Linear regulated 150V supply

In article <41a1d879$0$78772$e4fe514c@news.xs4all.nl>,
Frank Bemelman <f.bemelmanx@xs4all.invalid.nl> wrote:
[...]
BTW, what are 'poles' and 'zeros'? I tried to find an
explanation of the meaning (in context of amplifiers etc),
but I can't seem to find any. All texts seem to assume
it is known to the reader. (blush).
Click to expand...

You can repressent linear systems as the "Laplace transform" of the
"transfer function"

Transfer function: If you divide the output by the input, treating gain
and phase shift and magnetude and angle of complex numbers, you get the
"transfer function" F(S). You can think of "S" as a generalized form of
frequency that includes exponential growth/decay.

Transfer functions can look something like this:



(AS+1)
-------
(BS+1)


The (AS+1) part defines a zero. If you make S just the right value it
comes out to zero.

The (BS+1) part defines a pole. If you dare to put just the right value
into S, your calculator will talk in higher and higher pitched voice and
then explode. The firemen will have to slide down their "pole", jump in
the fire truck and put the fire out.


When a system is a closed loop, G(S) is usually used for the gain going
through it and H(S) is used for the gain of the feedback. The gain with
the loop closed is:



G(S)
---------------
1 + G(S)H(S)

That 1+ bit in the denominator can mean that the poles in the whole system
are in a different place than in either G(S) or H(S).

--
--
kensmith@rahul.net forging knowledge
 
Rich Grise wrote...
Would "from the top of R10" work? R1, R2, R3, R7/R8? I've just tried to
simulate the circuit in my head, and have realized that it's like any
oscillator - you can't start the analysis until it's already running. ;-)
Click to expand...
Hey, you aren't implying this is an oscillator are you? Well, at
least it would be a high-voltage power oscillator! You don't see
one of those everyday. :>)

And, don't FETs have some leakage at zero Vgs?
Click to expand...
In practise, not significantly. Not enough to insure starting the
circuit anyway. For example, an IRF9620 is spec'd at 100uA max at
200V with Vgs = 0, but if measured you'd probably see something in
the low nA range. Maybe I'll set one up later and get a reading.



--
Thanks,
- Win
 
John Fields wrote...
4W
////////
+170>--+------+------D S--------+----------+---->+150
| | G | |
| [R] | [R] [R]
| | | | |
| +--------+ +---+-----+ |
| | | | | |
|+ | | [R] | |
[BFC] | | | +| |
| D /-|--+ [C] |
| G--+--[R]---< | | | |
| S | \+|--|-----|----+
| | R <-----, | |+ | |
| | | | | |+ | |
| | [C]<-, | | [REF] | [R]
| | | | | | | | |
GND>---+------+----+---|---|--+---+-----+----+---->GND
| |
| |
But,-------> CONSIDERATION |
.. |
Click to expand...
Hey, help it out a bit ------'


--
Thanks,
- Win
 
On Sun, 21 Nov 2004 23:12:53 GMT, artie <artie.m@gmail.com> wrote:

In article <41a10a09$0$28028$5402220f@news.sunrise.ch>, Rene
Tschaggelar <none@none.net> wrote:

Is the LM723 still the regulator of choice for a
regulated 150V supply driven by a line transformer ?
The current to be supplied is moderate, say up too
200mA.
Or are there any alternatives with an external pass
transistor / FET ?

Rene

VR150?
Click to expand...

Not as a shunt reg at 200 mA! But a VR150 plus a fet/bipolar/toob
follower would be nice.


Doesn't somebody make a really high-voltage equivalent to an LM317? I
think I saw that somewhere recently.

John
 
On Mon, 22 Nov 2004 19:44:53 GMT, "Genome" <dna@nothere.net> wrote:

"Winfield Hill" <hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote in
message news:cnt54r02qs8@drn.newsguy.com...
Jim Thompson wrote...
Genome wrote:
Winfield Hill wrote ...

Estimating the gain going around the loop, we'll start with the
R9/R8 divider, which will have a gain of about 5/150 = 1/30.

Oh come on Win!

The junction of R9/R8 is at the summing node of the Op-Amp.
You're doing a small signal AC analysis.

ROTFLMAO!

