Maintaining a Vbe Multiplier's bias value

Jim Thompson
Guest

Fri Feb 12, 2010 5:54 pm

On Fri, 12 Feb 2010 08:23:36 -0800, John Larkin
<jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:

Quote:

On Fri, 12 Feb 2010 10:23:24 -0500, Phil Hobbs
pcdhSpamMeSenseless_at_electrooptical.net> wrote:

On 2/12/2010 9:52 AM, Jim Thompson wrote:
On Thu, 11 Feb 2010 23:13:24 -0500, Phil Hobbs
pcdhSpamMeSenseless_at_electrooptical.net> wrote:
[snip]

IIRC the LM395 is basically an LM309 with the voltage reference removed.
Emitter-follower regulators are nearly bulletproof unless you
discharge a cap into the output.

Cheers

Phil Hobbs

Close, but no cigar, LM395 = LM317 with some metal rearrangements.

I was thinking 317 but then I decided that the 395 was older than that.
I guess not.

I did this analysis for ICE back in 1980:

http://analog-innovations.com/SED/ICE-LM195-LM117.pdf

Additionally: Amusing myself with the thoughts of complementary-
follower-style power amplifiers made from LM317/LM337 pairs, it fails
because both, internally, are NPN's pass devices, so the LM337 has GBW
and stability issues plus it needs substantial idle load to stay
_vaguely_ stable.

Non-LDO three-terminal regulators are so trouble-free that it's easy to
confuse them with Newton's laws. ;)

Cheers

Phil Hobbs

LM1117 is an "MDO" regulator. It has an NPN pass transistor but a bit
lower dropout voltage than an LM317. Its ideal as a 3.3-to-1.25 volt
FPGA core voltage source... no resistors! My purchasing notes say "Do
not buy Fairchild per JL" but I can't recall why.

As with all vregs, one has to be careful about the output capacitors.

John

I'm puzzled why the big semiconductor houses don't turn out discrete
LDO's on a CMOS process. I do it all the time on complex CMOS
ASIC's... like a PLL chip, fundamental power fed from +3.3V, but
internal regulators producing +2.5V and +1.8V... at hundreds of mA !!!

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

George Herold
Guest

Fri Feb 12, 2010 6:16 pm

On Feb 12, 9:52 am, Jim Thompson <To-Email-Use-The-Envelope-I...@My-
Web-Site.com> wrote:

Quote:

On Thu, 11 Feb 2010 23:13:24 -0500, Phil Hobbs

pcdhSpamMeSensel...@electrooptical.net> wrote:
On 2/11/2010 10:35 PM, George Herold wrote:
On Feb 11, 1:14 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 11 Feb 2010 10:08:38 -0600, "Tim Williams"

tmoran...@charter.net> wrote:
"Jim Thompson"<To-Email-Use-The-Envelope-I...@My-Web-Site.com> wrote in
messagenews:r358n59g5vkv4brn2vc795lhoineb2jvhd_at_4ax.com...
And a 2.5V "dead-band", but it _is_ precisely known, and temperature
stable. Interesting thought if you have high enough power supplies.

Bonus: the dead band allows you to use that TL431 "Vbe" mentinoed earlier.

Too bad they're so slow (hardly capable for audio). Does anyone make "fast"
regulators (without being stupid LDOs)?

Tim

If you drive both adjust pins with the signal input, the 317 output is
Vin+1.25 and the 337 output is Vin-1.25. Connect them to the output
through a couple of resistors, valued to set the idle current. Where's
the deadband?

Or you can take the output from the 317 output pin, with the 337 now
acting like a constant-current sink to the 317.

I like to use LM1117s as power emitter followers, inside the loop of
an opamp. That makes a cheap, well protected power driver, for load
cell excitation and such. I did a bunch of tests to see whether
flailing the adj pin can damage the regulator, and never managed to
break one.

John

Cool! I think I got it... though if I try it in the future and let
the smoke out of something... then I might have questions.

George H.

IIRC the LM395 is basically an LM309 with the voltage reference removed.
Emitter-follower regulators are nearly bulletproof unless you
discharge a cap into the output.

Cheers

Phil Hobbs

Close, but no cigar, LM395 = LM317 with some metal rearrangements.

I did this analysis for ICE back in 1980:

http://analog-innovations.com/SED/ICE-LM195-LM117.pdf

Additionally: Amusing myself with the thoughts of complementary-
follower-style power amplifiers made from LM317/LM337 pairs, it fails
because both, internally, are NPN's pass devices, so the LM337 has GBW
and stability issues plus it needs substantial idle load to stay
_vaguely_ stable.

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon athttp://www.analog-innovations.com| 1962 |

I love to cook with wine. Sometimes I even put it in the food.- Hide quoted text -

- Show quoted text -

Thanks Jim, I always wondered why there wasn't a compliment to the
LM395. (And why the negative voltage regulators would 'wig out' if
you didn't follow the cap reccomendations.)

George H.

George Herold
Guest

Fri Feb 12, 2010 6:27 pm

On Feb 12, 10:03 am, Phil Hobbs
<pcdhSpamMeSensel...@electrooptical.net> wrote:

Quote:

On 2/12/2010 9:36 AM, Jim Thompson wrote:

On Thu, 11 Feb 2010 19:37:27 -0800, Jon Kirwan
j...@infinitefactors.org> wrote:

On Thu, 11 Feb 2010 19:23:04 -0800 (PST), George Herold
ggher...@gmail.com> wrote:

On Feb 11, 3:24 am, Jon Kirwan<j...@infinitefactors.org> wrote:
On Wed, 10 Feb 2010 19:55:44 -0800 (PST), George Herold

ggher...@gmail.com> wrote:
I'm perhaps more of a novice than you...

Somehow, I doubt that. I barely rate "hobbyist."

but I find opamp circuits complicated enough....

Don't sweat it. While some opamps leave _some_ issues nearly
ignorable, there is always some tough problem at that scale
that makes it non-trivial and interesting to work on, I
imagine. Each macroscale view has it's own complexity.
Telescoping levels, where the complexity at one stage doesn't
take away from interesting complexity at another level.

