I need some help with a Power Supply

[Scott Wiper]

I am designing this and I want to keep it simple So I have devised this in
paper only I would like some help with issues that may be present here.

I am useing trasistors in a pulsing format to shunt high current set by
zener diodes on there bases taking into fact that there are .7 volt drops on
the transistor when they are set in darlington pairs.

The schmatic is here: http://wwwtravel-net.com/~swiper/switch.gif



--
My cat Tigger says every morning...
"Before my morning coffee... I might as well be a dog!"
To contact follow the link below.
http://www.travel-net.com/~swiper/email.htm

 

[Jan Panteltje]

On a sunny day (Wed, 19 May 2004 17:29:31 -0400) it happened "Scott Wiper"
<nobody@devnull.spamcop.met> wrote in <1lQqc.43307$kc2.650168@nnrp1.uunet.ca>:

I am designing this and I want to keep it simple So I have devised this in
paper only I would like some help with issues that may be present here.

I am useing trasistors in a pulsing format to shunt high current set by
zener diodes on there bases taking into fact that there are .7 volt drops on
the transistor when they are set in darlington pairs.

The schmatic is here: http://wwwtravel-net.com/~swiper/switch.gif
dot after the www

Also put fuse in primary
Although it may work, what is missing here is an inductor in series with
the switch.
The idea of a switcher is that the transistor is either on or off.
In both these cases it dissipates very little.
In your diagram current flows while voltage is across the transistor.
Also the right way to drive a switch is so the transistor saturates, and
you lose about 0.2 V.
This could be done with a drive transformer, or using a PNP the other way
around (emittor on +) or putting the NPN in the ground side...
It's a bit late, does this make sense?
Do you have and or know how to calculate such an inductor?
JP

 

[Scott Wiper]

I have sent this to you in privateas well I hope you do not mind. Because of
the invaluable help I have received from this group I am relying on you for
help and the understanding of swicher type power supplys

I will not attach this schmactic but you can abtain it at:
http://www.travel-net.com/~swiper/switch.gif


Also put fuse in primary

I have placed the fuse on T1 primary

Although it may work, what is missing here is an inductor in series with
the switch.

Can you explain this and show it on the diagramme that I have provided?

The idea of a switcher is that the transistor is either on or off.
In both these cases it dissipates very little.

This is why I am going to sustained frequency so I can shunt high currents
at 13.8vdc @ 3.5A
This is also to charge 11AHR of lead acid storage so that the 5VDC is
maintained when mains are lost.

This could be done with a drive transformer, or using a PNP the other way
around (emittor on +) or putting the NPN in the ground side...

I have changed Q1 so that is is a pnp that has it's emitter tied to + and
drives Q2 with the voltage referance D3 and series limit resistor. As for
the inductor. Is this placed in series on the collector Q3?

 

[Jan Panteltje]

On a sunny day (Thu, 20 May 2004 10:48:38 -0400) it happened "Scott Wiper"
<nobody@devnull.spamcop.met> wrote in <ez3rc.43684$kc2.655010@nnrp1.uunet.ca>:

I will not attach this schmactic but you can abtain it at:
http://www.travel-net.com/~swiper/switch.gif


Also put fuse in primary
Great, at least you will have electricity, and the rest of the states

too if something happens.

Although it may work, what is missing here is an inductor in series with
the switch.
Yes

Can you explain this and show it on the diagramme that I have provided?
OK, I will try.

You have 35.6 V DC on one side of teh transistor
You have (or want!) 12 V DC on the other side.
You have a big electrolytic capacitor of 4700uF acorss the 12 V.
Now think switch (as in normal mechanical one).
Think it is 'on'.
Now the 35.6V is directly connected to the 12 V, and guess what,
teh 12V will ALSO become 35 V (or there about), the electrolytic
cap will charge and things will go BOOm on the firs t'on' time of the
switch (transistor).
Current wil be limited a bit by the transistors, because of the drive
and its gain, etc, but if not boom the fist pulse on, then a later one.

The idea of a switcher is that the transistor is either on or off.
In both these cases it dissipates very little.

This is why I am going to sustained frequency so I can shunt high currents
at 13.8vdc @ 3.5A
This is also to charge 11AHR of lead acid storage so that the 5VDC is
maintained when mains are lost.

OK, lets talk about switchmode a bit.
Why would we want to switch?
To avoid losing a lot of power in a series regulator.

How does it work?

\ S1 L1
35V------------0 0-------------^^^^^^^^---------- 12V 3A
| |
| |
=== ===
/ \ D1 --- C1
| |
| |
----------------------------------------------- GND


This diagram shows a basic switching system, used to get a lower
output voltage.
I just put some values there for voltage and current as an example.