Not, so silly at all. To simplify for Frank, who professed some
intimidation, I broke the circuit up into a bunch of small pieces,
one of which is those two resistors NOT connected to the opamp.


The inputs draw no current.
Feedback acts to maintain the inputs at the same voltage.

My point was that together they represent a substantial loss of
signal, rather than a gain (to be compared to the high gain of
the power MOSFET stage moments later). Well, perhaps that point
and its significance was lost on my readers. Genome would prefer
to skip that and advance straight to saying the small-signal gain
is the impedance of R6 and C2, divided by R8.

Bullfuckingshit

Fine, that's OK too.


No, it's not fucking OK.

It's bastard fucking wrong There's a big cunting fuck off and die
difference.

I learnt part of this shit from you and now you're acting like a second
grade pantyboy



--
Thanks,
- Win

Idiot.

DNA
Click to expand...
ROTFLMAO! AGAIN! Genome, you're keeping "technician-boy" crawling!

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
Spehro Pefhany wrote...
the renowned John Larkin wrote:

Doesn't somebody make a really high-voltage equivalent to
an LM317? I think I saw that somewhere recently.

The Supertex LR8 is good for 450V, but only tens of mA out.
The TI TL783 is good for hundreds of mA, but only 125V Vin-Vout.
Click to expand...
Right. There was also the Harris HIP5600, good for 400V and 40mA
in a TO-220 case. It's still available from surplus sources.


--
Thanks,
- Win
 
On Mon, 22 Nov 2004 16:31:25 -0500, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

On Mon, 22 Nov 2004 12:35:16 -0800, the renowned John Larkin
jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sun, 21 Nov 2004 23:12:53 GMT, artie <artie.m@gmail.com> wrote:

In article <41a10a09$0$28028$5402220f@news.sunrise.ch>, Rene
Tschaggelar <none@none.net> wrote:

Is the LM723 still the regulator of choice for a
regulated 150V supply driven by a line transformer ?
The current to be supplied is moderate, say up too
200mA.
Or are there any alternatives with an external pass
transistor / FET ?

Rene

VR150?


Not as a shunt reg at 200 mA! But a VR150 plus a fet/bipolar/toob
follower would be nice.


Doesn't somebody make a really high-voltage equivalent to an LM317? I
think I saw that somewhere recently.

John

The Supertex LR8 is good for 450V, but only tens of mA out.

The TI TL783 is good for hundreds of mA, but only 125V Vin-Vout.
Click to expand...
I think it was somebody with a weirder name than Supertex; I'll try to
find it. Brain cells low on critical enzyme "chocolate."


John
 
Genome wrote...
"OK by you, Genome?"

No, not very good at all.
Click to expand...
Well, come on man, talk to me, I respect your opinions.
Drop all the R8 R9 stuff? Drop R9 from consideration?


--
Thanks,
- Win
 
Rich Grise wrote...
I get to correct Win Hill! Yee-Ha!

"that'll dominate", or "that'll be dominant".
Click to expand...
Excellent.


--
Thanks,
- Win
 
On 22 Nov 2004 02:41:46 -0800, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

budgie wrote...

I guess it depends where you'd reather spend your time and money -
in the regulator or before it. I personally gave up 723's when
I first had occasion to use a 317.

In giving up the 723 you gave up the ability to set a reasonable
current limit, and to use foldback current limiting (yes, thermal
limiting is nice, but often one would rather prevent the power
element from overheating rather than simply limit the overheating
to a rather high temperature). You also gave up the quiet voltage
zener reference, and the ability to further filter the reference.
And you gave up the superior output voltage stability that comes
from separating the reference from the power element, keeping the
high pass-transistor die temperature away from the reference zener.
Click to expand...
and gained incredible simplicity and lower parts count. You may have entirely
valid arguments, but I'm not going back ;-)
 
Jim Thompson wrote...
Winfield Hill wrote:

Genome wrote...

"OK by you, Genome?"

No, not very good at all.

Well, come on man, talk to me, I respect your opinions.
Drop all the R8 R9 stuff? Drop R9 from consideration?

Well, duh! "R9/R8 divider" is what got you in trouble
in the first place.
Click to expand...
Well, I liked it. Certainly my overall understanding wasn't
in trouble, even if it's clear to others my choice of "divide
and conquer" explanatory circuit segments was badly chosen,
pun intended. That's in trouble, and not useful, judging from
you blokes. Thanks for the heads up. OK, drop the dividers!