(I would say more but I'm keeping in mind your warning about
long-windedness and will now muzzle myself.)

And tend to stick transistors
only on the edges of things. (Mostly on the output side... on the
input you have to 'know more' than the guys who designed the opamp...
hard to do for a novice.)

I do the same things except that I enjoy math and BJTs give
me an excuse, perhaps. Maybe that's the only difference.

I guess if I was designing an audio amp I'd figure on an opamp driving
some sort of FET output stage. The question of how to bias the output
stage is interesting. And also of how all the NFB works.

Might as well just get a power opamp like the OPA502 and be
done with it. Give it two rails, feed the input, and just
drive the hell out of a speaker. Or get two of them and do a
bridge amplifier. But where is the enjoyment in that? Or
the learning? Someone else already did most of the fun stuff
and there's nothing really left to do except some hook up and
heat sinking. It's not at all satisfying to me, anyway.

An audio amplifier is basically a power opamp. Using an
opamp to make one feels to me like building a car by first
buying a car without the tires, selecting and installing some
tires, and then saying you designed and built yourself a car.

Jon

Big Grins!

Yeah I applaud your effort, I wait for further posts.

For me, I’m building electronics to either detect something or drive
something that’s detecting something. So the fun is in making good
detectors or drivers.

George H.

Well, I am wanting, eventually, to build something I need.
Something I cannot buy in the market because the need is
unique.

This divides into two parts. Design and build. Since the
item is unique, I can't just go out and buy it. And getting
the features I need cannot just be "hacked" into existing
designs without at least knowing _some_ stuff, first. I
might as well turn the "design" part into a fair learning
experience, as a separate project of its own. Get past that
and when it comes time to build what I want I'll be able to
build on what I learned and add what I need and then do a
modest hobbyist level whack at actually making what I want to
make.

If someone else were to do this for me (hire a designer),
they'd get all the fun of learning on the job and taking my
money with it. They get the money, they get to further their
own education, and I get a tool. One tool. Once. Next
time, I get to pay someone else to learn for me.

It almost feels like paying someone to go do your exercising
for you. No satisfaction and no weight loss. They get all
the _real_ benefits.

Part of the fun isn't the destination itself but it is what
you see and enjoy while getting there, too. You take a plane
when all you need is to "get there" quick, but you drive when
you want to enjoy stops along the way. I used to fly to
Burbank every week for a year and a half. Slept in a hotel
for 3 nights a week, worked day and night in between, flew
home. Barely saw anything but hotel room walls, cubical
walls, a few cement roads, pollution so thick you couldn't
see the Burbank hills from the Lockheed center, and not much
else. The destination was important, of course. Paid the
bills and I enjoyed the work, too. But there is a lot more
to see in the 1000 miles from here to there.

Anyway, I'm driving this time, not flying.

Besides, I'd rather _keep_ the money and _keep_ the education
for myself. That way it pays off, again and again.

Jon

The fish/fish rule ?:-)

...Jim Thompson

Build a man a fire, you keep him warm for a night. Set a man afire, you
keep him warm for the rest of his life. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs at electrooptical dot nethttp://electrooptical.net- Hide quoted text -

- Show quoted text -

"> Build a man a fire, you keep him warm for a night. Set a man
afire, you

Quote:

keep him warm for the rest of his life. Wink

"

Can't help being reminded of "The Cremation of Sam McGee".

And there sat Sam, looking cool and calm, in the heart of the furnace
roar;
And he wore a smile you could see a mile, and he said: "Please close
that door.
It's fine in here, but I greatly fear you'll let in the cold and storm
—
Since I left Plumtree, down in Tennessee, it's the first time I've
been warm."

Taken from here,
"http://en.wikisource.org/wiki/The_Cremation_of_Sam_McGee"

George h.

Jon Kirwan
Guest

Fri Feb 12, 2010 8:02 pm

On Fri, 12 Feb 2010 07:36:52 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon_at_My-Web-Site.com> wrote:

Quote:

On Thu, 11 Feb 2010 19:37:27 -0800, Jon Kirwan
jonk_at_infinitefactors.org> wrote:

snip

Well, I am wanting, eventually, to build something I need.
Something I cannot buy in the market because the need is
unique.

This divides into two parts. Design and build. Since the
item is unique, I can't just go out and buy it. And getting
the features I need cannot just be "hacked" into existing
designs without at least knowing _some_ stuff, first. I
might as well turn the "design" part into a fair learning
experience, as a separate project of its own. Get past that
and when it comes time to build what I want I'll be able to
build on what I learned and add what I need and then do a
modest hobbyist level whack at actually making what I want to
make.

If someone else were to do this for me (hire a designer),
they'd get all the fun of learning on the job and taking my
money with it. They get the money, they get to further their
own education, and I get a tool. One tool. Once. Next
time, I get to pay someone else to learn for me.

It almost feels like paying someone to go do your exercising
for you. No satisfaction and no weight loss. They get all
the _real_ benefits.

Part of the fun isn't the destination itself but it is what
you see and enjoy while getting there, too. You take a plane
when all you need is to "get there" quick, but you drive when
you want to enjoy stops along the way. I used to fly to
Burbank every week for a year and a half. Slept in a hotel
for 3 nights a week, worked day and night in between, flew
home. Barely saw anything but hotel room walls, cubical
walls, a few cement roads, pollution so thick you couldn't
see the Burbank hills from the Lockheed center, and not much
else. The destination was important, of course. Paid the
bills and I enjoyed the work, too. But there is a lot more
to see in the 1000 miles from here to there.

Anyway, I'm driving this time, not flying.

Besides, I'd rather _keep_ the money and _keep_ the education
for myself. That way it pays off, again and again.

Jon

The fish/fish rule ? Smile

...Jim Thompson

Almost. It's more like: I'd like to get to Fresno Flats
(aka Oakhurst, near Yosemite) from Portland, but rather than
just fly there (okay, so I have to parachute out) I'd like to
enjoy the trip this time, too, and detour around a bit to
some of the sights along the way.

