a lot of switching mode power supply design questions

[[email protected]]

here is the circuits picture.

http://www.flickr.com/photos/26914086@N05/2519852045/

I build a deadbug circuits of it. However, the MOSFET is not fully on
or off, the lower drain waveform is shown on the right bottom. it
should be an inversed gate waveform.

load is a photodiode with voltage drop of 2V / 10A.

Is't caused by some parasitic inductance? I couldn't figure out the
parameters. Here I use these equations to calculate the L

V=L di/dt
dt = D/f
di = current ripple
V = Vcc-Vds_on-Vload

but the original circuits use two inductors, why?
Why here are two anti parallel shottkey diodes? I don't think it's
necessary. one is protection and the other is "free wheeling"? Is this
high-side driven a "buck" configuration? I removed the shottkey and
the waveform doesn't change too much.

Why there are two capacitors? local voltage reservoir? can I pick any
large capacitors? or I need to use equation
i=Cdv/dt
dt=D/f
dv=di*Cesr
i=10A?

from IXDD414, the output is shorted by 1k to ground, I don't think
it's necessary.

I am totally confused by the TVL4112 comparator and two FET current
mirror(?). How does is work? I know the comparator show be high or low
but here it will follow the input voltage until the + and - are equal.
How does the feedback work? on the top the current sensing gain is 20
if I remembered. how to calculate the current here? apparently a LM324
won't work as a comparator.


Thank you guys.

Have a nice holiday.

 

[Jamie]

here is the circuits picture.

http://www.flickr.com/photos/26914086@N05/2519852045/

I build a deadbug circuits of it. However, the MOSFET is not fully on
or off, the lower drain waveform is shown on the right bottom. it
should be an inversed gate waveform.

load is a photodiode with voltage drop of 2V / 10A.

Is't caused by some parasitic inductance? I couldn't figure out the
parameters. Here I use these equations to calculate the L

V=L di/dt
dt = D/f
di = current ripple
V = Vcc-Vds_on-Vload

but the original circuits use two inductors, why?
Why here are two anti parallel shottkey diodes? I don't think it's
necessary. one is protection and the other is "free wheeling"? Is this
high-side driven a "buck" configuration? I removed the shottkey and
the waveform doesn't change too much.

Why there are two capacitors? local voltage reservoir? can I pick any
large capacitors? or I need to use equation
i=Cdv/dt
dt=D/f
dv=di*Cesr
i=10A?

from IXDD414, the output is shorted by 1k to ground, I don't think
it's necessary.

I am totally confused by the TVL4112 comparator and two FET current
mirror(?). How does is work? I know the comparator show be high or low
but here it will follow the input voltage until the + and - are equal.
How does the feedback work? on the top the current sensing gain is 20
if I remembered. how to calculate the current here? apparently a LM324
won't work as a comparator.


Thank you guys.

Have a nice holiday.

If my symbols aren't failing me, I would say those are non enhanced
MosFets, which means they are on and need to be reversed biased to
be off. In this case, pulling it common I don't think is going to cut it.
Are we using the correct components? Maybe an Enhanced NMOS is what you
want?

 

[MooseFET]

here is the circuits picture.

http://www.flickr.com/photos/26914086@N05/2519852045/

I build a deadbug circuits of it. However, the MOSFET is not fully on
or off, the lower drain waveform is shown on the right bottom. it
should be an inversed gate waveform.

Are you sure that the MOSFET is really doing what you think it is.
The slow rise in voltage of the drain of the MOSFET could be caused by
a large capacitance on the drain.

The circuit looks to me to be basically this:

D1
----------!<-----------------------
! !
Vin ---+---/\/\---)))))----[LOAD]--))))----+--
R1 L1 L2 !
!!-
---!! Q1
!!-- GND


Barring miswires: The drain node of Q1 has the capacitance of L2, D1
and Q1 on it.
The only likely source for enough capacitance to be a problem at your
claimed 10A
is the L2 inductor if it is way the wrong kind of inductor.

load is a photodiode with voltage drop of 2V / 10A.

What sort of photo diode is that???????

Is't caused by some parasitic inductance? I couldn't figure out the
parameters. Here I use these equations to calculate the L

V=L di/dt
dt = D/f
di = current ripple
V = Vcc-Vds_on-Vload

but the original circuits use two inductors, why?

Who knows. Can you ask the guy who did it.

Why here are two anti parallel shottkey diodes?