--
Thanks,
- Win
 
On Mon, 22 Nov 2004 19:26:06 -0800, Winfield Hill wrote:

budgie wrote...

Winfield Hill wrote:

budgie wrote...

I guess it depends where you'd reather spend your time and money -
in the regulator or before it. I personally gave up 723's when
I first had occasion to use a 317.

In giving up the 723 you gave up the ability to set a reasonable
current limit, and to use foldback current limiting (yes, thermal
limiting is nice, but often one would rather prevent the power
element from overheating rather than simply limit the overheating
to a rather high temperature). You also gave up the quiet voltage
zener reference, and the ability to further filter the reference.
And you gave up the superior output voltage stability that comes
from separating the reference from the power element, keeping the
high pass-transistor die temperature away from the reference zener.

and gained incredible simplicity and lower parts count. You may
have entirely valid arguments, but I'm not going back ;-)

Yeah, well I haven't used a 723 more than twice in the last 20 years
either. <sigh> But I've seen a lot of them in commercial linear
supplies, where they are dominant; twenty cents a pop is appealing.
Click to expand...
I've noticed that practically every Power-One and Lambda supply I've ever
seen uses them.

Cheers!
Rich
 
"Winfield Hill" <hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote in
message news:cnu9t70l2e@drn.newsguy.com...
Jim Thompson wrote...

Winfield Hill wrote:

Genome wrote...

"OK by you, Genome?"

No, not very good at all.

Well, come on man, talk to me, I respect your opinions.
Drop all the R8 R9 stuff? Drop R9 from consideration?

Well, duh! "R9/R8 divider" is what got you in trouble
in the first place.

Well, I liked it. Certainly my overall understanding wasn't
in trouble, even if it's clear to others my choice of "divide
and conquer" explanatory circuit segments was badly chosen,
pun intended. That's in trouble, and not useful, judging from
you blokes. Thanks for the heads up. OK, drop the dividers!


--
Thanks,
- Win
Click to expand...
Sorry, I was being a pissed up asshole.

Your method is right, it's the way I do it and... I learnt it from you.

Break the circuit down into its separate blocks. Work out the gain of each
block then multiply them together to get the overall gain.

It's sometimes better to remove the error amplifier from the circuit and
work out the control to output gain, that would be from the base of the
transistor to the output at the smoothing capacitor.

Then you decide what gain you require from the amplifier to get a stable
loop.

BUT.......

The mistake you are making, and it's a common one, is to include both R9 and
R8 in the loop as a gain term when they are connected to the 'inverting
input' of the error amplifier.

Bear in mind that what you are doing is a 'small' signal analysis.

A change in voltage at the output of the error amplifier, dVvea, produces a
change in current in the transistor which gives a change in voltage across
its collector resistor which is a change in the gate voltage. That gives a
change in current through the mosfet which is converted to a change in
voltage across the output capacitor.

OK up to now. BUT then you say that that change in output voltage is
attenuated by R8/R9 to become a change in voltage at the input to the error
amplifier.

It isn't because the inverting terminal is fixed, by feedback, at Vref.

The change in output voltage results in a change in current through R8, none
of which flows in R9. It flows through C2 and R6 to appear at the output of
the op-amp as a result of feedback.

I think you are reaching that conclusion yourself but you're weedling about
the place in an effort to keep R9 in the picture. In this case, small signal
AC, it isn't there.

If you were feeding back to the non-inverting terminal of the op-amp then it
would be.

DNA
 
On Mon, 22 Nov 2004 01:40:45 GMT, Robert Baer
<robertbaer@earthlink.net> wrote:

John Fields wrote:

On Sun, 21 Nov 2004 22:31:49 +0100, Rene Tschaggelar <none@none.net
wrote:

Is the LM723 still the regulator of choice for a
regulated 150V supply driven by a line transformer ?
The current to be supplied is moderate, say up too
200mA.
Or are there any alternatives with an external pass
transistor / FET ?

---
How about something like this?