First off, maybe to the Opal Creek Wilderness area to see
some of the 800 year old cedar groves; then maybe south to a
short visit at White's Electronics in Sweet Home; then to
Crater Lake; maybe then to Oregon Caves; off perhaps to do
some panning for gold in Jackson County while also taking a
hike in the Siskiyous; then on over to hwy 101 at Brookings
to do some whale watching; down to Crescent City to visit the
Jedediah Smith and Del Norte parks and see what's left of the
redwoods there; ....

Well, might as well make a thing of it this time 'round.
Oakhurst won't go away while I'm on the road.

Jon

John Larkin
Guest

Fri Feb 12, 2010 9:18 pm

On Fri, 12 Feb 2010 08:16:34 -0800 (PST), George Herold
<ggherold_at_gmail.com> wrote:

Quote:

On Feb 12, 9:52 am, Jim Thompson <To-Email-Use-The-Envelope-I...@My-
Web-Site.com> wrote:
On Thu, 11 Feb 2010 23:13:24 -0500, Phil Hobbs

pcdhSpamMeSensel...@electrooptical.net> wrote:
On 2/11/2010 10:35 PM, George Herold wrote:
On Feb 11, 1:14 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 11 Feb 2010 10:08:38 -0600, "Tim Williams"

tmoran...@charter.net> wrote:
"Jim Thompson"<To-Email-Use-The-Envelope-I...@My-Web-Site.com> wrote in
messagenews:r358n59g5vkv4brn2vc795lhoineb2jvhd_at_4ax.com...
And a 2.5V "dead-band", but it _is_ precisely known, and temperature
stable. Interesting thought if you have high enough power supplies.

Bonus: the dead band allows you to use that TL431 "Vbe" mentinoed earlier.

Too bad they're so slow (hardly capable for audio). Does anyone make "fast"
regulators (without being stupid LDOs)?

Tim

If you drive both adjust pins with the signal input, the 317 output is
Vin+1.25 and the 337 output is Vin-1.25. Connect them to the output
through a couple of resistors, valued to set the idle current. Where's
the deadband?

Or you can take the output from the 317 output pin, with the 337 now
acting like a constant-current sink to the 317.

I like to use LM1117s as power emitter followers, inside the loop of
an opamp. That makes a cheap, well protected power driver, for load
cell excitation and such. I did a bunch of tests to see whether
flailing the adj pin can damage the regulator, and never managed to
break one.

John

Cool! I think I got it... though if I try it in the future and let
the smoke out of something... then I might have questions.

George H.

IIRC the LM395 is basically an LM309 with the voltage reference removed.
Emitter-follower regulators are nearly bulletproof unless you
discharge a cap into the output.

Cheers

Phil Hobbs

Close, but no cigar, LM395 = LM317 with some metal rearrangements.

I did this analysis for ICE back in 1980:

http://analog-innovations.com/SED/ICE-LM195-LM117.pdf

Additionally: Amusing myself with the thoughts of complementary-
follower-style power amplifiers made from LM317/LM337 pairs, it fails
because both, internally, are NPN's pass devices, so the LM337 has GBW
and stability issues plus it needs substantial idle load to stay
_vaguely_ stable.

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon athttp://www.analog-innovations.com| 1962 |

I love to cook with wine. Sometimes I even put it in the food.- Hide quoted text -

- Show quoted text -

Thanks Jim, I always wondered why there wasn't a compliment to the
LM395. (And why the negative voltage regulators would 'wig out' if
you didn't follow the cap reccomendations.)

George H.

There are lots of positive LDOs around these days that have PNP
(namely high impedance) outputs. They tend to be very picky about
output capacitance ESR and such, with appropriately evasive
datasheets. I had one powering an FPGA core, putting a 0.2 volt p-p,
50 KHz triangle into the poor chip. That caused a lot of jitter.

John

Jim Thompson
Guest

Fri Feb 12, 2010 9:30 pm

On Fri, 12 Feb 2010 12:18:33 -0800, John Larkin
<jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:

Quote:

On Fri, 12 Feb 2010 08:16:34 -0800 (PST), George Herold
ggherold_at_gmail.com> wrote:

On Feb 12, 9:52 am, Jim Thompson <To-Email-Use-The-Envelope-I...@My-
Web-Site.com> wrote:
On Thu, 11 Feb 2010 23:13:24 -0500, Phil Hobbs

pcdhSpamMeSensel...@electrooptical.net> wrote:
On 2/11/2010 10:35 PM, George Herold wrote:
On Feb 11, 1:14 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 11 Feb 2010 10:08:38 -0600, "Tim Williams"

tmoran...@charter.net> wrote:
"Jim Thompson"<To-Email-Use-The-Envelope-I...@My-Web-Site.com> wrote in
messagenews:r358n59g5vkv4brn2vc795lhoineb2jvhd_at_4ax.com...
And a 2.5V "dead-band", but it _is_ precisely known, and temperature
stable. Interesting thought if you have high enough power supplies.

Bonus: the dead band allows you to use that TL431 "Vbe" mentinoed earlier.

Too bad they're so slow (hardly capable for audio). Does anyone make "fast"
regulators (without being stupid LDOs)?

Tim

If you drive both adjust pins with the signal input, the 317 output is
Vin+1.25 and the 337 output is Vin-1.25. Connect them to the output
through a couple of resistors, valued to set the idle current. Where's
the deadband?

Or you can take the output from the 317 output pin, with the 337 now
acting like a constant-current sink to the 317.

I like to use LM1117s as power emitter followers, inside the loop of
an opamp. That makes a cheap, well protected power driver, for load
cell excitation and such. I did a bunch of tests to see whether
flailing the adj pin can damage the regulator, and never managed to
break one.

John

Cool! I think I got it... though if I try it in the future and let
the smoke out of something... then I might have questions.

George H.

