power with the n-type MOSFET at 15.6kHz. The variable resistor(R5) is used to adjust the voltage and the circuit is intended to charge to roughly 100V. I got this design online where it was intended to power NIXIE tubes using 180 volts(with a different diode, MOSFET and NPN transistor). Unfortunately, my circuit only outputs a max 75 volts. I've build multiple designs with different frequencys and still have only gotten to 75 volts. I want it to charge my cap at least to 100 volts. I would also like to know if my problem is the speed of my diode or transistor. I have heard of "reverse recovery time". Could my transistor or diode have too high of a recovery time or does my circuit have something else holding it back? I know my MOSFET is rated for only 60 volts so I am going to get a higher rated one but, should I look for a specific type?

The circuit I am following:
http://www.dos4ever.com/flyback/flyback.html

 

Do you happen to have a datasheet for your MOSFET?

 

Do you happen to have a datasheet for your MOSFET?

I found one online here:
https://pdf1.alldatasheet.com/datasheet-pdf/view/52370/FAIRCHILD/FQP30N06L.html

I bought mine at sparkfun, which is a FQP30N06L. I have all the serial codes for the transistors and diode in the picture above.

 

Thank you for the datasheet.

That MOSFET is not the best candidate for this project, mainly because of its large input capacitance.
Incidentally, I have calculated the frequency of the NE555 by two different means and come up with 33.3 kHz.

You are right about reverse recovery. That 1N4001 looks like a short circuit at 31 kHz, and on top of that it is only a 50 volt diode -the reverse breakdown of the diode might be what always limits your voltage. But there is more:

The pulse width (MOSFET on time) would ideally be close to 16.7 microseconds However, with that on time if the battery were holding at 9 volts and the MOSFET didn't come out of saturation, with that value for the inductor the peak drain current would be near 9 amps!

Measure the voltage on your 9 volt battery while your circuit is operating. If it is heavily loaded down that would be confirmation that the inductor needs to be much higher.

For this kind of thing - low power in the 100 to 150 volt range I prefer to use a small bipolar transistor (KSP44, 2N5551) and an inductance in the millihenry range.


The 1 mH inductor across the base-emitter of the transistor helps it turn off quickly, which is useful for the generation of high voltage.

If you feel that you really need to use a MOSFET consider one with a lower input capacitance like the IRF310.

 

Thank you for the datasheet.

That MOSFET is not the best candidate for this project, mainly because of its large input capacitance.
Incidentally, I have calculated the frequency of the NE555 by two different means and come up with 33.3 kHz.

You are right about reverse recovery. That 1N4001 looks like a short circuit at 31 kHz, and on top of that it is only a 50 volt diode -the reverse breakdown of the diode might be what always limits your voltage. But there is more:

The pulse width (MOSFET on time) would ideally be close to 16.7 microseconds However, with that on time if the battery were holding at 9 volts and the MOSFET didn't come out of saturation, with that value for the inductor the peak drain current would be near 9 amps!

Measure the voltage on your 9 volt battery while your circuit is operating. If it is heavily loaded down that would be confirmation that the inductor needs to be much higher.

For this kind of thing - low power in the 100 to 150 volt range I prefer to use a small bipolar transistor (KSP44, 2N5551) and an inductance in the millihenry range.

View attachment 156975
The 1 mH inductor across the base-emitter of the transistor helps it turn off quickly, which is useful for the generation of high voltage.

If you feel that you really need to use a MOSFET consider one with a lower input capacitance like the IRF310.

Oops, I have a typo. My inductor is 17 mH not 17 uH and I am using a variable power supply which is drawing a nice 0.6 Amps.

How did you get 33.3 kHz as the frequency? I used the formula 1.44/(R1*2R2)*C1 to get my frequency 1.44/((1000*20000)0.0000000022 = 32.7 Hz (oops again) Also how dos this capacitance on the mosfet affect the frequency? Dos it increase the period of the frequency to 16.7 ms and how did you calculate that?

I am not trying to power a neon tube but just want to charge a big capacitor and would like to keep my circuit as similar to my original if possible. The one you posted has a couple more peaces to it and some of it i don't understand but, I am interested in learning if you could explain the purposes of the extra components. Also why would adding a inductor across the base emitter help it turn off quicker?

Sorry for all the questions. I have have little classes in electrics and I am mostly self taught. I love electronics and hope to become an electrical engineer one day though

 

V = L di/dt where "di" means "change in", in other words the voltage across the inductor is proportional to the amps per second change through the inductor times the number of Henrys. This applies to both how quickly current builds up in the inductor when the MOSFET is on and how much voltage appears across the inductor when the MOSFET turns off.
See Inductors and Calculus

The large input capacitance slows down turn-off which can hinder generation of high voltage.

If your inductor is 17 millihenries the interwinding capacitance and the inductance might be causing the self-resonant frequency to be so low that it keeps the current from changing fast enough to develop your desired voltage.I say "might", and now the 1N4001 is suspect #1.

You need a fast rectifier with a PIV rating of more than your output voltage. Notice that I used a UF4007 -with UF meaning "Ultra Fast". The faster the better.

This power supply is for a pulse generator. I put neon lamps in power supplies over 100 volts to remind me to keep my hands away when its on. The 33 Meg resistor is a bleeder resistor so that the voltage across the 2.2 uf drops to a safe level within a few minutes after turning the power supply. The button in series with the 10k resistor is to quickly discharge the 2.2 uf before working on the circuit or the pulse generator.

