Amplification factor for common emitter amplifier

[George Herold]

On Tuesday, August 13, 2019 at 4:14:11 PM UTC-4, Steve Wilson wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain
is Gm * Rload. That magically collapses to Vgain = 40 times the DC
voltage drop in the collector resistor.

Transistors are actually pretty simple.

Here are two circuits with identical operating points. Your formula sets
the gain at 40 * 5 = 40.

Circuit A has a gain of 0.535. Circuit B has a gain of 176.7.

Your equation is wrong.

I'm not sure, but I think that's straight out of AoE... It's one
of the 'in chapter problems' in first two editions.

George H.

Version 4
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TEXT 720 -352 Left 2 !.tran 0 10m 0 1u
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TEXT 1016 288 Left 2 ;Vg = 1.767/ 10e-3 = 176.7
TEXT 552 256 Left 2 ;A
TEXT 1096 248 Left 2 ;B

 

[John Larkin]

On Tue, 13 Aug 2019 15:26:09 -0700, John Larkin
<jlarkin@highlandSNIPMEtechnology.com> wrote:

On Tue, 13 Aug 2019 20:14:05 GMT, Steve Wilson <no@spam.com> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain
is Gm * Rload. That magically collapses to Vgain = 40 times the DC
voltage drop in the collector resistor.

Transistors are actually pretty simple.

Here are two circuits with identical operating points. Your formula sets
the gain at 40 * 5 = 40.

You might ask someone to check that multiply for you.


Circuit A has a gain of 0.535. Circuit B has a gain of 176.7.

Your equation is wrong.

George specified

(assuming no/small resistance from emitter to ground.)

Why did you snip that?

176 is close to 200. A little Early effect will drop the gain some
from an ideal transistor. The 0.025 factor isn't exact either. And the
estimate is of course small-signal.

And there are some real, ohmic resistances too, wire bonds and
contact/spreading resistances in the silicon of a real transistor.

All that reduces the gain a bit from the theory.

 

[Steve Wilson]

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

On Tue, 13 Aug 2019 20:14:05 GMT, Steve Wilson <no@spam.com> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain
is Gm * Rload. That magically collapses to Vgain = 40 times the DC
voltage drop in the collector resistor.

Transistors are actually pretty simple.

Here are two circuits with identical operating points. Your formula sets
the gain at 40 * 5 = 40.

You might ask someone to check that multiply for you.

Already posted the fix. Also included Legg's formula Vg = Zc/Ze, which you
ignored. Here's my post:

Steve Wilson <no@spam.com> wrote:

Here are two circuits with identical operating points. Your formula sets
the gain at 40 * 5 = 40.

40 * 5 = 200

Legg is right.

Vg = Zc / Ze

Circuit A has a gain of 0.535. Circuit B has a gain of 176.7.

Your equation is wrong.

George specified

(assuming no/small resistance from emitter to ground.)

Why did you snip that?

You have to include the impedance in the emitter.

176 is close to 200. A little Early effect will drop the gain some
from an ideal transistor. The 0.025 factor isn't exact either. And the
estimate is of course small-signal.

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded. You need some way to stabilize the bias. Also,
there is usually some load on the collector, which you ignore. You could
also have transformer coupling or a resonant circuit in the collector,
which would produce a low dc resistance and consequently low voltage drop.
This would produce low gain in your formula.

Legg's formula includes the impedance on both the collector and emitter,
and includes loading due to the following stage and bypassing on the
emitter.

Legg's formula can also consider the DC case by omitting the load impedance
and emitter bypass. So you can get both the DC gain and high frequency
gain.

Your equation is only for DC gain, and ignores the resistance in the
emitter. It does not include the load impedance on the collector or
degeneration in the emitter. It is useless for real-world circuits.

 

On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

> You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

 

On Wed, 14 Aug 2019 10:09:49 GMT, Steve Wilson <no@spam.com> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

On Tue, 13 Aug 2019 20:14:05 GMT, Steve Wilson <no@spam.com> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain
is Gm * Rload. That magically collapses to Vgain = 40 times the DC
voltage drop in the collector resistor.

Transistors are actually pretty simple.

Here are two circuits with identical operating points. Your formula sets
the gain at 40 * 5 = 40.

You might ask someone to check that multiply for you.

Already posted the fix. Also included Legg's formula Vg = Zc/Ze, which you
ignored. Here's my post:

Steve Wilson <no@spam.com> wrote:

Here are two circuits with identical operating points. Your formula sets
the gain at 40 * 5 = 40.

40 * 5 = 200

Legg is right.

Vg = Zc / Ze

That's infinite with no emitter resistor, right?

Circuit A has a gain of 0.535. Circuit B has a gain of 176.7.

Your equation is wrong.