IIRC the LM395 is basically an LM309 with the voltage reference removed.
Emitter-follower regulators are nearly bulletproof unless you
discharge a cap into the output.

Cheers

Phil Hobbs

Close, but no cigar, LM395 = LM317 with some metal rearrangements.

I did this analysis for ICE back in 1980:

http://analog-innovations.com/SED/ICE-LM195-LM117.pdf

Additionally: Amusing myself with the thoughts of complementary-
follower-style power amplifiers made from LM317/LM337 pairs, it fails
because both, internally, are NPN's pass devices, so the LM337 has GBW
and stability issues plus it needs substantial idle load to stay
_vaguely_ stable.

...Jim Thompson
[snip]
- Show quoted text -

Thanks Jim, I always wondered why there wasn't a compliment to the
LM395. (And why the negative voltage regulators would 'wig out' if
you didn't follow the cap reccomendations.)

George H.

There are lots of positive LDOs around these days that have PNP
(namely high impedance) outputs. They tend to be very picky about
output capacitance ESR and such, with appropriately evasive
datasheets. I had one powering an FPGA core, putting a 0.2 volt p-p,
50 KHz triangle into the poor chip. That caused a lot of jitter.

John

bg
Guest

Fri Feb 12, 2010 9:43 pm

big snip -

This is an example of a common emitter voltage amplifier. It might be one of
the easiest stages to design. Normaly, I would start off by knowing what I
need for the stage, for example I need a voltage gain of 20, an input
impedance of 10K, Z out of ?? etc ----

An important thing to notice is that I do not labor over any tedious
calculations. All of the part values are based on simple resistor ratios.
Knowing how the ratios work, and how they affect each other, allows me to
adust just about any parameter with a quick mental estimate.

Example - - - -
The circuit is based on the 2n3904 again, and there are probably hundreds if
not thousands of other transistors that could fit in this circuit and work
just as well.
The voltage gain of a common emitter stage is reduced to RL/ RE. The
emitter resistor, RE , has to provide enough feedback for this to be true. I
can't Make RE = 1 ohm and Rl = 1megohm and have a gain of 1 million. There
are limits.

If I intended to run a CE amp with a 1 ma collector Q point current, and
with a 15 volt supply, I might choose 7.5K for RL which would put Vc at
Vcc/2. I know there will be an emitter resistor for feedback, and because
there is a voltage drop across RE, I will want to raise the collector Q
point voltage a bit higher to achieve the largest signal swing possible. For
a starting point, I'll make RL = 6.8k and RE 680 ohms. When the transistor
is cut off Vc = 15 volts. When the transistor is saturated RL and RE form a
voltage divider across the 15 volt supply, therefore Vc = 1.4 volts. The
midpoint between 15v and 1.4v comes out to 8.2 volts.
Notice I neglected Vce when the transistor is saturated. The overall picture
is more important at this point than details. I now can expect a voltage
gain of 10, a collector Q point of 8.2v at 1ma, and an output resistance of
about 6.8k.
Up to this point, here is the sum total of calculations -
The 1ma collector current was a choice because I know than a 2n3904 is
intended to operate at a collector current in that range.
I want a voltage gain of ten, no particluar reason, again it was a choice.
Re = RL/10 I do that in my head.
I did use a calculator to arrive at the desired Q point voltage.

The next step -
The resistor RB which connects between the base and ground needs to be small
compared to the input impedance of the transistor. I know from experience ,
and also because I read the book "transistor circuit design" , that RB
should typically be about 10 to 20 times larger than RE. So I'll go with
6.8K. I can do this because I know that the input of the transistor will
appear as an open circuit compared to RB. It isn't any different than a
simple voltage divider. For example if I had a divider using two 1K
resistors, I know that any load of 10K or higher isn't going to
significantly change the divider ratio. I'll still get a division of about
1/2 right? As the load dips below 10K, the output of the divider will be
less. Gut feel, and how much accuracy I'm willing to loose, determines if I
need to crunch some numbers.
The only calculation I made for RB is - - RB equals 10 to 20 RE

The next step, is to select the forward bias resistor RA that connects
between the base and Vcc.
For this resistor I made an educated guess and hit the exact value on the
very first shot. I knew that the ratio of Rb to Ra would be close to 10 to
1.
68k is a standard value so I chose that and expected to tweak it later. But
in all fairness, this is how I arrived at that guess.
The voltage across RB should be Vbe + Vre. We know Vre is 1ma through 680
ohms, and notice that I neglected the base current flowing through RE
becuase IE probably equals Ic + 1%Ic, the one percent doesn't matter. Vbe is
a tough one to call. Typically it ranges from .5 to .7 volts. Seeing as I
will probably miss the target, and I will want to use a standard value, I'll
do the math, see where I come out and tweak values if neccessary. The math
tells me that the current through RA is the same current as through RB +
Ib. The current through RB is Vbe + Vre / Rb = about 1.6 volts/6.8k = 235
micro amp. Notice that I neglected Ib, and I can because I expect it to
insignificant.
Therefore RA drops 15 volts - 1.6 volts at 235 micro amp which works out to
57k.

The next step
If you put all of this into circuitmaker , the collector Q point comes in at
6.2 volts if Ra = 57k. An obvious tweak is to raise Ra to the next standard
value, but for the model I used, Ra works out to be 68k for a collector Q
point of 8.37v which was my first guess.

The beauty of this is that I can scale Ra and Rb up or down so that standard
values can be used. As long as I keep Rb around 10RE.

Circuitmaker shows that the input impedance is 5.8k up until the input
capacitance becomes an issue. My 6.8k Rb is being shunted by Ra and the
transistor input impedance so I would expect the input Z to be less than
6.8K. The parallel equivalent of Ra and Rb is a hair under 6.8k, therefore
the majority of the shunting effect is due to the transistor. I could quite
easily lower Ra and Rb to make the transistor input Z even less noticeable.
Circuit maker also confirms that the output resistance is 6.8k and the
voltage gain is 19.571db (10x would be 20db). The peak to peak output signal
clips around 13 volts.