4W
////////
+170>--+------+------D S--------+----------+---->+150
| | G | |
| [R] | [R] [R]
| | | | |
| +--------+ +---+-----+ |
| | | | | |
|+ | | [R] | |
[BFC] | | | +| |
| D /-|--+ [C] |
| G--+--[R]---< | | | |
| S | \+|--|-----|----+
| | | | |+ | |
| | [C] | [REF] | [R]
| | | | | | |
GND>---+------+----+----------+---+-----+----+---->GND

--
John Fields

Oh, my! No consideration for all that extra gain (and likely
oscillation)!
Click to expand...
---
It's just a topology, Robert, not a finished design.

4W
////////
+170>--+------+------D S--------+----------+---->+150
| | G | |
| [R] | [R] [R]
| | | | |
| +--------+ +---+-----+ |
| | | | | |
|+ | | [R] | |
[BFC] | | | +| |
| D /-|--+ [C] |
| G--+--[R]---< | | | |
| S | \+|--|-----|----+
| | | | |+ | |
| | [C]<---+ | [REF] | [R]
| | | | | | | |
GND>---+------+----+-----|----+---+-----+----+---->GND
|
|
But,--------------> CONSIDERATION


--
John Fields
 
Frank Bemelman wrote...
Winfield Hill wrote...
Spehro Pefhany wrote...
Robert Baer wrote:
John Fields wrote:
Rene Tschaggelar wrote:

Is the LM723 still the regulator of choice for a regulated
150V supply driven by a line transformer? The current to
be supplied is moderate, say up too 200mA. Or are there
any alternatives with an external pass transistor / FET ?

How about something like this?

4W
////////
+170>--+------+------D S--------+----------+---->+150
| | G | |
| [R] | [R] [R]
| | | | |
| +--------+ +---+-----+ |
| | | | | |
|+ | | [R] | |
[BFC] | | | +| |
| D /-|--+ [C] |
| G--+--[R]---< | | | |
| S | \+|--|-----|----+
| | | | |+ | |
| | [C] | [REF] | [R]
| | | | | | |
GND>---+------+----+----------+---+-----+----+---->GND

Oh, my! No consideration for all that extra gain (and
likely oscillation)!

Likely? Guaranteed. Three poles in the feedback loop, no zeros.

Being simple minded, I would have thought that the R-C network
at the opamp's output would prevent oscillation.

BTW, what are 'poles' and 'zeros'? I tried to find an explanation
of the meaning (in context of amplifiers etc), but I can't seem to
find any. All texts seem to assume it is known to the reader.
Click to expand...
The breakpoint associated with a high frequency-response rolloff
is called a pole, and at the pole the signal path experiences a
45-degree phase lag. At higher frequencies this pole's phase lag
increases to 90 degrees. A negative feedback loop is already at
-180 degrees (hence the negative feedback), and an additional 90
makes -270. Add another full 90 degrees and the loop will become
unstable with -360 degrees of phase shift, which is the equivalent
of 0 degrees or no shift at all, reinforcing the signal endlessly,
making an oscillation. Now add a third 90 degree pole...

Let's look at the pole-zero loop compensation in my design.

.. 20W
.. ///////
.. +170>--+---+---+--+-R1--+--S D----+-----+----+---> +150V
.. | | | | | G | | | 200mA
.. | | | E R2 | | | |
.. + | | R4 B---+----|--R3--' | _|_+
.. BFC | | C | | --- C3
.. | R10 +--+----------' | | large
.. | | | ,-----+ |
.. gnd +---|-----------, C2 | | gnd
.. | | ,--R6--|-||-, R7 R8
.. Z1 | | |_ | | |
.. | C | / -|--+---|-----+
.. gnd B--+----< | | |
.. E \_+|------+ R9
.. | | |+ |
.. R5 | REF gnd
.. | gnd |
.. gnd gnd

Estimating the gain going around the loop, we'll start with the R9/R8
divider, which will have a gain of about 5/150 = 1/30. We can also
estimate the gain of the R4/R5 level-shifting stage at unity, because
that protects the FET gate from damage (the opamp saturates at 12V).

The P-channel FET pass element has gain, given by the FET's g_m times
the output load, which is primarily capacitive at high frequencies.
This is a high-gain stage with a pole at a relatively low frequency.
For example an IRF9630 FET in linear mode might have a g_m of 10A/V
at 200mA. Say the current limit resistor R1 is 2.2 ohms, that'll be
dominate over 1/g_m. Assuming our specified 200mA load at 150V acts
like a 750-ohm resistor, the FET stage gain is 750/2.2 = 340 at DC.
Let's say C3 = 10uF, then considering the 750-ohm load, that'll make
a gain pole at 21Hz.

