Reply by Paul Taylor October 3, 20112011-10-03
On Oct 2, 4:36=A0pm, Fred Bloggs <bloggs.fredbloggs.f...@gmail.com>
wrote:
> On Oct 1, 12:14=A0pm, Daku <dakup...@gmail.com> wrote: > > > > > > > Could some electronics guru please clarify the following ? > > The following is the basic structure of a boost converter: > > > DC_in -> inductor --> diode --> capacitor --> output > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0switch =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 GND =A0 =A0 =A0=
=A0 =A0 =A0 =A0 =A0 =A0GND
> > The diode is forward biased in the sense that when > > the switch is closed, the diode is reverse-biased and > > prevents the capacitor from discharging. The question > > is: > > How is the maximum output voltage set, i.e., how is > > maximum output voltage set ? Is it done by the reverse > > breakdown voltage of the diode, since once that is > > exceeded, the capacitor will try to discharge, when > > the switch is on. > > The output voltage is set by the duty cycle, D, of switch which is > defined as Ton/(Ton+Toff) where Ton is the switch ON time and Toff is > the switch OFF time. This follows from the fact that in steady state > the average voltage across an ideal inductance is zero. During the ON > time the voltage across the inductor is DC_in, and for the OFF time > the voltage is DC_in-Vout (making the simplification that diode > forward drop is 0), so that the average is 0=3DDC_in x Ton + (DC_in- > Vout) x Toff. Solving for Vout yields Vout=3D DC_in x (Ton+Toff)/Toff =3D > DC_in/(1-D).- Hide quoted text - > > - Show quoted text -
This explains things a little, might be of help. http://ww1.microchip.com/downloads/en/AppNotes/00980a.pdf
Reply by Nico Coesel October 2, 20112011-10-02
Daku <dakupoto@gmail.com> wrote:

>Could some electronics guru please clarify the following ? >The following is the basic structure of a boost converter: > >DC_in -> inductor --> diode --> capacitor --> output > | | > | | > switch | > | | > GND GND >The diode is forward biased in the sense that when >the switch is closed, the diode is reverse-biased and >prevents the capacitor from discharging. The question >is: >How is the maximum output voltage set, i.e., how is >maximum output voltage set ? Is it done by the reverse >breakdown voltage of the diode, since once that is >exceeded, the capacitor will try to discharge, when >the switch is on.
Just google it and you'll find a formula which says that the input/ouput voltage ratio is determined by the duty cycle. This goes as long as the minimum load current is satisfied. -- Failure does not prove something is impossible, failure simply indicates you are not using the right tools... nico@nctdevpuntnl (punt=.) --------------------------------------------------------------
Reply by Fred Bloggs October 2, 20112011-10-02
On Oct 1, 12:14=A0pm, Daku <dakup...@gmail.com> wrote:
> Could some electronics guru please clarify the following ? > The following is the basic structure of a boost converter: > > DC_in -> inductor --> diode --> capacitor --> output > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0switch =A0 =A0 =A0=
=A0 =A0 =A0 =A0 =A0 =A0 =A0|
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 GND =A0 =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0GND
> The diode is forward biased in the sense that when > the switch is closed, the diode is reverse-biased and > prevents the capacitor from discharging. The question > is: > How is the maximum output voltage set, i.e., how is > maximum output voltage set ? Is it done by the reverse > breakdown voltage of the diode, since once that is > exceeded, the capacitor will try to discharge, when > the switch is on.
The output voltage is set by the duty cycle, D, of switch which is defined as Ton/(Ton+Toff) where Ton is the switch ON time and Toff is the switch OFF time. This follows from the fact that in steady state the average voltage across an ideal inductance is zero. During the ON time the voltage across the inductor is DC_in, and for the OFF time the voltage is DC_in-Vout (making the simplification that diode forward drop is 0), so that the average is 0=3DDC_in x Ton + (DC_in- Vout) x Toff. Solving for Vout yields Vout=3D DC_in x (Ton+Toff)/Toff =3D DC_in/(1-D).
Reply by Fred Bloggs October 2, 20112011-10-02
On Oct 2, 10:05=A0am, Jamie
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
> Daku wrote: > > On Oct 1, 12:57 pm, Ecnerwal > > <MyNameForw...@ReplaceWithMyVices.Com.invalid> wrote: > > >>In article > >><bb74a697-6fcd-403d-b060-41e0e9381...@j20g2000vby.googlegroups.com>, > > >>Ain't no guru. > > >>In the given example (your basic joule thief circuit, with diode and ca=
p in
> >>place of the LED, and not showing the trigger coil and transistor that =
are
> >>the switch in that case), it's not "set", though it will come to a sett=
ing
> >>depending on the input voltage and the real characteristics of the indu=
ctor.