The drawing isn't totally clear but one schottky appears to be across
the load to prevent reverse voltage on it. During the power down or
power up of the system, the LC circuit could ring and make a reverse
voltage on the load. This may be bad.

I don't think it's
necessary. one is protection and the other is "free wheeling"? Is this
high-side driven a "buck" configuration? I removed the shottkey and
the waveform doesn't change too much.

Why there are two capacitors? local voltage reservoir? can I pick any
large capacitors? or I need to use equation

You need a low ESR and low ESL and high ripple current. Pick them
based one those numbers more than the capacitance.

 

[Jamie]

Where do you get "non-enhanced" mode MOSFET's in the discrete world?
They are even rare in my integrated circuit world.

...Jim Thompson

well, that's the common symbol used late time I checked.

http://www.rocw.raifoundation.org/c...igitalcircuits-I/lecture-notes/lecture-10.pdf

and after looking, I would say if this information is correct, he's
using a P-channel which also could be an issue here.

But then again, maybe incorrect footprints are being used.

http://webpages.charter.net/jamie_5"

 

[[email protected]]

this circuits has minor problems but basically it can turn on-off
mosfet. my deadbug doesn't work.

sorry for the wrong symbol, this is a n-chan mosfet IRL3803 with 1v
Vgth. "D" connected to high voltage thus it's n channel.

vcc is 12 v coz i want to charge Cg quickly.

load photo diode is a high power led.

I am still confused about the feedback mechanism here, is there other
way to do a close loop for smps?

 

[Paul E. Schoen]

here is the circuits picture.

http://www.flickr.com/photos/26914086@N05/2519852045/

I build a deadbug circuits of it. However, the MOSFET is not fully on
or off, the lower drain waveform is shown on the right bottom. it
should be an inversed gate waveform.

load is a photodiode with voltage drop of 2V / 10A.

Is't caused by some parasitic inductance? I couldn't figure out the
parameters. Here I use these equations to calculate the L

V=L di/dt
dt = D/f
di = current ripple
V = Vcc-Vds_on-Vload

but the original circuits use two inductors, why?
Why here are two anti parallel shottkey diodes? I don't think it's
necessary. one is protection and the other is "free wheeling"? Is this
high-side driven a "buck" configuration? I removed the shottkey and
the waveform doesn't change too much.

Why there are two capacitors? local voltage reservoir? can I pick any
large capacitors? or I need to use equation
i=Cdv/dt
dt=D/f
dv=di*Cesr
i=10A?

from IXDD414, the output is shorted by 1k to ground, I don't think
it's necessary.

I am totally confused by the TVL4112 comparator and two FET current
mirror(?). How does is work? I know the comparator show be high or low
but here it will follow the input voltage until the + and - are equal.
How does the feedback work? on the top the current sensing gain is 20
if I remembered. how to calculate the current here? apparently a LM324
won't work as a comparator.

It seems that the MOSFET is turning on and off properly, and the drain
waveform is exactly as would be expected, depending on the actual frequency
and the values of the inductors.

When Q1 turns ON, the drain voltage goes to zero, as it should, and the
inductors are storing energy. If you look at the curent waveform, you
should see the current rise linearly. If the current (or the MOSFET
voltage) starts to rise sharply, the inductors are saturating, but this
does not appear to be the case.

When Q1 turns OFF, the inductors will release their stored energy into the
load, through the diode to Vcc. This will be seen as an exponential
waveform on the drain as you show in your sketch. The current will be
decreasing during this time, and may reach zero, at which point the Vd will
reach Vcc as you seem to expect.

This appears to be a PWM LED brightness control, with inductance to reduce
the power dissipation that would otherwise occur with a limiting resistor.
The circuit appears to be needlessly complicated. There are many LED driver
ICs that should meet your needs with fewer components.

The usual means for wide range brightness control is to set up a high
frequency (~100kHz-1MHz) switching circuit to provide the maximum current
through the LED, and then use PWM at a lower frequency (~100-200 Hz) to
achieve brightness modulation with a range of up to 1000:1.

Look at devices made by www.linear.com, www.ti.com, www.national.com,
www.maxim-ic.com, and www.zetex.com.

Here are more links:

http://www.electronicsweekly.com/Ar...etex-power-led-driver-chips-how-they-work.htm

http://www.zetex.com/3.0/pdf/zxld1360.pdf


Paul

 

[legg]

Jim Thompson wrote:

well, that's the common symbol used late time I checked.

http://www.rocw.raifoundation.org/c...igitalcircuits-I/lecture-notes/lecture-10.pdf

and after looking, I would say if this information is correct, he's
using a P-channel which also could be an issue here.