You might have noticed that I never mentioned Beta. Over the range of
collector current that this stage operates , I can expect beta to be alot
higher than my target voltage gain. I have an open loop current gain
probably 10 times higher than my voltage gain with feedback. If I consider
the equation A' = A/ 1+AB, I could solve for B or just plug numbers in to
see that the stage gain is controlled by the feedback, and as long as Beta
remains fairly high, it is not a factor. From experience, I know that my
stage voltage gain will be much lower (due to feedback), than the transistor
current gain. Therefore I can ignore the transistor current gain.

As a voltage amplifier for low frequencies, I would expect that my
preceeding stage will need an output resistance about 10 times lower than
this stage's input resistance. The preceeding collector load resistor would
need to be 680 ohms or less. The following stage would need an input
impedance of 10 times this stage's output impedance or 68k. Those values are
what I would be aiming for if I was putting this stage between two voltage
amplifiers.

Summary -
#1
Select a transistor that is meant to operate at the collector current for
which you intend to operate.
The collector load resistor IS the output impedance. That value combined
with the collector Q point current determines the Q point voltage. Any value
can be varied to zero in on the best compramise of Zout, Ic, RL, Vc etc.

#2
Select your voltage gain with, Gain = RL/RE
Again vary any value to achieve whatever compramise best suits your
situation.

#3
Consider what the peak to peak output signal needs to be. Make sure that
your Vcc and Q point can accomodate that signal swing.
Your peak to peak output signal is found by considering the collector
voltage when the transistor is open, and then when it is shorted. To avoid
clipping at the signal peaks, Vcc should be high enough so that the signal
does not need to swing to those extremes. In other words , leave some
headroom. More headroom will avoid distorting at the signal peaks. It is not
uncommon for Vcc to be 50% or higher than the peak to peak signal swing. The
only limit is that the transistor's maximum power dissapation or voltage
rating might be exceeded.

#4
Revisit steps one through three. These steps determine the voltage gain and
the maximum output signal that can be ahieved

#5
Select a value for Rb. Rb should be low compared to Re. Rb will be close to
the input impedance of your amp so consider that too. As you raise the value
of Rb, the transistor shunting effect becomes more significant and the input
Z of the amp becomes less than Rb. Again it is a compramise. There is no
limit for how low you go with Rb, but there would be benifits to using
something like a common base amplifier instead of a common emitter amplifier
if Rb needs to be very low.

#6
Select a value for Ra. This is best done by considering the bias voltage
needed across Rb. If an exact figure for VBE is not known, and it seldom is
you can always assume .6 volts, tweak later. The voltage across RB is .6
volts + the voltage across the emitter resistor. Ohms law will get you the
current in Rb and from there Ra will need to drop Vcc - Erb at Irb amps.
Bear in mind that ending up with standard values is a plus. Rb and Ra can be
tweaked to get the best compramise. As those values are varied to standard
values, the collector Q point voltage will shift, as will the input
impedance.

There are other compramises that affect frequency response, noise, power
dissapation, maximum ratings and so on. This is meant to show only how to
juggle values around without resorting to tedious calculations, to achieve
the DC operating point, set input and output z's, voltage gain and signal
swing peak levels.

Jim Thompson
Guest

Fri Feb 12, 2010 9:44 pm

On Fri, 12 Feb 2010 13:43:37 -0700, "bg" <bg_at_nospam.com> wrote:

Quote:

big snip -

This is an example of a common emitter voltage amplifier. It might be one of
the easiest stages to design. Normaly, I would start off by knowing what I
need for the stage, for example I need a voltage gain of 20, an input
impedance of 10K, Z out of ?? etc ----

An important thing to notice is that I do not labor over any tedious
calculations. All of the part values are based on simple resistor ratios.
Knowing how the ratios work, and how they affect each other, allows me to
adust just about any parameter with a quick mental estimate.

Example - - - -
The circuit is based on the 2n3904 again, and there are probably hundreds if
not thousands of other transistors that could fit in this circuit and work
just as well.
The voltage gain of a common emitter stage is reduced to RL/ RE. The
emitter resistor, RE , has to provide enough feedback for this to be true. I
can't Make RE = 1 ohm and Rl = 1megohm and have a gain of 1 million. There
are limits.

If I intended to run a CE amp with a 1 ma collector Q point current, and
with a 15 volt supply, I might choose 7.5K for RL which would put Vc at
Vcc/2. I know there will be an emitter resistor for feedback, and because
there is a voltage drop across RE, I will want to raise the collector Q
point voltage a bit higher to achieve the largest signal swing possible. For
a starting point, I'll make RL = 6.8k and RE 680 ohms. When the transistor
is cut off Vc = 15 volts. When the transistor is saturated RL and RE form a
voltage divider across the 15 volt supply, therefore Vc = 1.4 volts. The
midpoint between 15v and 1.4v comes out to 8.2 volts.
Notice I neglected Vce when the transistor is saturated. The overall picture
is more important at this point than details. I now can expect a voltage
gain of 10, a collector Q point of 8.2v at 1ma, and an output resistance of
about 6.8k.
Up to this point, here is the sum total of calculations -
The 1ma collector current was a choice because I know than a 2n3904 is
intended to operate at a collector current in that range.
I want a voltage gain of ten, no particluar reason, again it was a choice.
Re = RL/10 I do that in my head.
I did use a calculator to arrive at the desired Q point voltage.