> >>Assuming you don't exceed the breakdown of the capacitor or diode, and =
no
> >>current is drawn from the output, you'll land somewhere, and the easies=
t
> >>way to tell where is experiment. Likewise if you do draw current from t=
he
> >>output, but the voltage will be lower - as it's not regulated. > > >>In a regulated DC/DC converter, there will be feedback from the output =
side
> >>which controls the switch (transistor, normally) to maintain a particul=
ar
> >>voltage (or in some cases, current) on the output side. > > >>This is fairly comprehensive, if slightly pompous, explanation of the > >>Joule Thief that your circuit almost is: > > >>http://www.talkingelectronics.com/projects/LEDTorchCircuits/LEDTorchC..=
.
> > >>This has a better simplified explanation, and many variants if you poke > >>around the blog: > > >>http://watsonseblog.blogspot.com/2011/08/2011-aug-01-joule-thief-simp..=
.
> > >>-- > >>Cats, coffee, chocolate...vices to live by > >>Please don't feed the trolls. Killfile and ignore them so they will go =
away.
> > > Thanks for the explanation. I have created a working SPICE model of 5x > > DC-DC boost converter in the past which uses PWM for switching. The > > reason I asked is that I am thinking of some modifications. > > Start with this video, this guy seems to be doing a good job explaining > things.http://www.youtube.com/watch?v=3DGc2eMBbfk7o&feature=3Drelated > > Jamie- Hide quoted text - > > - Show quoted text -
He has huh? Who TF can stay awake through that incredibly boring, lifeless , monotone, rambling....ehhhhh....uhhhhh....ahhhhh
Reply by Jamie October 2, 20112011-10-02
Daku wrote:

> On Oct 1, 12:57 pm, Ecnerwal > <MyNameForw...@ReplaceWithMyVices.Com.invalid> wrote: > >>In article >><bb74a697-6fcd-403d-b060-41e0e9381...@j20g2000vby.googlegroups.com>, >> >> >> > > >>Ain't no guru. >> >>In the given example (your basic joule thief circuit, with diode and cap in >>place of the LED, and not showing the trigger coil and transistor that are >>the switch in that case), it's not "set", though it will come to a setting >>depending on the input voltage and the real characteristics of the inductor. >>Assuming you don't exceed the breakdown of the capacitor or diode, and no >>current is drawn from the output, you'll land somewhere, and the easiest >>way to tell where is experiment. Likewise if you do draw current from the >>output, but the voltage will be lower - as it's not regulated. >> >>In a regulated DC/DC converter, there will be feedback from the output side >>which controls the switch (transistor, normally) to maintain a particular >>voltage (or in some cases, current) on the output side. >> >>This is fairly comprehensive, if slightly pompous, explanation of the >>Joule Thief that your circuit almost is: >> >>http://www.talkingelectronics.com/projects/LEDTorchCircuits/LEDTorchC... >> >>This has a better simplified explanation, and many variants if you poke >>around the blog: >> >>http://watsonseblog.blogspot.com/2011/08/2011-aug-01-joule-thief-simp... >> >>-- >>Cats, coffee, chocolate...vices to live by >>Please don't feed the trolls. Killfile and ignore them so they will go away. > > > Thanks for the explanation. I have created a working SPICE model of 5x > DC-DC boost converter in the past which uses PWM for switching. The > reason I asked is that I am thinking of some modifications.
Start with this video, this guy seems to be doing a good job explaining things. http://www.youtube.com/watch?v=Gc2eMBbfk7o&feature=related Jamie
Reply by Jamie October 2, 20112011-10-02
Daku wrote:

> On Oct 1, 12:57 pm, Ecnerwal > <MyNameForw...@ReplaceWithMyVices.Com.invalid> wrote: > >>In article >><bb74a697-6fcd-403d-b060-41e0e9381...@j20g2000vby.googlegroups.com>, >> >> >> > > >>Ain't no guru. >> >>In the given example (your basic joule thief circuit, with diode and cap in >>place of the LED, and not showing the trigger coil and transistor that are >>the switch in that case), it's not "set", though it will come to a setting >>depending on the input voltage and the real characteristics of the inductor. >>Assuming you don't exceed the breakdown of the capacitor or diode, and no >>current is drawn from the output, you'll land somewhere, and the easiest >>way to tell where is experiment. Likewise if you do draw current from the >>output, but the voltage will be lower - as it's not regulated. >> >>In a regulated DC/DC converter, there will be feedback from the output side >>which controls the switch (transistor, normally) to maintain a particular >>voltage (or in some cases, current) on the output side. >> >>This is fairly comprehensive, if slightly pompous, explanation of the >>Joule Thief that your circuit almost is: >> >>http://www.talkingelectronics.com/projects/LEDTorchCircuits/LEDTorchC... >> >>This has a better simplified explanation, and many variants if you poke >>around the blog: >> >>http://watsonseblog.blogspot.com/2011/08/2011-aug-01-joule-thief-simp... >> >>-- >>Cats, coffee, chocolate...vices to live by >>Please don't feed the trolls. Killfile and ignore them so they will go away. > > > Thanks for the explanation. I have created a working SPICE model of 5x > DC-DC boost converter in the past which uses PWM for switching. The > reason I asked is that I am thinking of some modifications.
