bootstrap circuit.. how it works...

[noviceguy]

Hi all...

I am very new to circuit design.. and is currently looking at some
application notes available on the net.. about DC-DC flyback
converters... now i'm looking at Renesas's HA16384 PWMIC.. inside it i
saw a sample application circuit..
however, there is a particular circuit that i am having a hard time to
understand....
maybe its my overall understanding that is lacking..
i search the internet but.. i'm still confused about it...
i am really hoping that someone could advice me on the issue...

my problem lies in the boot-strap circuit to provide initial power to
the PWM IC...
i'm trying to understand why they are using a 220kohm resistor and a
10uF capacitor (rating 50V). The resistor is connected directly to the
rectified high voltage input of the transformer.
how does this circuit function?
it is also connected to the Auxilary winding of the transformer.. but
initially, there would not be any power from this as the PWM is still
not operating right?
so the PWM will get its power from the resistor-capacitor path.. is
this correct?
but this path there will be very high voltage..
does the resistor and capacitor combination reduces it to the PWM IC
safe operating voltages?the IC max operating voltage is 30V.
i'm really confused....i've search the internet... and the say this is
a simple boot-strap circuit... but i still dont get it...
how they come out with the 220kohm and 10uF values...
what kind of calculation involved?

also.. after the Auxilary winding starts operating and supply Vcc to
the PWM IC, what happen to the 220kohm resistor connection?

i try to represent the circuit below..
141V (from example)
rectified voltage -------------|--------------------- To TF
|
|
R = 220kohm (1/4W)
|
other cct -----------------|------Diode------TF Auxilary Winding
|
PWM IC Vcc------------------|
|
C = 10uF (50V)
|
|
GND

i hope my question make sense..

thanks in advance for any advices...

Best Regards

 

[John Popelish]

Hi all...

I am very new to circuit design.. and is currently looking at some
application notes available on the net.. about DC-DC flyback
converters... now i'm looking at Renesas's HA16384 PWMIC.. inside it i
saw a sample application circuit..
however, there is a particular circuit that i am having a hard time to
understand....
maybe its my overall understanding that is lacking..
i search the internet but.. i'm still confused about it...
i am really hoping that someone could advice me on the issue...

my problem lies in the boot-strap circuit to provide initial power to
the PWM IC...
i'm trying to understand why they are using a 220kohm resistor and a
10uF capacitor (rating 50V). The resistor is connected directly to the
rectified high voltage input of the transformer.
how does this circuit function?
it is also connected to the Auxilary winding of the transformer.. but
initially, there would not be any power from this as the PWM is still
not operating right?
so the PWM will get its power from the resistor-capacitor path.. is
this correct?
but this path there will be very high voltage..
does the resistor and capacitor combination reduces it to the PWM IC
safe operating voltages?the IC max operating voltage is 30V.
i'm really confused....i've search the internet... and the say this is
a simple boot-strap circuit... but i still dont get it...
how they come out with the 220kohm and 10uF values...
what kind of calculation involved?

also.. after the Auxilary winding starts operating and supply Vcc to
the PWM IC, what happen to the 220kohm resistor connection?

i try to represent the circuit below..
141V (from example)
rectified voltage -------------|--------------------- To TF
|
|
R = 220kohm (1/4W)
|
other cct -----------------|------Diode------TF Auxilary Winding
|
PWM IC Vcc------------------|
|
C = 10uF (50V)
|
|
GND

i hope my question make sense..

thanks in advance for any advices...

Best Regards

First of all, I doubt very much that the 220k 1/4 watt resistor is
rated for continuous rectified line voltage. I would use a 1/2 watt,
just for the voltage rating.

The idea is that the control chip remains off until the capacitor
charges up to some predefined operating voltage via the 220k resistor.
Then the control circuit is switched on and it has a millisecond or
so to operate the supply before the capacitor is discharged to the
point that the control circuit is switched back to a dormant state.
If the diode from the auxiliary winding contributes power before that
happens, it takes over as the power supply to the capacitor and the
start up current from the 220k is incidental.

The trick is to have a clean shut down of all loads on the capacitor
(to well below the current supplied by the 220k resistor at low line
voltage) except for a very low current voltage measurement that wakes
everything up, cleanly. And there must be a considerable hysteresis
in the voltage sensing, so that once switched on, the capacitor
voltage can decay quite a bit before everything is switched off to
wait for another start charge cycle.

