Hello,

I think that the needed current is very low.
At the given page is stated that the current through the gel may lead to heating of the gel:



I think you do not need much power for you inverter.

Bertus

Thanks.

I think you wanted to typed booster instead.

I know it requires low current, about less than 10mA. Even when I connect 5 pieces of 9V batteries in series to get 45V, the measured current used is only 5.3mA. But, there are problems when i connect the intended load to my circuit.

For example,

1. The voltage initially is 100V (at this part, resistor is the load). Then, when i connect to intended load, it becomes 80V.

2. The inductor is hot as well as my mosfet when intended load is connected.

3. Hiss sound seems like appearing in low frequency at higher watt resistor(0.5watt)

4. Lower inductance provides higher maximum output voltage. For example, 1mH provided 100V;10mH = 50V. Supposed, the main factor for the output voltage is switching frequency and its duty cycle,right?

Most of my problems are related to my inductor and yet I don't find any solution to solve it. What kind of rating or charateristic of the inductor should I choose, I am still not very sure with it. So, I'm forced to choose some high ampere rating inductor to do "trial and errors". which i think it's a bad.

 

Hello,

I think the used diode (RU2) is causing the problem.
At higher switching frequencies it will probably be not fast enough.
Better try a faster diode.

I have attached the datasheets of the RU2 and the ES1G for comparison.

Bertus

 

In order for the manufacturer to get a high inductance in a very small space (say, a 10mm drum form), they have to use a very small diameter wire, which you have confirmed. So, your 3mH inductor's wire had a high DC resistance to begin with (likely over 1 Ohm), and you were driving it very hard with 12v and a long duty cycle. It's no wonder that it melted.

I have a couple of similar-sized 180uH inductors; they measure roughly 0.3 Ohms on my inexpensive DMM.
My best guess is that they are part number ELC09D181.
Datasheet: http://docs-asia.electrocomponents.com/webdocs/0d90/0900766b80d90b31.pdf
The datasheet indicates that this inductor has a nominal resistance of 360m Ohms, and a maximum DC current of 0.61A. In a simulation I ran ~20kHz startup, ~48kHz when 77v reached) with a 10mA load, output at 77v, the average current exceeded the limit during start-up, but decreased to 131mA once 77v was reached.

In that same datasheet, a 3.3mH inductor has a resistance of 7 Ohms, and a maximum current of 0.14A. Power dissipation in that inductor would be much higher (~19.5 times higher) than the 180uH inductor due to the high DC resistance, and saturation would be a big problem at low frequencies - which you've already discovered.

The RU2 diode has a ~400nS reverse recovery time; that's a bit long for a fast recovery diode. In the simulation, I'm using a UPSC600 Schottky diode rated for 1A, 600v. Schottky diodes have very fast recovery times, and lower forward voltages than standard silicon diodes.

 

One problem with the basic design is that there is no provision for output current limiting; if the load is too great, the inductor and/or MOSFET will self-destruct.

Adding a transistor and a couple of resistors to the load return path to limit maximum load current via the CNTL input could help that considerably; it could also be used to limit start-up current by adding a capacitor.

 

Re: doing a PCB once the testing is done

You really need to make the entire circuit much more compact while you are testing. The breadboard/protoboard and long wiring is adding lots of stray inductance and capacitance.

You could build the circuit on stripboard/veroboard to help avoid many problems by eliminating many parasitic elements. The very long wires between the MOSFET and the 555 timer is a very big problem.

 

Hello,

I think the used diode (RU2) is causing the problem.
At higher switching frequencies it will probably be not fast enough.
Better try a faster diode.

I have attached the datasheets of the RU2 and the ES1G for comparison.

Bertus

But, in my design, the frequency would never higher than 23kHz. I thought RU2 is good enough.

Previously, I was using 1N4148 but it doesn't allow larger current and that's why I choose RU2 instead.