But then again, maybe incorrect footprints are being used.

http://webpages.charter.net/jamie_5"

Consult whoever drew the schematic.

Assuming he meant NMOS, what part did you actually throw in there?
This could be a very low-current curcuit, in which case anything in a
to220 body may be overkill. Not possible to comment without frequency
and component value info (unreadable).

Can't make out what's on the other end of the load, either, from the
drawing text resolution; some kind of Maxim part.

RL

 

[MooseFET]

Consult whoever drew the schematic.

Assuming he meant NMOS, what part did you actually throw in there?
This could be a very low-current curcuit, in which case anything in a
to220 body may be overkill. Not possible to comment without frequency
and component value info (unreadable).

Can't make out what's on the other end of the load, either, from the
drawing text resolution; some kind of Maxim part.

Well tbat at least suggests the Maxim has really shipped a part.
Maybe it doesn't really work as the OP thinks it does.

 

[[email protected]]

maxim one is high-side current sensing with 20x gain.
it does carry 10A but I list my questions above.
thanks.

 

[MooseFET]

this circuits has minor problems but basically it can turn on-off
mosfet. my deadbug doesn't work.

sorry for the wrong symbol, this is a n-chan mosfet IRL3803 with 1v
Vgth. "D" connected to high voltage thus it's n channel.

vcc is 12 v coz i want to charge Cg quickly.

load photo diode is a high power led.

That clears that up.

I am still confused about the feedback mechanism here, is there other
way to do a close loop for smps?

You still haven't really stated the goal of the circuit. It is, I
assume, to make a constant current in the high powered LED.

I will start with an over simplified design. If you want to go
further I will come back with more. [LOAD] below refers to a circuit
with the LED in it.

A bit of ASCII art in two parts:

D1
-------!<-----------
! !
Vin ---+--[LOAD]---((((-----+
L1 !
!
!!-
Drive------!! Q1
!!-+--------- Sense
!
\
/ R1
\ 0.01R
!
GND


Vcc
! Vcc Vcc
\ ! !
LM339 R2 / 100K \ !/ NPN 2N4401
U1A \ R3 / ---! Q2
Sense ---!-\ ! U1B \ ! !\
! >---+-+-----------!+\ ! ! !
0.1V--!+/ ! ! >--+----+ +---Drive
C2 === Vcc/2--!-/ ! !
100p ! ! !/
GND ----! PNP 2N4403
!\ Q3
!
GND

When the current in the R1 tries to pass 10A, U1A discharges C2 to
ground. U1B sees the voltage on C2 go below Vcc/2 and yanks its
output down which via Q3 turns off the MOSFET.

The time it takes for the MOSFET to get turned off is very short but
not so short that U1A stops pulling down on C2 before it gets down to
ground.

C2 starts charging back up towards Vcc as soon as Q1 is turned off.
Follows the charging curve of:

-t/(100p*100K)
V = Vcc * (1 - e )

and hits 1/2 Vcc after 0.69 time constants or

0.69 * 100p * 100K =6.9us

When it gets to 1/2 Vcc, U1B uses Q2 to pull the gate of Q1 up again
turning it back on and the cycle repeats.

The peak current and not the average current in L1 is what is being
regulated but that should not be a big issue if the Vcc is fairly well
known because the ripple will have a known amplitude.

The LOAD circuit may contain a second inductor and a capacitor if you
want to have no ripple current in the LED.

 

[[email protected]]

the goal is 0-10A current source for led driving
max chip: high side current sensing, 20x gain
ucc chip: pwm, 400khz, width varies with input voltage
ixdd chip: mosfet driver
opamp: current amplifier

this circuits has minor problems but basically it can turn on-off
mosfet. my deadbug doesn't work.

sorry for the wrong symbol, this is a n-chan mosfet IRL3803 with 1v
Vgth. "D" connected to high voltage thus it's n channel.

vcc is 12 v coz i want to charge Cg quickly.

load photo diode is a high power led.

That clears that up.

I am still confused about the feedback mechanism here, is there other
way to do a close loop for smps?

You still haven't really stated the goal of the circuit. It is, I
assume, to make a constant current in the high powered LED.

I will start with an over simplified design. If you want to go
further I will come back with more. [LOAD] below refers to a circuit
with the LED in it.