The next step -
The resistor RB which connects between the base and ground needs to be small
compared to the input impedance of the transistor. I know from experience ,
and also because I read the book "transistor circuit design" , that RB
should typically be about 10 to 20 times larger than RE. So I'll go with
6.8K. I can do this because I know that the input of the transistor will
appear as an open circuit compared to RB. It isn't any different than a
simple voltage divider. For example if I had a divider using two 1K
resistors, I know that any load of 10K or higher isn't going to
significantly change the divider ratio. I'll still get a division of about
1/2 right? As the load dips below 10K, the output of the divider will be
less. Gut feel, and how much accuracy I'm willing to loose, determines if I
need to crunch some numbers.
The only calculation I made for RB is - - RB equals 10 to 20 RE

The next step, is to select the forward bias resistor RA that connects
between the base and Vcc.
For this resistor I made an educated guess and hit the exact value on the
very first shot. I knew that the ratio of Rb to Ra would be close to 10 to
1.
68k is a standard value so I chose that and expected to tweak it later. But
in all fairness, this is how I arrived at that guess.
The voltage across RB should be Vbe + Vre. We know Vre is 1ma through 680
ohms, and notice that I neglected the base current flowing through RE
becuase IE probably equals Ic + 1%Ic, the one percent doesn't matter. Vbe is
a tough one to call. Typically it ranges from .5 to .7 volts. Seeing as I
will probably miss the target, and I will want to use a standard value, I'll
do the math, see where I come out and tweak values if neccessary. The math
tells me that the current through RA is the same current as through RB +
Ib. The current through RB is Vbe + Vre / Rb = about 1.6 volts/6.8k = 235
micro amp. Notice that I neglected Ib, and I can because I expect it to
insignificant.
Therefore RA drops 15 volts - 1.6 volts at 235 micro amp which works out to
57k.

The next step
If you put all of this into circuitmaker , the collector Q point comes in at
6.2 volts if Ra = 57k. An obvious tweak is to raise Ra to the next standard
value, but for the model I used, Ra works out to be 68k for a collector Q
point of 8.37v which was my first guess.

The beauty of this is that I can scale Ra and Rb up or down so that standard
values can be used. As long as I keep Rb around 10RE.

Circuitmaker shows that the input impedance is 5.8k up until the input
capacitance becomes an issue. My 6.8k Rb is being shunted by Ra and the
transistor input impedance so I would expect the input Z to be less than
6.8K. The parallel equivalent of Ra and Rb is a hair under 6.8k, therefore
the majority of the shunting effect is due to the transistor. I could quite
easily lower Ra and Rb to make the transistor input Z even less noticeable.
Circuit maker also confirms that the output resistance is 6.8k and the
voltage gain is 19.571db (10x would be 20db). The peak to peak output signal
clips around 13 volts.

You might have noticed that I never mentioned Beta. Over the range of
collector current that this stage operates , I can expect beta to be alot
higher than my target voltage gain. I have an open loop current gain
probably 10 times higher than my voltage gain with feedback. If I consider
the equation A' = A/ 1+AB, I could solve for B or just plug numbers in to
see that the stage gain is controlled by the feedback, and as long as Beta
remains fairly high, it is not a factor. From experience, I know that my
stage voltage gain will be much lower (due to feedback), than the transistor
current gain. Therefore I can ignore the transistor current gain.

As a voltage amplifier for low frequencies, I would expect that my
preceeding stage will need an output resistance about 10 times lower than
this stage's input resistance. The preceeding collector load resistor would
need to be 680 ohms or less. The following stage would need an input
impedance of 10 times this stage's output impedance or 68k. Those values are
what I would be aiming for if I was putting this stage between two voltage
amplifiers.

Summary -
#1
Select a transistor that is meant to operate at the collector current for
which you intend to operate.
The collector load resistor IS the output impedance. That value combined
with the collector Q point current determines the Q point voltage. Any value
can be varied to zero in on the best compramise of Zout, Ic, RL, Vc etc.

#2
Select your voltage gain with, Gain = RL/RE
Again vary any value to achieve whatever compramise best suits your
situation.

#3
Consider what the peak to peak output signal needs to be. Make sure that
your Vcc and Q point can accomodate that signal swing.
Your peak to peak output signal is found by considering the collector
voltage when the transistor is open, and then when it is shorted. To avoid
clipping at the signal peaks, Vcc should be high enough so that the signal
does not need to swing to those extremes. In other words , leave some
headroom. More headroom will avoid distorting at the signal peaks. It is not
uncommon for Vcc to be 50% or higher than the peak to peak signal swing. The
only limit is that the transistor's maximum power dissapation or voltage
rating might be exceeded.

#4
Revisit steps one through three. These steps determine the voltage gain and
the maximum output signal that can be ahieved

#5
Select a value for Rb. Rb should be low compared to Re. Rb will be close to
the input impedance of your amp so consider that too. As you raise the value
of Rb, the transistor shunting effect becomes more significant and the input
Z of the amp becomes less than Rb. Again it is a compramise. There is no
limit for how low you go with Rb, but there would be benifits to using
something like a common base amplifier instead of a common emitter amplifier
if Rb needs to be very low.

#6
Select a value for Ra. This is best done by considering the bias voltage
needed across Rb. If an exact figure for VBE is not known, and it seldom is
you can always assume .6 volts, tweak later. The voltage across RB is .6
volts + the voltage across the emitter resistor. Ohms law will get you the
current in Rb and from there Ra will need to drop Vcc - Erb at Irb amps.
Bear in mind that ending up with standard values is a plus. Rb and Ra can be
tweaked to get the best compramise. As those values are varied to standard
values, the collector Q point voltage will shift, as will the input
impedance.

There are other compramises that affect frequency response, noise, power
dissapation, maximum ratings and so on. This is meant to show only how to
juggle values around without resorting to tedious calculations, to achieve
the DC operating point, set input and output z's, voltage gain and signal
swing peak levels.

Bob Monsen
Guest

Fri Feb 12, 2010 10:21 pm

"Jim Thompson" <To-Email-Use-The-Envelope-Icon_at_My-Web-Site.com> wrote in
message news:0m1bn5ln6ag64hdqnmr4te1qjeiec22i02_at_4ax.com...