So we have one serious pole so far; fine, and that's a stable loop.
The opamp stage has lots of gain and a pole, whose frequency we pick
with C2. However we get to cancel the opamp pole with R6. A good
use of that is to cancel the phase shift from the FET-stage pole.

There's a last significant pole, from the FET gate input capacitance
and R5. If R5 = 3.3k and Ciss = 900pF this pole will be at 53kHz,
but that'll not give us much trouble if the loop's overall bandwidth
is limited to below 50kHz. Alternately another zero can be added to
cancel this pole, for example with a 1000pF cap around R5. That will
allow us to design a faster feedback loop as we chose C2 and R6.

Another possibility is a cap around R8, but that's dangerous because
an instantaneous output short could cause a spike damaging the opamp.

See how that works? It's not so hard.


--
Thanks,
- Win
 
On Mon, 22 Nov 2004 06:09:22 -0800, Winfield Hill wrote:

The P-channel FET pass element has gain, given by the FET's g_m times
the output load, which is primarily capacitive at high frequencies.
This is a high-gain stage with a pole at a relatively low frequency.
For example an IRF9630 FET in linear mode might have a g_m of 10A/V
at 200mA. Say the current limit resistor R1 is 2.2 ohms, that'll be
^^^^^^^^^^
dominate over 1/g_m. Assuming our specified 200mA load at 150V acts
^^^^^^^^
like a 750-ohm resistor, the FET stage gain is 750/2.2 = 340 at DC.
Let's say C3 = 10uF, then considering the 750-ohm load, that'll make
a gain pole at 21Hz.
Click to expand...

I get to correct Win Hill! Yee-Ha!

"that'll dominate", or
"that'll be dominant".

:-D :-D :-D

OBTW - Absolutely AWESOME exposition - Thanks!

Cheers!
Rich Grise, self-appointed chief,
Grammar Police
 
Genome wrote...
Genome would prefer to skip that and advance straight to saying
the small-signal gain is the impedance of R6 and C2, divided by R8.

Bullfuckingshit

Fine, that's OK too.

No, it's not fucking OK.

It's bastard fucking wrong There's a big cunting fuck off and die
difference.

I learnt part of this shit from you and now you're acting like a
second grade pantyboy
Click to expand...
I didn't mean anything in a negative way, just the opposite. But
I don't get your point. If you'd calm down a bit (and slow down
for my poor slow brain), and explain in detail, I'd appreciate it.
I'm sure you have an excellent point (or I made a dumb mistake),
but I don't see it. (but please see below first)

For convenience here's the drawing again (with Tony's correction).

. 20W
. ///////
. +170>--+---+---+--+-R1--+--S D----+-----+----+---->+150
. | | | | | G | | |
. |+ | | E R2 | | | |
. BFC | R4 B---+----|--R3--' | _|_+
. | R10 | C | | --- C3
. gnd | +--+----------' | | large
. +---|-----------+--R11--+--R7--+ |
. | | | C2 | | gnd
. Z1 | ,--R6--|-||-, | R8
. | | | |_ | | |
. gnd C | / -|--+--|------+
. B--+----< | | |
. E \_+|-----+ R9
. | | |+ |
. R5 | REF gnd
. | gnd |
. gnd gnd
Click to expand...
Taking my indirect way, and leaving the R8 R9 divider separately in
place for pedagogy (for example it's necessary to prevent blowing up
the opamp, and to set the DC output), we have a gain of R9/(R8+R9),
and an output impedance of R8 R9/(R8+R9). Applying this to the opamp,
we get an ac current of 1/R8, showing that R9 is irrelevant - I assume
that was your point? Fine, OK, very good, but I still feel better
including R9 in the discussion for any newcomers. Bad idea, by you?

On the other hand, when following a signal through a summing junction,
for purposes of back-of-the-envelope calculations, I always ignore any
miscellany whatever from the summing junction to ground or from other
non-signal nodes. Later one has to come back and consider this stuff,
to evaluate the opamp's task - high frequency gain, stability, offsets,
etc. OK by you, Genome?


--
Thanks,
- Win
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top