problem with boost converter

[majorloo]

hi every body
 iam trying to designed a 12-120 Dc-DC converter with 300mA output but it has voltage drop.
is there any body can help me?

[Zero999]

hi every body
 iam trying to designed a 12-120 Dc-DC converter with 300mA out out but it has voltage drop.
is there any body can help me?

You forgot to post the schematic, bill of materials and pictures of the actual circuit.

[majorloo]

excuse me i forgot that.

[Zero999]

Oh dear, that schematic isn't very clear. I can hardly make out half of the component values.

The first thing which strikes me, is the MOSFET is only being turned off by a pull-down resistor with an illegible value which could cause it to be slow to turn off.

[majorloo]

excuse me i will upload a better one in this post. what shold i do for mosfet switching?

[Zero999]

Why not upload the image here?

You should also crop it, so only the part with the drawing is on the file, rather than a huge amount of white space. Increasing the contrast makes it more clear and changing to monochrome and using PNG, rather than JPG, reduces the file size.

The 120R pull-down is unlikely to be the problem here. What inductor are you using? Are you sure it's rated to the peak current? Have you measured the input current to the regulator? Does it increase more than expected, when the 300mA load is connected.

[majorloo]

hi
I really appreciate your guidance i finally can solved voltage droped, but i have a new problem. when my circuit is under load It starts to warm up and after about 40~50 second finally its Burns. please help me.

[Zero999]

The MOSFET doesn't have a high enough current rating or doesn't have a good enough heat sink?

[Kelbit]

Can also be inductor saturation - seen that before when the inductor saturation current is less than the peak current of the boost converter. You'll slam against saturation current and then the inductor starts acting like a dead short. That generates some heat.

[majorloo]

my circuit is 36 Watt it means that my input current is about 3 Amper. irf640n is a 200v, 11Amper and 125 Watt  transistor and i use a heat seank. Am i have a meastake in choseing the  transistor.

[majorloo]

how can i understand that my inductor is saturated or not. and What parameters depend on the saturation of the inductor?

[Zero999]

my circuit is 36 Watt it means that my input current is about 3 Amper. irf640n is a 200v, 11Amper and 125 Watt  transistor and i use a heat seank. Am i have a meastake in choseing the  transistor.

The input power, therefore the current, will be higher than the output power, as there will be losses. No form of energy conversion is 100% efficient.

how can i understand that my inductor is saturated or not

If the inductor is saturating the current will suddenly rise and it will heat up, as the inductance will drop dramatically. Measure the input current, first with no load, then slowly increase the load current. Assuming everything else is equal, the input current should be roughly proportional to the output current. If the input current dramatically rises, after an increase in load current, then the inductor is reaching saturation.

What parameters depend on the saturation of the inductor?

The core material of the inductor and the number of amps*turns.

[KL27x]

my circuit is 36 Watt it means that my input current is about 3 Amper. irf640n is a 200v, 11Amper and 125 Watt

The 11 amps is constant current rating, with heatsinking to keep the FET at specified test temperature. During switching is where the heat builds up. Faster frequency = more heat. This is why you want to drive it hard. I think better to use push pull driver than a pulldown. But maybe the inductor is the weak link. 
 

[Zero999]

my circuit is 36 Watt it means that my input current is about 3 Amper. irf640n is a 200v, 11Amper and 125 Watt

The 11 amps is constant current rating, with heatsinking to keep the FET at specified test temperature. During switching is where the heat builds up. Faster frequency = more heat. This is why you want to drive it hard. I think better to use push pull driver than a pulldown. But maybe the inductor is the weak link. 

I agree about using a push-pull driver, although I thought 120R should have been enough to switch the MOSFET off quickly enough. If not, a couple of transistors, configured as emitter followers could be used to make a push-pull driver. The pull down resistor, R10, at the input of the buffer could then be reduced to save power.


https://electronics.stackexchange.com/questions/74465/how-to-reduce-mosfet-turn-off-delay

This can also be done with one diode and transistor. Q2 & Q3 are the transistors inside the TL494, shown for clarity.

[ocset]

12v to 120vdc requires a high duty cycle, maybe the tl494 cannot supply that..
for booster, (1-D) = Vin/Vout
So (1-D) = 0.1
Therefore D = 0.9 minimum needed.
Unless you do some trick with discontinuous mode.

[ocset]

Fear not though, as i have  an answer for you as in the cascaded boost converter..attached is an ltspice simulation of it...ltspice is free download

[Zero999]

Fear not though, as i have  an answer for you as in the cascaded boost converter..attached is an ltspice simulation of it...ltspice is free download

Very elaborate and probably not very efficient.

How about using a single stage boost but with a transformer, rather than a single inductor? It will enable a lower duty cycle to be used and be simple and more efficient than cascading more than one stage.

[majorloo]

hi every body
I am very close to solved the problem. the most of the problem was about the coil core. it was ferrit but not a good ferrit and the core was saturated. i will send the finally schematic in this post.

[majorloo]

hi
this is the finally circuits. but i have a new question, what should i do when my coil is in saturated. i was changed my feedback to have 12-36 volt with 1.5 amper out put it was worked but my input current was about 13 amper i think my coil is saturated and change my coil 4 or 5 time but the input current just change between 7 to 13 amper but it shoud be about 5 to 6 amper. what shoud i do??

[ocset]

some ferrite torroids have non linear cores, so the inductance that you get at higher current is much less than the inductance at low current...in other words, the inductance that you measure with a meter may not be what you get when youve got big current floiwng in it

[xavier60]

I used the UC3843 for a similar project, much simpler and has high current Gate drive.
I omitted the Source current sensing resistor, so it was necessary to apply slope compensation to the Isense pin.
I had trouble with the diode. Schottky diodes would occasionally fail, excessive Dv/Dt I think. Silicon diodes would produce a nasty reverse recovery current spike. A ferrite bead made it better.