 

The datasheet indicates that this inductor has a nominal resistance of 360m Ohms, and a maximum DC current of 0.61A. In a simulation I ran ~20kHz startup, ~48kHz when 77v reached) with a 10mA load, output at 77v, the average current exceeded the limit during start-up, but decreased to 131mA once 77v was reached.

Do you mean that, the frequency is increasing when the output is increased?

In that same datasheet, a 3.3mH inductor has a resistance of 7 Ohms, and a maximum current of 0.14A. Power dissipation in that inductor would be much higher (~19.5 times higher) than the 180uH inductor due to the high DC resistance, and saturation would be a big problem at low frequencies - which you've already discovered.

Actually I am still not very sure with the saturation of ind current? In brief, is it meaning that the inductor has reached the maximum energy it could store? And low frequency makes the ON TIME to be too long and saturates the inductor.

The RU2 diode has a ~400nS reverse recovery time; that's a bit long for a fast recovery diode. In the simulation, I'm using a UPSC600 Schottky diode rated for 1A, 600v. Schottky diodes have very fast recovery times, and lower forward voltages than standard silicon diodes.

I will try to find other type of fast switching schottky diodes.

 

One problem with the basic design is that there is no provision for output current limiting; if the load is too great, the inductor and/or MOSFET will self-destruct.

Adding a transistor and a couple of resistors to the load return path to limit maximum load current via the CNTL input could help that considerably; it could also be used to limit start-up current by adding a capacitor.

Sorry that I don't clear/sure with where are them. Is it pin5 of the 555timer? Could you make it more clearly?

Adding a capacitor to which part?

 

http://docs-asia.electrocomponents.com/webdocs/046c/0900766b8046c62e.pdf

http://docs-asia.electrocomponents.com/webdocs/0928/0900766b809285fd.pdf

http://docs-asia.electrocomponents.com/webdocs/0928/0900766b809285fd.pdf

These are the diode that I found, are they okay in my circuit?

Apart from that, while choosing the inductor, which part of the characteristic of the inductor I should pay attention to? I think choose a right inductor is crucial in my circuit.

 

Do you mean that, the frequency is increasing when the output is increased?

Yes, it actually does. That's the function of the transistor who's collector is connected to CTRL (pin 5) of the 555 timer. Along with the frequency increase, the duty cycle is decreased.

Actually I am still not very sure with the saturation of ind current? In brief, is it meaning that the inductor has reached the maximum energy it could store?

That is exactly right! It cannot store any more electrical energy, so the current flow through the inductor increases rapidly up to the the point where it is only the resistance of the wire and the MOSFET Rds(on) that limits the current. That is why the inductor got hot and melted.

And low frequency makes the ON TIME to be too long and saturates the inductor.

Yes. That is why I had you decrease C1 to reduce the maximum amount of ON time by increasing the frequency, and increase R1 to decrease the duty cycle.

I will try to find other type of fast switching Schottky diodes.

OK. Make certain that that it is rated for at least 100v, and at least 1A. A UF4002, UF4003 or UF4004 would work just fine. These are ultrafast recovery diodes that are rated for 1A, and 100v, 200v and 400 volts. They have a Vf of 1V @ 1A, and a reverse recovery time of 50nS; about 8x faster than the RU2 diodes you are using.

A high-voltage Schottky diode might have a lower Vf, but will likely be more expensive.

 

http://docs-asia.electrocomponents.com/webdocs/046c/0900766b8046c62e.pdf

http://docs-asia.electrocomponents.com/webdocs/0928/0900766b809285fd.pdf

These are the diode that I found, are they okay in my circuit?

Apart from that, while choosing the inductor, which part of the characteristic of the inductor I should pay attention to? I think choose a right inductor is crucial in my circuit.

You posted the 2nd link twice, but that's OK.
The BYT03-400 is faster than the UF540x series diodes, which is better - but that or the UF5402 thru UF5404 would work fine. The UF5405 through UF5408 would not be as good, due to their higher recovery times and higher Vf.

As far as the inductors, those you've already ordered can probably be made to work. The small inductors with the large values and high internal resistance won't work well.