Now let's first look at what we need for the switch.
25V in and 12 V out, we need the switch to be on for 12 / 35 = 34%
of the time.
We know tha tthe ciurrent in an inductor rises linear with time
(if no saturation effects).
---- ----
| | | |
--- ------------ -------------- switch drive

/\ /\
/ \ / \
/ \ / \
/ \ / \
\ / current in L1



If we have 3A output, then that is teh AVERAGE output current.

If the switch is only on 35 %, then in the time the switch is on,
the average current in the switch (it has all to go through the switch)
is (100 / 35) * 3 = 8.57A
We know the switch current rizes linear, so the peak current is then
17.14 A.
The 2N3055 is specified as 10A peak (I think) so it is out.
But we can use a heavier transistor.

Now the first diagram.
If the switch comes on, on one side of L1 is 35V, and
on the other side 12 V (assuming all is working and C1 is charged already).

Say you switch frequency is 68 kHz (I really do not know, but from the
diagram).
Then 1 period takes 14.7 uS
If the switch is on for 35% of the time, it will be on for about 5uS.
We know we want the current ot be 17.14 A after this 5uS.
So from i = t / L we see that L = t / i, so 5E-6 / 17.14 =
L = 17.14 / 5E-6 = 0.29 uH
Not a whole lot!
And a 2N3055 does not switch very well at 68kHz....
Consider going down in frequency, and a transistor that can handle more
current.
maybe 16 kHz -> 1.23 uH
Why is D1?
When the switch opens again, a back-emf will occur at L1, it has negative
direction, and is caused by the energy stored in the magnetic field in L1.
D1 will provide a path to ground, and the coil will deliver energy to the
output during this time too.
Why the coil?
because it stores and then releases energy, so we lose little in heat.
A series transistor or series resistor would lose energy in heat.
These sort of coils can be wound or bought, but these are just some gross
calculations, to give SOME idea what to expect,
The waveforms are more complex.
You would likely like to play with the free LT spice, especially
made for switchers
http://www.linear.com/software/
and look at the real waveforms!
I hope I did not screw up on the math and stuff, does this give you some
idea?
JP

 

[Jan Panteltje]

On a sunny day (Thu, 20 May 2004 10:48:38 -0400) it happened "Scott Wiper"
<nobody@devnull.spamcop.met> wrote in <ez3rc.43684$kc2.655010@nnrp1.uunet.ca>:

I will not attach this schmactic but you can abtain it at:
http://www.travel-net.com/~swiper/switch.gif


Also put fuse in primary
Great, at least you will have electricity, and the rest of the states

too if something happens.

Although it may work, what is missing here is an inductor in series with
the switch.
Yes

Can you explain this and show it on the diagramme that I have provided?
OK, I will try.

You have 35.6 V DC on one side of teh transistor
You have (or want!) 12 V DC on the other side.
You have a big electrolytic capacitor of 4700uF acorss the 12 V.
Now think switch (as in normal mechanical one).
Think it is 'on'.
Now the 35.6V is directly connected to the 12 V, and guess what,
teh 12V will ALSO become 35 V (or there about), the electrolytic
cap will charge and things will go BOOm on the firs t'on' time of the
switch (transistor).
Current wil be limited a bit by the transistors, because of the drive
and its gain, etc, but if not boom the fist pulse on, then a later one.

The idea of a switcher is that the transistor is either on or off.
In both these cases it dissipates very little.

This is why I am going to sustained frequency so I can shunt high currents
at 13.8vdc @ 3.5A
This is also to charge 11AHR of lead acid storage so that the 5VDC is
maintained when mains are lost.

OK, lets talk about switchmode a bit.
Why would we want to switch?
To avoid losing a lot of power in a series regulator.

How does it work?

\ S1 L1
35V------------0 0-------------^^^^^^^^---------- 12V 3A
| |
| |
=== ===
/ \ D1 --- C1
| |
| |
----------------------------------------------- GND


This diagram shows a basic switching system, used to get a lower
output voltage.
I just put some values there for voltage and current as an example.

Now let's first look at what we need for the switch.
25V in and 12 V out, we need the switch to be on for 12 / 35 = 34%
of the time.
We know tha tthe ciurrent in an inductor rises linear with time
(if no saturation effects).
---- ----
| | | |
--- ------------ -------------- switch drive

/\ /\
/ \ / \
/ \ / \
/ \ / \
\ / current in L1



If we have 3A output, then that is teh AVERAGE output current.