http://www.youtube.com/watch?v=5JSr19x7kco something to look at.. and a few others there. Jamie
Reply by Daku October 2, 20112011-10-02
On Oct 2, 2:00=A0am, NT <meow2...@care2.com> wrote:
ischarge, when
> > the switch is on. > > If you dont add any further bits than you've drawn, there is no > voltage control. Voltage control is achieved by not switching the > switch on when sufficient V exists in the capacitor. > > Note to get adequate V control you need a capacitor that stores far > more energy than the coil does. If, for experimental reasons, you were > to use a small cap and a big coil, one single switching cycle would > put a large voltage increase into the cap, making regulation > unachievable. > > We all had to learn some time ya know. > > NT
I am afraid the simple diagram I had in my initial post was lifted from the Maxim Web site. My own design is a lot complicated.
Reply by NT October 2, 20112011-10-02
On Oct 1, 5:14=A0pm, Daku <dakup...@gmail.com> wrote:
> Could some electronics guru please clarify the following ? > The following is the basic structure of a boost converter: > > DC_in -> inductor --> diode --> capacitor --> output > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0switch =A0 =A0 =A0=
=A0 =A0 =A0 =A0 =A0 =A0 =A0|
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 GND =A0 =A0 =A0 =
=A0 =A0 =A0 =A0 =A0 =A0GND
> The diode is forward biased in the sense that when > the switch is closed, the diode is reverse-biased and > prevents the capacitor from discharging. The question > is: > How is the maximum output voltage set, i.e., how is > maximum output voltage set ? Is it done by the reverse > breakdown voltage of the diode, since once that is > exceeded, the capacitor will try to discharge, when > the switch is on.
If you dont add any further bits than you've drawn, there is no voltage control. Voltage control is achieved by not switching the switch on when sufficient V exists in the capacitor. Note to get adequate V control you need a capacitor that stores far more energy than the coil does. If, for experimental reasons, you were to use a small cap and a big coil, one single switching cycle would put a large voltage increase into the cap, making regulation unachievable. We all had to learn some time ya know. NT
Reply by October 2, 20112011-10-02
On Sat, 1 Oct 2011 09:14:41 -0700 (PDT), Daku <dakupoto@gmail.com>
wrote:

>Could some electronics guru please clarify the following ? >The following is the basic structure of a boost converter: > >DC_in -> inductor --> diode --> capacitor --> output > | | > | | > switch | > | | > GND GND >The diode is forward biased in the sense that when >the switch is closed, the diode is reverse-biased and >prevents the capacitor from discharging. The question >is: >How is the maximum output voltage set, i.e., how is >maximum output voltage set ? Is it done by the reverse >breakdown voltage of the diode, since once that is >exceeded, the capacitor will try to discharge, when >the switch is on.
Check the position of your diode. I think you'll find that it needs to be moved. boB
Reply by Daku October 2, 20112011-10-02
On Oct 1, 12:57=A0pm, Ecnerwal
<MyNameForw...@ReplaceWithMyVices.Com.invalid> wrote:
> In article > <bb74a697-6fcd-403d-b060-41e0e9381...@j20g2000vby.googlegroups.com>, > > >
> Ain't no guru. > > In the given example (your basic joule thief circuit, with diode and cap =
in
> place of the LED, and not showing the trigger coil and transistor that ar=
e
> the switch in that case), it's not "set", though it will come to a settin=
g
> depending on the input voltage and the real characteristics of the induct=
or.
> Assuming you don't exceed the breakdown of the capacitor or diode, and no > current is drawn from the output, you'll land somewhere, and the easiest > way to tell where is experiment. Likewise if you do draw current from the > output, but the voltage will be lower - as it's not regulated. > > In a regulated DC/DC converter, there will be feedback from the output si=
de
> which controls the switch (transistor, normally) to maintain a particular > voltage (or in some cases, current) on the output side. > > This is fairly comprehensive, if slightly pompous, explanation of the > Joule Thief that your circuit almost is: > > http://www.talkingelectronics.com/projects/LEDTorchCircuits/LEDTorchC... > > This has a better simplified explanation, and many variants if you poke > around the blog: > > http://watsonseblog.blogspot.com/2011/08/2011-aug-01-joule-thief-simp... > > -- > Cats, coffee, chocolate...vices to live by > Please don't feed the trolls. Killfile and ignore them so they will go aw=
ay. Thanks for the explanation. I have created a working SPICE model of 5x DC-DC boost converter in the past which uses PWM for switching. The reason I asked is that I am thinking of some modifications.