Quote:

On Fri, 12 Feb 2010 08:23:36 -0800, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 12 Feb 2010 10:23:24 -0500, Phil Hobbs
pcdhSpamMeSenseless_at_electrooptical.net> wrote:

On 2/12/2010 9:52 AM, Jim Thompson wrote:
On Thu, 11 Feb 2010 23:13:24 -0500, Phil Hobbs
pcdhSpamMeSenseless_at_electrooptical.net> wrote:
[snip]

IIRC the LM395 is basically an LM309 with the voltage reference
removed.
Emitter-follower regulators are nearly bulletproof unless you
discharge a cap into the output.

Cheers

Phil Hobbs

Close, but no cigar, LM395 = LM317 with some metal rearrangements.

I was thinking 317 but then I decided that the 395 was older than that.
I guess not.

I did this analysis for ICE back in 1980:

http://analog-innovations.com/SED/ICE-LM195-LM117.pdf

Additionally: Amusing myself with the thoughts of complementary-
follower-style power amplifiers made from LM317/LM337 pairs, it fails
because both, internally, are NPN's pass devices, so the LM337 has GBW
and stability issues plus it needs substantial idle load to stay
_vaguely_ stable.

Non-LDO three-terminal regulators are so trouble-free that it's easy to
confuse them with Newton's laws. ;)

Cheers

Phil Hobbs

LM1117 is an "MDO" regulator. It has an NPN pass transistor but a bit
lower dropout voltage than an LM317. Its ideal as a 3.3-to-1.25 volt
FPGA core voltage source... no resistors! My purchasing notes say "Do
not buy Fairchild per JL" but I can't recall why.

As with all vregs, one has to be careful about the output capacitors.

John

I'm puzzled why the big semiconductor houses don't turn out discrete
LDO's on a CMOS process. I do it all the time on complex CMOS
ASIC's... like a PLL chip, fundamental power fed from +3.3V, but
internal regulators producing +2.5V and +1.8V... at hundreds of mA !!!

ST Arm Cortex-M3 has an internal 1.8V regulator for the core, and can take
any input voltage from 2.0 to 3.8V.

Regards,
Bob Monsen

bg
Guest

Fri Feb 12, 2010 10:25 pm

Jim Thompson wrote in message ...

Quote:

On Fri, 12 Feb 2010 13:43:37 -0700, "bg" <bg_at_nospam.com> wrote:

big snip -

Son of a gun. Yesterday you couldn't even spell "engineer", now you
are one (;-0

I finally did something right? Wow!!!

Jim Thompson
Guest

Fri Feb 12, 2010 10:38 pm

On Fri, 12 Feb 2010 14:25:06 -0700, "bg" <bg_at_nospam.com> wrote:

Quote:

Jim Thompson wrote in message ...
On Fri, 12 Feb 2010 13:43:37 -0700, "bg" <bg_at_nospam.com> wrote:

big snip -

Son of a gun. Yesterday you couldn't even spell "engineer", now you
are one (;-0

I finally did something right? Wow!!!

Actually you found the "art" part needed to rough-in a design. Now go
back and fix the flaws Wink

Phil Hobbs
Guest

Fri Feb 12, 2010 10:42 pm

On 2/12/2010 11:24 AM, John Larkin wrote:

Quote:

On Fri, 12 Feb 2010 10:03:11 -0500, Phil Hobbs
pcdhSpamMeSenseless_at_electrooptical.net> wrote:

Build a man a fire, you keep him warm for a night. Set a man afire, you
keep him warm for the rest of his life. ;)

Cheers

Phil Hobbs

Got snow?

John

About a foot, day before yesterday. 45 and sunny yesterday and today.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs at electrooptical dot net
http://electrooptical.net

Phil Hobbs
Guest

Fri Feb 12, 2010 10:43 pm

On 2/12/2010 11:27 AM, George Herold wrote:

Quote:

On Feb 12, 10:03 am, Phil Hobbs
pcdhSpamMeSensel...@electrooptical.net> wrote:
On 2/12/2010 9:36 AM, Jim Thompson wrote:

On Thu, 11 Feb 2010 19:37:27 -0800, Jon Kirwan
j...@infinitefactors.org> wrote:

On Thu, 11 Feb 2010 19:23:04 -0800 (PST), George Herold
ggher...@gmail.com> wrote:

On Feb 11, 3:24 am, Jon Kirwan<j...@infinitefactors.org> wrote:
On Wed, 10 Feb 2010 19:55:44 -0800 (PST), George Herold

ggher...@gmail.com> wrote:
I'm perhaps more of a novice than you...

Somehow, I doubt that. I barely rate "hobbyist."

but I find opamp circuits complicated enough....

Don't sweat it. While some opamps leave _some_ issues nearly
ignorable, there is always some tough problem at that scale
that makes it non-trivial and interesting to work on, I
imagine. Each macroscale view has it's own complexity.
Telescoping levels, where the complexity at one stage doesn't
take away from interesting complexity at another level.

(I would say more but I'm keeping in mind your warning about
long-windedness and will now muzzle myself.)

And tend to stick transistors
only on the edges of things. (Mostly on the output side... on the
input you have to 'know more' than the guys who designed the opamp...
hard to do for a novice.)

I do the same things except that I enjoy math and BJTs give
me an excuse, perhaps. Maybe that's the only difference.

I guess if I was designing an audio amp I'd figure on an opamp driving
some sort of FET output stage. The question of how to bias the output
stage is interesting. And also of how all the NFB works.

Might as well just get a power opamp like the OPA502 and be
done with it. Give it two rails, feed the input, and just
drive the hell out of a speaker. Or get two of them and do a
bridge amplifier. But where is the enjoyment in that? Or
the learning? Someone else already did most of the fun stuff
and there's nothing really left to do except some hook up and
heat sinking. It's not at all satisfying to me, anyway.

An audio amplifier is basically a power opamp. Using an
opamp to make one feels to me like building a car by first
buying a car without the tires, selecting and installing some
tires, and then saying you designed and built yourself a car.

Jon

Big Grins!

Yeah I applaud your effort, I wait for further posts.

For me, I’m building electronics to either detect something or drive
something that’s detecting something. So the fun is in making good
detectors or drivers.

George H.

Well, I am wanting, eventually, to build something I need.
Something I cannot buy in the market because the need is
unique.