If the switch is only on 35 %, then in the time the switch is on,
the average current in the switch (it has all to go through the switch)
is (100 / 35) * 3 = 8.57A
We know the switch current rizes linear, so the peak current is then

17.14 A.
The 2N3055 is specified as 10A peak (I think) so it is out.

Actually this is wrong, the current would be close to 4 A if all is stable,
so 2N3055 would work.
But there is a catch, you would need to switch of your switch if Ic approaches
Icmax of the transistor.
Else the 3055 may still be in big trouble when switching on.
In the start up case you have 35 V acorss the coil when the swithc is first on.
'Softstart' may help too, (in soft start the pulse width is gradually increased
from zero).
JP

 

[Scott Wiper]

"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:c8iqdu$im8$1@news.epidc.co.kr...
On a sunny day (Thu, 20 May 2004 10:48:38 -0400) it happened "Scott Wiper"
<nobody@devnull.spamcop.met> wrote in
<ez3rc.43684$kc2.655010@nnrp1.uunet.ca>:

I will not attach this schmactic but you can abtain it at:
http://www.travel-net.com/~swiper/switch.gif

Although it may work, what is missing here is an inductor in series with
the switch.
Yes

Can you explain this and show it on the diagramme that I have provided?
OK, I will try.

You have 35.6 V DC on one side of the transistor
Yes I do

You have (or want!) 12 V DC on the other side.
I want 13.8 volts to charge the battery and maintain its charge when the

mains are present.

You have a big electrolytic capacitor of 4700uF across the 12 V.
Now think switch (as in normal mechanical one).
First of all C2 the large electrolytic capacitor is connected only to the

emitter of Q3 and should only charge to 14.0VDC. That is 16 - 1.4 (Q2/Q3) -
0.6 (D4) = 14.0VDC

This is why I am going to sustained frequency so I can shunt high currents
at 13.8vdc @ 3.5A
This is also to maintain charge on the lead acid battery @ 11AHR so that the
5VDC is
maintained when mains are lost.

OK, lets talk about switchmode a bit.
Why would we want to switch?
To avoid losing a lot of power in a series regulator.

How does it work?

\ S1 L1
35V------------0 0-------------^^^^^^^^---------- 13.8V 5A
| |
| |
=== ===
/ \ D1 --- C1
| |
| |
----------------------------------------------- GND


The above part I understand I have changed the voltage to 13.8 at 5A

If we have 3A output, then that is the AVERAGE output current.

I want 13.8 volts @ 5A so the switch drive should be at 39% of 35V, so (100
/ 39)*5 = 6.57A


The sustained frenquency on U1 and U1A is not 68 kilohutrz it is actually
68.6 hurtz the reason I have selected this because the 2N3055 is not a high
speed switch. If 68.6 hurtz is to slow I can increase to 288 hurtz by
removing the 10k resistor on the astables on the 556 IC and replace with a
2.0K resistor. If you print the image that I have provided at
http://www.travel-net.com/~swiper/switch.gif you will see that C4 and C6
are 1UF not 1nF. I understand that L1 would be a limitor on voltage in the
swicher. Now I know how they reduce the 170vdc from the mains in real
transformerless switchers by useing the inductors I see in them. You have
taught me something. If you can send your the math formulas on how to to
reduce 170VDC to 38VDC to this address
(http://www.travel-net.com/~swiper/email.htm) privately I would apreciate
it.

I have just aquired a 25vac stepdown transformer rated at 100VA (That is 100
watts 4A)

You would likely like to play with the free LT spice, especially
made for switchers
http://www.linear.com/software/
and look at the real waveforms!
I hope I did not screw up on the math and stuff, does this give you some
idea?
JP

 

[Scott Wiper]

"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:c8ircc$j8t$1@news.epidc.co.kr...
On a sunny day (Thu, 20 May 2004 10:48:38 -0400) it happened "Scott Wiper"
<nobody@devnull.spamcop.met> wrote in ez3rc.43684$kc2.655010@nnrp1.uunet.ca:

I use a program called expresspcb. That you can dowload at this location
http://www.expresspcb.com

I have the sheet here for you to download here
http://www.travel-net.com/swiper/switch.sch

This so we can trade sheet files directly (I do not allow e-mail attachments
at this time) So you will have to use a www server to trade.

You can print this image in ladscape and fit to sheet
http://www.travel-net.com/swiper/switch.gif

 

[Jan Panteltje]

On a sunny day (Thu, 20 May 2004 20:32:57 -0400) it happened "Scott Wiper"
<nobody@devnull.spamcop.met> wrote in <17crc.43988$kc2.658929@nnrp1.uunet.ca>:

First of all C2 the large electrolytic capacitor is connected only to the
emitter of Q3 and should only charge to 14.0VDC. That is 16 - 1.4 (Q2/Q3) -
0.6 (D4) = 14.0VDCOK, I see your point.