A bit of ASCII art in two parts:

D1
-------!<-----------
! !
Vin ---+--[LOAD]---((((-----+
L1 !
!
!!-
Drive------!! Q1
!!-+--------- Sense
!
\
/ R1
\ 0.01R
!
GND

Vcc
! Vcc Vcc
\ ! !
LM339 R2 / 100K \ !/ NPN 2N4401
U1A \ R3 / ---! Q2
Sense ---!-\ ! U1B \ ! !\
! >---+-+-----------!+\ ! ! !
0.1V--!+/ ! ! >--+----+ +---Drive
C2 === Vcc/2--!-/ ! !
100p ! ! !/
GND ----! PNP 2N4403
!\ Q3
!
GND

When the current in the R1 tries to pass 10A, U1A discharges C2 to
ground. U1B sees the voltage on C2 go below Vcc/2 and yanks its
output down which via Q3 turns off the MOSFET.

The time it takes for the MOSFET to get turned off is very short but
not so short that U1A stops pulling down on C2 before it gets down to
ground.

C2 starts charging back up towards Vcc as soon as Q1 is turned off.
Follows the charging curve of:

-t/(100p*100K)
V = Vcc * (1 - e )

and hits 1/2 Vcc after 0.69 time constants or

0.69 * 100p * 100K =6.9us

When it gets to 1/2 Vcc, U1B uses Q2 to pull the gate of Q1 up again
turning it back on and the cycle repeats.

The peak current and not the average current in L1 is what is being
regulated but that should not be a big issue if the Vcc is fairly well
known because the ripple will have a known amplitude.

The LOAD circuit may contain a second inductor and a capacitor if you
want to have no ripple current in the LED.

 

[Paul E. Schoen]

the goal is 0-10A current source for led driving
max chip: high side current sensing, 20x gain
ucc chip: pwm, 400khz, width varies with input voltage
ixdd chip: mosfet driver
opamp: current amplifier

You will have problems using a proto-board at 400 kHz. And the IRL3803,
with 230 nSec rise time, will very likely overheat due to switching losses.
The main reason for using high frequencies is to reduce the size of the
inductive and capacitive components. If you have room, you would do much
better to run this at about 50 kHz and use something like a 100 uH
inductor. At that frequency, you might be able to use a breadboard.

Paul

 

[[email protected]]

I do use lower frequency but ripple is way too large. I think high
frequency can lower ripple?

 

[Paul E. Schoen]

I do use lower frequency but ripple is way too large. I think high
frequency can lower ripple?

If the LED is used for lighting purposes, the human eye will integrate PWM
above about 100 Hz. If there is some reason you need to keep LED current
constant with low ripple, you can use a larger inductor, or add more
filtering to get it as low as you wish. But that is probably not necessary.
If you see any flicker, it is because there is some other instability in
the circuit, probably a poorly compensated feedback loop. This is one
reason why most brightness modulators run the LED at full brightness when
it is ON, and use a fixed PWM to adjust the on/off ratio in a 100-200 Hz
fixed frequency, as I posted previously.

Paul

 

[[email protected]]

This design aims to keep current adjustable yet constant with low
ripple. I couldn't use low freq. dimm. adjustment.
I wonder if I use a much bigger inductor, e.g. a 1000 uH compared to
a theoretical 330 uH, may I get much cleaner
current? Large inductor means large ESR, ESC. I'm using a homemade
toroid inductor. 1000 uH with 176 ohm and
26 uF @1k Hz.

 

[Jamie]

This design aims to keep current adjustable yet constant with low
ripple. I couldn't use low freq. dimm. adjustment.
I wonder if I use a much bigger inductor, e.g. a 1000 uH compared to
a theoretical 330 uH, may I get much cleaner
current? Large inductor means large ESR, ESC. I'm using a homemade
toroid inductor. 1000 uH with 176 ohm and
26 uF @1k Hz.

Trying to figure out what you're doing?
wouldn't a photo feed back work better like used in laser diode systems
to maintain a desired level of luminance? current monitoring may not
be such the idea.


http://webpages.charter.net/jamie_5"

 

[Paul E. Schoen]

Trying to figure out what you're doing?
wouldn't a photo feed back work better like used in laser diode systems
to maintain a desired level of luminance? current monitoring may not
be such the idea.

These figures don't make much sense, and the exact intent and purpose of
the circuit, and its complete design specifications, are still a mystery. A
toroidal inductor with 176 ohms ESR is nearly impossible unless it is wound
with Nichrome, and I don't know what ESC is (maybe equivalent series
capacitance?) You could use a 26 uF capacitor as part of a filter for the
output, but I don't know where your 1 kHz comes from. And also how do you
come up with a 330 uH theoretical inductance?