This divides into two parts. Design and build. Since the
item is unique, I can't just go out and buy it. And getting
the features I need cannot just be "hacked" into existing
designs without at least knowing _some_ stuff, first. I
might as well turn the "design" part into a fair learning
experience, as a separate project of its own. Get past that
and when it comes time to build what I want I'll be able to
build on what I learned and add what I need and then do a
modest hobbyist level whack at actually making what I want to
make.

If someone else were to do this for me (hire a designer),
they'd get all the fun of learning on the job and taking my
money with it. They get the money, they get to further their
own education, and I get a tool. One tool. Once. Next
time, I get to pay someone else to learn for me.

It almost feels like paying someone to go do your exercising
for you. No satisfaction and no weight loss. They get all
the _real_ benefits.

Part of the fun isn't the destination itself but it is what
you see and enjoy while getting there, too. You take a plane
when all you need is to "get there" quick, but you drive when
you want to enjoy stops along the way. I used to fly to
Burbank every week for a year and a half. Slept in a hotel
for 3 nights a week, worked day and night in between, flew
home. Barely saw anything but hotel room walls, cubical
walls, a few cement roads, pollution so thick you couldn't
see the Burbank hills from the Lockheed center, and not much
else. The destination was important, of course. Paid the
bills and I enjoyed the work, too. But there is a lot more
to see in the 1000 miles from here to there.

Anyway, I'm driving this time, not flying.

Besides, I'd rather _keep_ the money and _keep_ the education
for myself. That way it pays off, again and again.

Jon

The fish/fish rule ?:-)

...Jim Thompson

Build a man a fire, you keep him warm for a night. Set a man afire, you
keep him warm for the rest of his life. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs at electrooptical dot nethttp://electrooptical.net- Hide quoted text -

- Show quoted text -

"> Build a man a fire, you keep him warm for a night. Set a man
afire, you
keep him warm for the rest of his life. Wink

"

Can't help being reminded of "The Cremation of Sam McGee".

And there sat Sam, looking cool and calm, in the heart of the furnace
roar;
And he wore a smile you could see a mile, and he said: "Please close
that door.
It's fine in here, but I greatly fear you'll let in the cold and storm
—
Since I left Plumtree, down in Tennessee, it's the first time I've
been warm."

Taken from here,
"http://en.wikisource.org/wiki/The_Cremation_of_Sam_McGee"

George h.

Yep, that was quite a night on the marge of Lake LaBarge.

Cheers

Phil 'sourdough' Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs at electrooptical dot net
http://electrooptical.net

krw
Guest

Fri Feb 12, 2010 10:54 pm

On Fri, 12 Feb 2010 13:21:58 -0800, "Bob Monsen" <rcmonsen_at_gmail.com>
wrote:

Quote:

"Jim Thompson" <To-Email-Use-The-Envelope-Icon_at_My-Web-Site.com> wrote in
message news:0m1bn5ln6ag64hdqnmr4te1qjeiec22i02_at_4ax.com...
On Fri, 12 Feb 2010 08:23:36 -0800, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 12 Feb 2010 10:23:24 -0500, Phil Hobbs
pcdhSpamMeSenseless_at_electrooptical.net> wrote:

On 2/12/2010 9:52 AM, Jim Thompson wrote:
On Thu, 11 Feb 2010 23:13:24 -0500, Phil Hobbs
pcdhSpamMeSenseless_at_electrooptical.net> wrote:
[snip]

IIRC the LM395 is basically an LM309 with the voltage reference
removed.
Emitter-follower regulators are nearly bulletproof unless you
discharge a cap into the output.

Cheers

Phil Hobbs

Close, but no cigar, LM395 = LM317 with some metal rearrangements.

I was thinking 317 but then I decided that the 395 was older than that.
I guess not.

I did this analysis for ICE back in 1980:

http://analog-innovations.com/SED/ICE-LM195-LM117.pdf

Additionally: Amusing myself with the thoughts of complementary-
follower-style power amplifiers made from LM317/LM337 pairs, it fails
because both, internally, are NPN's pass devices, so the LM337 has GBW
and stability issues plus it needs substantial idle load to stay
_vaguely_ stable.

Non-LDO three-terminal regulators are so trouble-free that it's easy to
confuse them with Newton's laws. ;)

Cheers

Phil Hobbs

LM1117 is an "MDO" regulator. It has an NPN pass transistor but a bit
lower dropout voltage than an LM317. Its ideal as a 3.3-to-1.25 volt
FPGA core voltage source... no resistors! My purchasing notes say "Do
not buy Fairchild per JL" but I can't recall why.

As with all vregs, one has to be careful about the output capacitors.

John

I'm puzzled why the big semiconductor houses don't turn out discrete
LDO's on a CMOS process. I do it all the time on complex CMOS
ASIC's... like a PLL chip, fundamental power fed from +3.3V, but
internal regulators producing +2.5V and +1.8V... at hundreds of mA !!!

ST Arm Cortex-M3 has an internal 1.8V regulator for the core, and can take
any input voltage from 2.0 to 3.8V.

Same with the Altera Max-II CPLDs. They'll take 1.8V, 2.5V, or 3.3V.

krw
Guest

Fri Feb 12, 2010 11:07 pm

On Fri, 12 Feb 2010 08:24:24 -0800, John Larkin
<jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:

Quote:

On Fri, 12 Feb 2010 10:03:11 -0500, Phil Hobbs
pcdhSpamMeSenseless_at_electrooptical.net> wrote:

Build a man a fire, you keep him warm for a night. Set a man afire, you
keep him warm for the rest of his life. ;)

Cheers

Phil Hobbs

Got snow?

We have about 1" now. They cancelled work for today about 9:30 last
night. Since there was no snow this morning (it started about
10:00AM) I went into work. I was the only one there. My wife's
employer closed about 2:00, with less than 1" on the ground. The
streets are just now starting to get some slush on them. I guess it's
a good thing the locals are scared shitless of snow.