question about boost converter

Thread Starter

psoke0

Joined Mar 31, 2017
102
guys i dont know what is going on with my circuit. when i run it with 12 v its perfect its gives me 220 v in 2,2 uF cap. but when i run it at 5 v its only gives me 110 v. i tried use more turns on inductor and less turns nothing changed. is it possile that at 5v mosfet doesnt turn on fully and thats why its doing this ? itried this also it didnt work at all. and im using two differant power adaptors

1593775858619.png


i just tried this and its works this way too give 220 v.
so its not about gate voltage then what is it ?


1593776699349.png
 
Last edited:

Irving

Joined Jan 30, 2016
814
When you say you tried it at 5v, you mean in simulation not for real?

What 2.2u cap, I only see 100u there.

Anyway, stop and think about what's generating that higher voltage and what it relates to, and you'll see inductance isn't the driver...
 
Last edited:

BobTPH

Joined Jun 5, 2013
2,433
With no load and ideal components, the voltage should keep rising forever. My calculation shows that it should reach 280V in 1 second with 5V input.

How long did you run the simulation?

Bob
 

Thread Starter

psoke0

Joined Mar 31, 2017
102
When you say you tried it at 5v, you mean in simulation not for real?

What 2.2u cap, I only see 100u there.

Anyway, stop and think about what's generating that higher voltage and what it relates to, and you'll see inductance isn't the driver...
With no load and ideal components, the voltage should keep rising forever. My calculation shows that it should reach 280V in 1 second with 5V input.

How long did you run the simulation?

Bob

Guys this is not simulation i build the curcuit in real life. this curuit i show is to just show you the configuration of my real curcuit
 

Thread Starter

psoke0

Joined Mar 31, 2017
102
i was using this diode 1n5408 and it was doing that. now i changed it to byc8 600 diode and its gives around 190 V which is good . but what is the differance between these diodes ? both are fast switching
 

Irving

Joined Jan 30, 2016
814
1n5408 isn't a fast switching diode, its a general purpose rectifier and quite slow compared to the byc8600.

The output voltage is determined by the back-emf from the inductor when the transistor switches off, and that's directly related to the current flowing in the inductor. But the key parameter is the L/R time constant and and since resistance is determined by length and inductance by #turns which is also directly related to length L/R is constant for that inductor. Since its quite large and probably quite high resistance the current and therefore back-emf is mostly controlled by the input voltage.

What wire size are you using for the inductor, and how many turns on what core?

Also you have no load on that so you are seeing open-circuit volts. What are you expecting to power?
 

Thread Starter

psoke0

Joined Mar 31, 2017
102
1n5408 isn't a fast switching diode, its a general purpose rectifier and quite slow compared to the byc8600.

The output voltage is determined by the back-emf from the inductor when the transistor switches off, and that's directly related to the current flowing in the inductor. But the key parameter is the L/R time constant and and since resistance is determined by length and inductance by #turns which is also directly related to length L/R is constant for that inductor. Since its quite large and probably quite high resistance the current and therefore back-emf is mostly controlled by the input voltage.

What wire size are you using for the inductor, and how many turns on what core?

Also you have no load on that so you are seeing open-circuit volts. What are you expecting to power?
the diode semms to fixed it i will use no load its just for charging the cap . when i use less inductor turns its lowers the voltage can we make much less turns and get the same voltage somehow ? i dont want to use big inductor
 

Papabravo

Joined Feb 24, 2006
13,939
the diode semms to fixed it i will use no load its just for charging the cap . when i use less inductor turns its lowers the voltage can we make much less turns and get the same voltage somehow ? i dont want to use big inductor
Only if we apply magical thinking. It seems to be a popular pastime nowadays.
 

Irving

Joined Jan 30, 2016
814
You need a smaller inductor with less resistance to improve L/R ratio. This will increase the peak current in the inductor resulting in a larger back-emf and therefore higher output voltage. It will allow you to increase switching speed and reduce losses.

How did you make the current inductor (or part #)
 

Bordodynov

Joined May 20, 2015
2,628
The proposed scheme is bad. It cannot form shutdown of MOS transistor and therefore cannot work. Apparently you didn't draw the circuit completely. You need a resistor from shutter to minus power. For good inductive ejection, the transistor needs to be shut down quickly.
 

Thread Starter

psoke0

Joined Mar 31, 2017
102
what wire size and what core is that?
i removed that core from old cfl lamp and wire size is about 0.3 mm

The proposed scheme is bad. It cannot form shutdown of MOS transistor and therefore cannot work. Apparently you didn't draw the circuit completely. You need a resistor from shutter to minus power. For good inductive ejection, the transistor needs to be shut down quickly.
in real circuit there is pull down resistor i just didnt show it here
 
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