I found this Maxim circuit that can drive 30 amps of LED load with 5000:1
dimming, but it still uses PWM and will have high ripple:
http://www.maxim-ic.com/quick_view2.cfm/qv_pk/5266

If this is a laser diode, there are many factors to consider, and I don't
think it is possible to adjust output over more than a limited range with a
low ripple DC current. This seems helpful:
http://www.ilxlightwave.com/appnotes/overview_laser_diode_characteristics.pdf

Here is a very extensive overview of laser diodes, from a Wiki link:
http://www.repairfaq.org/sam/laserdio.htm

I really don't think this is a laser diode, as it would be about 20 watts
(2V at 10 amps), which is generally only possible for extremely fast pulses
or very high-end (and expensive) industrial lasers. And if it is, it must
be controlled with photo feedback, and the output light power varies from
0-100% over a narrow range of current, and is highly variable with respect
to temperature and composition.

If this is a high power visible LED, the highest currents I know of for
single units is about 1 ampere continuous. If this is a cluster of ten or
more such LEDs in parallel, then the mismatch in Vf and variations in
temperature will result in current hogging and destruction of several of
the devices. Standard practice is to put the LEDs in series and generate a
higher voltage to drive the entire string.

Paul

 

[MooseFET]

On May 27, 11:35 am, [email protected] wrote:
Top posting fixed. Please put answers after the questions not above.

the goal is 0-10A current source for led driving

That much is part of the goal. How much ripple can you have on the
LED? What is the nominal forward drop of the LED? What is the range
of input voltage?

max chip: high side current sensing, 20x gain
ucc chip: pwm, 400khz, width varies with input voltage
ixdd chip: mosfet driver
opamp: current amplifier

Those are all your attempted method to get to the goal. Why do you
want the high side current sensing? Why do you think a gain of 20x is
a good idea? Why did you select 400KHz?

The high side current sensing doesn't really have any advantage that I
can see. The gain of 20x seems on the low side if you are bothering
to have gain in the current sense part at all. 400KHz forces you to
use high frequency cores which have low mu values and hence low
inductance per unit size. It also means that the MOSFET spends a lot
of its time somewhere between on and off.

I will start with an over simplified design. If you want to go
further I will come back with more. [LOAD] below refers to a circuit
with the LED in it.

A bit of ASCII art in two parts:

D1
-------!<-----------
! !
Vin ---+--[LOAD]---((((-----+
L1 !
!
!!-
Drive------!! Q1
!!-+--------- Sense
!
\
/ R1
\ 0.01R
!
GND

Vcc
! Vcc Vcc
\ ! !
LM339 R2 / 100K \ !/ NPN 2N4401
U1A \ R3 / ---! Q2
Sense ---!-\ ! U1B \ ! !\
! >---+-+-----------!+\ ! ! !
0.1V--!+/ ! ! >--+----+ +---Drive
C2 === Vcc/2--!-/ ! !
100p ! ! !/
GND ----! PNP 2N4403
!\ Q3
!
GND

When the current in the R1 tries to pass 10A, U1A discharges C2 to
ground. U1B sees the voltage on C2 go below Vcc/2 and yanks its
output down which via Q3 turns off the MOSFET.

The time it takes for the MOSFET to get turned off is very short but
not so short that U1A stops pulling down on C2 before it gets down to
ground.

C2 starts charging back up towards Vcc as soon as Q1 is turned off.
Follows the charging curve of:

-t/(100p*100K)
V = Vcc * (1 - e )

and hits 1/2 Vcc after 0.69 time constants or

0.69 * 100p * 100K =6.9us

When it gets to 1/2 Vcc, U1B uses Q2 to pull the gate of Q1 up again
turning it back on and the cycle repeats.

The peak current and not the average current in L1 is what is being
regulated but that should not be a big issue if the Vcc is fairly well
known because the ripple will have a known amplitude.

The LOAD circuit may contain a second inductor and a capacitor if you
want to have no ripple current in the LED.

 

[MooseFET]

I do use lower frequency but ripple is way too large. I think high
frequency can lower ripple?

More filtering can lower the ripple.

Clever design can also lower the ripple.


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Copy the part between begin and paste it into a simple editor then
save it as led.asc. Open led.asc with LTSpice.

 

[[email protected]]

1 ohm is the load or in series with coil?
today I tried to drive a power laser diode but failed. is there a
chip to do this work? 0-10A current source, ripple as low as possible?