Notice that this is an open loop power supply in which the load is nearly constant and the output voltage is proportional to the input voltage, therefore the output sage is powered by an adjustable voltage regulator. The NE555 has since been replaced with an integrated circuit switching voltage regulator (MC34063) and a KSP42 so the output is regulated.

When base drive is removed from a bipolar switching transistor base current continues to flow and it dissipates slowly by recombination. This slows down the turn-off (makes T in the formula larger) and reduces the ability to generate high voltage, it also makes the circuit inefficient because the slowly turning off transistor is drawing current while its collector voltage is rising. With the inductor between the base and emitter, when base drive is removed, the inductor tends to continue to draw current from the base, which results in a reverse in base-emitter voltage which I picture as "sucking" the charge out of the base region. This helps the transistor turn off more quickly.

Sometimes you see a "speed up capacitor" across a transistor's base resistor.

This does almost the same thing but it can also cause excessive base current at turn-on.

 

.... The variable resistor(R5) is used to adjust the voltage and the circuit is intended to charge to roughly 100V......

Nope, the feedback is from the wrong point. R4 should be connected at the output (100 V ). Else, there is no feedback.

 

V = L di/dt where "di" means "change in", in other words the voltage across the inductor is proportional to the amps per second change through the inductor times the number of Henrys. This applies to both how quickly current builds up in the inductor when the MOSFET is on and how much voltage appears across the inductor when the MOSFET turns off.
See Inductors and Calculus

The large input capacitance slows down turn-off which can hinder generation of high voltage.

If your inductor is 17 millihenries the interwinding capacitance and the inductance might be causing the self-resonant frequency to be so low that it keeps the current from changing fast enough to develop your desired voltage.I say "might", and now the 1N4001 is suspect #1.

You need a fast rectifier with a PIV rating of more than your output voltage. Notice that I used a UF4008 -with UF meaning "Ultra Fast". The faster the better.

This power supply is for a pulse generator. I put neon lamps in power supplies over 100 volts to remind me to keep my hands away when its on. The 33 Meg resistor is a bleeder resistor so that the voltage across the 2.2 uf drops to a safe level within a few minutes after turning the power supply. The button in series with the 10k resistor is to quickly discharge the 2.2 uf before working on the circuit or the pulse generator.

Notice that this is an open loop power supply in which the load is nearly constant and the output voltage is proportional to the input voltage, therefore the output sage is powered by an adjustable voltage regulator. The NE555 has since been replaced with an integrated circuit switching voltage regulator (MC34063) and a KSP42 so the output is regulated.

When base drive is removed from a bipolar switching transistor base current continues to flow and it dissipates slowly by recombination. This slows down the turn-off (makes T in the formula larger) and reduces the ability to generate high voltage, it also makes the circuit inefficient because the slowly turning off transistor is drawing current while its collector voltage is rising. With the inductor between the base and emitter, when base drive is removed, the inductor tends to continue to draw current from the base, which results in a reverse in base-emitter voltage which I picture as "sucking" the charge out of the base region. This helps the transistor turn off more quickly.

Sometimes you see a "speed up capacitor" across a transistor's base resistor.
View attachment 156981
This does almost the same thing but it can also cause excessive base current at turn-on.

How would i choose the frequency of the 555 timer? I've heard 60-900Hz is good but that was for transformers. ultimately, how would I know the best frequency to switch the transistor? Calculating the frequency in your schematic (1.44/((R1*2R2)*C) was 0.0032 Hz which seems way to slow.

It seems i will get a faster diode but, will the UF4007 work too? It only has 17pF which is close to the UF4008. Also my circuit draws 0.6 A. The KSP42 has a max current rating of 0.5 A. Wont that blow?

 

Nope, the feedback is from the wrong point. R4 should be connected at the output (100 V ). Else, there is no feedback.

Oh, i see. I messed that part up in following the other person's design, but I just connected the resistor to the 100 voltage point and the circuit outputted only 15 volts instead of the usual 75 volts I got... weird

 

Hello,

R4 has a to low value.
You are using 220 in stead of 220K.


Bertus

 

In my schematic R1 = R2 = 15k, C = 1000 picofarads (or .001 uf). 48 kHz. (Edited to correct a serious typo)
Table of metric prefixes

The best frequency is probably going to be between 10 kHz and 100 kHz. It depends a lot on which switching device you are using and how you intend to drive it. More modern circuits switch in the MegaHertz range.

The UF4007 would work. Rated at an amp it should be fine for this an a variety of converters.

 

Hello,

In my schematic R1 = R2 = 15k, C = 1000 picofarads (or .01 uf). 48 kHz.

1000p is not .01uF but 0.001 uF

Bertus

 

Thank you for correcting my typo.

 

See

 

Thanks for all the help! I used a 22k Ohm resistor instead of the 220 Ohm one and got a max 35 volts. Better but still much lower then the 75 volts earlier. I will use the faster diode and design recommended by DickCapple and see where that gets me. I am ordering the parts now but, when they come in, I will let you know what happens and may have other questions.

Thanks again!

 

The FET needs to have a voltage rating greater than the desired output voltage. The FET being used is rated at 60 V, so at some voltage greater than that the body diode will avalanche and limit the output voltage - if the diode doesn't fail due to excessive power.

 

The FET needs to have a voltage rating greater than the desired output voltage. The FET being used is rated at 60 V, so at some voltage greater than that the body diode will avalanche and limit the output voltage - if the diode doesn't fail due to excessive power.

Looks like I needed a higher voltage rated MOSFET after all! thanks ebp and others for the help. My boost converter works great now!