This is why I am going to sustained frequency so I can shunt high currents
at 13.8vdc @ 3.5A
What is, according to you, the difference?

You have no current limit in the darlington formed by Q2 Q3, so switching
that on (via Q1), will charge C2 to 35 V!
The actual current is about ( (35 - 12) / R2 ) * beta Q2 * beta Q3
for a beta of say 50 (worst case) 2500 x 23mA = 57 Ampere!
(just an estimate).

I want 13.8 volts @ 5A so the switch drive should be at 39% of 35V, so (100
/ 39)*5 = 6.57A


The sustained frenquency on U1 and U1A is not 68 kilohutrz it is actually
68.6 hurtz the reason I have selected this because the 2N3055 is not a high
speed switch. If 68.6 hurtz is to slow I can increase to 288 hurtz by
Correct.

removing the 10k resistor on the astables on the 556 IC and replace with a
2.0K resistor. If you print the image that I have provided at
http://www.travel-net.com/~swiper/switch.gif you will see that C4 and C6
are 1UF not 1nF. I understand that L1 would be a limitor on voltage in the
swicher. Now I know how they reduce the 170vdc from the mains in real
transformerless switchers by useing the inductors I see in them. You have
taught me something. If you can send your the math formulas on how to to
reduce 170VDC to 38VDC to this address
(http://www.travel-net.com/~swiper/email.htm) privately I would apreciate
it.
I think you better get the spice program, and build a small LOW VOLTAGE

test setup, and take it from there.
As pointed out, you need PWM modulation, a level compare against a
reference, (error amplifier), a loop filter (so it is stable), a max current
trip (to prevent thing fron going poop), and a slow start circuit likely.
You have to make your inductor such that it does not saturate (in that
case, when the core is 100% magnetized, the current will rize exponentially).
Switchmode design is not for the faint of heart, and one working directly
on the mains is better done in a different way (you will want a transformer
with 2 windings, to get separation from the mains).
I have done this with a SEPP output, transformer, choke, electrolytics,
using HV fast transistors.
You still need all the trics I mentioned.
There are books on switchmode design, some experts in this group too,
so if you want to go all the way from the mains you can ask here.

I have just aquired a 25vac stepdown transformer rated at 100VA (That is 100
watts 4A)
Cool, that way you will be safe experimenting.

You would likely like to play with the free LT spice, especially
made for switchers
http://www.linear.com/software/
and look at the real waveforms!
JP

 

[Jan Panteltje]

On a sunny day (Thu, 20 May 2004 20:32:57 -0400) it happened "Scott Wiper"
<nobody@devnull.spamcop.met> wrote in <17crc.43988$kc2.658929@nnrp1.uunet.ca>:

This is why I am going to sustained frequency so I can shunt high currents
at 13.8vdc @ 3.5A
What is, according to you, the difference?

You have no current limit in the darlington formed by Q2 Q3, so switching
that on (via Q1), will charge C2 to 35 V!
The actual current is about ( (35 - 12) / R2 ) * beta Q2 * beta Q3
for a beta of say 50 (worst case) 2500 x 23mA = 57 Ampere!
(just an estimate).

Oops, correction, that will work to 12 V did not notice that zener the first
time, by why pulse it (and then 1000uF filter cap)?
JP

 

[Jan Panteltje]

On a sunny day (Thu, 20 May 2004 21:21:38 -0400) it happened "Scott Wiper"
<nobody@devnull.spamcop.met> wrote in <DQcrc.44000$kc2.659160@nnrp1.uunet.ca>:

"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:c8ircc$j8t$1@news.epidc.co.kr...
On a sunny day (Thu, 20 May 2004 10:48:38 -0400) it happened "Scott Wiper"
nobody@devnull.spamcop.met> wrote in ez3rc.43684$kc2.655010@nnrp1.uunet.ca:

I use a program called expresspcb. That you can dowload at this location
http://www.expresspcb.com
Well, OK, I use eagle, and Linux pcb...

I have the sheet here for you to download here
http://www.travel-net.com/swiper/switch.sch
It is ~swiper

I have it in the browser, and then I have to scroll constantly to see the
whole diagram.
In a gif viewer (xv) I cannot read the values.
Better is diagrams in a vector format (pdf, postcript).
Well, I will help, if I can, but I am not going to design your circuits
JP