 

Sorry that I don't clear/sure with where are them. Is it pin5 of the 555timer? Could you make it more clearly?

Adding a capacitor to which part?

See the attached simulation. It's similar to what you have now.

Rload should pass about 12mA current with Out being 77v.
R7 sets the continuous load current limit to roughly 0.66v/43 Ohms = 15.43mA.
If the voltage on the base of Q3 rises to 0.66v, the transistor conducts enough collector current to pull CTRL down a bit, which makes the 555 timer cycle faster with less ON time, limiting the current.

C3 is 1/100 the value of C5; if the voltage rise is too rapid on C5, C3 turns on Q3 which turns off the 555 timer for a moment, causing L1 to discharge. This helps to make the boost circuit a "soft start" circuit; it keeps the inductor current from ramping up too steeply on startup, which would cause saturation, and possible overshoot of the intended output voltage.

 

You posted the 2nd link twice, but that's OK.
The BYT03-400 is faster than the UF540x series diodes, which is better - but that or the UF5402 thru UF5404 would work fine. The UF5405 through UF5408 would not be as good, due to their higher recovery times and higher Vf.

As far as the inductors, those you've already ordered can probably be made to work. The small inductors with the large values and high internal resistance won't work well.

So, the main concern should be the internal resistance, am I right? Erm, I will try to buy some 220uH to 470uH inductor and try.

 

See the attached simulation. It's similar to what you have now.

Rload should pass about 12mA current with Out being 77v.
R7 sets the continuous load current limit to roughly 0.66v/43 Ohms = 15.43mA.
If the voltage on the base of Q3 rises to 0.66v, the transistor conducts enough collector current to pull CTRL down a bit, which makes the 555 timer cycle faster with less ON time, limiting the current.

C3 is 1/100 the value of C5; if the voltage rise is too rapid on C5, C3 turns on Q3 which turns off the 555 timer for a moment, causing L1 to discharge. This helps to make the boost circuit a "soft start" circuit; it keeps the inductor current from ramping up too steeply on startup, which would cause saturation, and possible overshoot of the intended output voltage.

I understand all the things except from the sentence i highlighted. How would a capacitor turn a Transistor ON? Or it's just because voltage at point R7 greater than threshold voltage and make the 555 off for a moment?

 

I want to buy 4 more types of inductor ELC16B,ELC12D,ELC11D and

http://malaysia.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=0228438

http://docs-asia.electrocomponents.com/webdocs/00ad/0900766b800ad9bc.pdf (744 70 20)

I wish to use these inductors to do comparisons and choose the best among them by considering all of the aspects like price and performance. Do you think there are good choices?

 

Let me re-word that to make it more clear.

"If the voltage rise is too rapid on C5, C3 couples the fast rising edge across the load, causing a rise in voltage at the junction of C3/R7/R8, which turns on Q3, turning off the 555 timer for a moment, causing L1 to discharge."

Does that make more sense now?

 

Let me re-word that to make it more clear.

"If the voltage rise is too rapid on C5, C3 couples the fast rising edge across the load, causing a rise in voltage at the junction of C3/R7/R8, which turns on Q3, turning off the 555 timer for a moment, causing L1 to discharge."

Does that make more sense now?

Ya, it is clear now. Thanks. I was confusing with your last post..

 

sadly, I bought a 4-pin-choke of 2.7mH with 4A rating. It meets all my required, but it is a 4-pin choke which cant be used in my circuit, right?

Is it possible I could use it in my circuit, I heard it is only for filtering use.

 

Are you going to link to the datasheet and tell us the part number that you purchased, or are we going to have to read through a whole bunch of stuff to find it?

It's very late here and I'm going to bed.

 

Are you going to link to the datasheet and tell us the part number that you purchased, or are we going to have to read through a whole bunch of stuff to find it?

It's very late here and I'm going to bed.

http://forum.allaboutcircuits.com/showthread.php?p=326431#post326431

I opened a new thread... It is for filtering function, right? erm, I think I got to buy another type of inductor.

Thanks.