20 years ago, I designed a circuit made of transistors, zener diodes
and resistors that performed this voltage sensing and power switching
function for the supply in a commercial printer. I wish I had a
schematic of it to show you.

 

[Tim Williams]

i'm trying to understand why they are using a 220kohm resistor and a
10uF capacitor (rating 50V). The resistor is connected directly to the
rectified high voltage input of the transformer.
how does this circuit function?
it is also connected to the Auxilary winding of the transformer.. but
initially, there would not be any power from this as the PWM is still
not operating right?
so the PWM will get its power from the resistor-capacitor path.. is
this correct?
but this path there will be very high voltage..

1. It takes some time for the voltage to rise (T = RC).
2. - What is the current draw when the voltage reaches, say, 10V? 30V?
What does this current do to the voltage across the 220k resistor?

the IC max operating voltage is 30V.

Just in case, there's probably a protection circuit. The UC3842 has an
internal zener diode. You could run it forever from the +320VDC rail
through a resistor, but that would be wasteful.

how they come out with the 220kohm and 10uF values...
what kind of calculation involved?

T=RC
R = V/I

also.. after the Auxilary winding starts operating and supply Vcc to
the PWM IC, what happen to the 220kohm resistor connection?

It's ignored. The aux. winding supplies more current to the chip, which it
requires once it comes online and starts moving things. Otherwise, this
current draw would pretty well consume the voltage supplied by the 220k and
it would not function very well.

The aux. winding also often supplies a figure of how much energy is going
through the transformer, important for maintaining regulation and isolation.

Tim

 

[noviceguy]

Thanks very much for the info John and Tim

i think i got a rough idea on how this circuit works..

you guys insight really help a lot...

again..thank you very much

Best regards

 

[Jasen Betts]

Hi all...


141V (from example)
rectified voltage -------------|--------------------- To TF
|
|
R = 220kohm (1/4W)
|
other cct -----------------|------Diode------TF Auxilary Winding
|
PWM IC Vcc------------------|
|
C = 10uF (50V)
|
|
GND

220K is quite a large resistor, with a supply of 141V only 0.64mA will flow
through it as long as the PWM IC is using atleast that much the capacitor
won't overcharge.

What I expect happens is that the PWM chip lies dormant until the capacitor
charges to about 12V (for example) then it turns on and with a pulse or two
to the driver transistors the whole powersupply wakes up, and from then on
most of the power to the PWM chip will be through the diode, but there'll
still be a little coming through the resistor.

becacuse the eaxilary winding has a lower voltage than the 50V capacitor's
limit it won't overflow the capacitor, and because the PWM chip is using
more than the 0.64mA the resistor can supply it won't either.

Bye.
Jasen

 

[noviceguy]

Thanks Jasen...

also.. i read in the HA16384 specs, got a characteristic that states
Stanby current of value 230uA (max)
will this current value come into consideration in selecting the
resistor?in addition to the capacitor charging time..

anyway.. i think in a week time or so.. i'll give this circuit a try..
well if i can get the transformer
still sourcing for it...
not so easy where i'm at unfortunately...

anyway.. thanks a lot for all you guys advices and insights..
really appreciate it...

best regards...

 

[Jasen Betts]

also.. i read in the HA16384 specs, got a characteristic that states
Stanby current of value 230uA (max)
will this current value come into consideration in selecting the
resistor?in addition to the capacitor charging time..

Yes, the standby current (or some portion of it when the capacitor is
depleted) will be flowing into that that chip while the capacitor is
charging up to the chips starting voltage.

230uA is well below thr 600-ish uA that will flow through the resistor,
leaving 370 or so to charge the capacitor.

anyway.. i think in a week time or so.. i'll give this circuit a try..
well if i can get the transformer
still sourcing for it...
not so easy where i'm at unfortunately...

something from inside an old PC powersupply?
when they fail it's almost always not the transformer.

anyway.. thanks a lot for all you guys advices and insights..
really appreciate it...

I appreciate your respoonse too... but don't rely on everything I post
I'm learning too.

Bye.
Jasen

 

[noviceguy]

no worries Jasen...

hmm.. PC power supply... duh!! why did i think of that....
thanks for the info... i always missed out on obvious things.. lol!

actually.. i'm trying to build a battery charger kind of thing..
so looking at some application notes on related stuff...

afterward.. maybe i'll try inverter circuit too

hmmm.. power supply stuff is quite fun...
who would have thought that.. lol!

well fun.. but dangerous as well..

anyway.. thanks again... for all the advices John, Tim and Jasen.

best regards