George specified

(assuming no/small resistance from emitter to ground.)

Why did you snip that?

You have to include the impedance in the emitter.

That's trivial. If you assume that the active emitter resistance is
only the resistor that you add external to the transistor, that's
wrong. The emitter diode has a dynamic resistance that matters, and
limits the gain to below infinite.

176 is close to 200. A little Early effect will drop the gain some
from an ideal transistor. The 0.025 factor isn't exact either. And the
estimate is of course small-signal.

Your formula is useless and misleading.

Merely correct. I can't help what people want to believe.

You normally never run a transistor
>with the emitter grounded.

Why not? It's done all the time.

>You need some way to stabilize the bias.

Lots of ways to do that. For example, bypass the emitter as you did in
your sim.



Also,
>there is usually some load on the collector, which you ignore.

Also trivial. Any real signal source drops when you load it. Do that
after you calculate the unloaded gain.



You could

also have transformer coupling or a resonant circuit in the collector,
which would produce a low dc resistance and consequently low voltage drop.
This would produce low gain in your formula.

There a million ways a smart person can wreck the gain of an
amplifier. Actually, a dumb person can do that too.

Legg's formula includes the impedance on both the collector and emitter,
and includes loading due to the following stage and bypassing on the
emitter.

Legg's formula can also consider the DC case by omitting the load impedance
and emitter bypass. So you can get both the DC gain and high frequency
gain.

Your equation is only for DC gain, and ignores the resistance in the
emitter.

Dead wrong. It is an estimate of the small-signal (that means AC)
gain.

The "resistance in the emitter" is the dynamic impedance of the b-e
junction. That's fundamental to understanding transistors.

It does not include the load impedance on the collector or

degeneration in the emitter. It is useless for real-world circuits.

It's an algebraic result of the actual transistor behavior in one
simple stated case. It's merely right.

 

On Wed, 14 Aug 2019 04:46:28 -0700 (PDT), tabbypurr@gmail.com wrote:

On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

The same simple concepts can be appied to a differential amplifier
biased by a current sink.

Suicide bias has poor control over Ic hence poor control over the
voltage drop in the collector resistor, hence less predictable voltage
gain.

For max gain and lowest noise in a single transistor amp, ground the
emitter.

 

[George Herold]

On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:

On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

GH

 

On Wed, 14 Aug 2019 08:24:28 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

GH

A resistor from Vcc to the base, with the emitter grounded. That's
basically a constant base bias current, so Ic is proportional to beta.

Early transistor circuits tended to do that. As do some cheap products
still.


I've done it in some special cases. One can now buy beta-graded
transistors, like BCX70J, that make the idea less silly.

This is cool:

https://www.dropbox.com/s/tqw2qe9mi4t8zmz/Suicide_Slicer.JPG?raw=1

I'm doing that as a CML-to-TTL converter.

 

On Wednesday, 14 August 2019 15:25:53 UTC+1, jla...@highlandsniptechnology.com wrote:

On Wed, 14 Aug 2019 04:46:28 -0700 (PDT), tabbypurr wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

The same simple concepts can be appied to a differential amplifier
biased by a current sink.

Suicide bias has poor control over Ic hence poor control over the
voltage drop in the collector resistor, hence less predictable voltage
gain.

V_Rc_quiesc is determined by Ib & beta. Beta varies, but within limits. Sometimes it's good enough. Often not. Suicide never happens if Ic is low, thus tr heating is low.

For max gain and lowest noise in a single transistor amp, ground the
emitter.

and max output swing.


NT

 

[piglet]

On 14/08/2019 4:24 pm, George Herold wrote:

On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

GH

If want it to be linear but emitter well and truly grounded one can
always de-suicide the bias by using feedback resistor to base from
collector. For max AC gain decouple it midway with a capacitor to ground.

piglet

 

>But you can also apply local resistive feedback from the >collector to the base

Sure. If you apply that to the bias divider fine, it will further stabilise the stage but will lower input Z. It may even clip off the top of the output if you really want the stability. The lower the Z of the bias the better Re works. I could put you together a stage that has near infinite voltage gain, but has one ohm input Z and 32 megohn output Z. Pretty much useless. (PRETTY MUCH, THERE ARE TIMES...)

In anything near that you don't have much real gain. (if any) It can be useful but if you get anywhere near that then you want to use a common base stage. In fact a real old time cascode stage would really do it. How much real gain does a cascode have ? It has voltage gain coming out of the cracks in the sky, but real gain is a whole different thing.

Conversely take the typical audio amp. At the output stages with just the drivers and outputs of course there is gain. But it is current gain, in fact it inherently has negative voltage gain. But there is a stage before that which does the voltage amplification and the output stage which is only current amplification serves to allow the voltage amplifier stage to run into a lighter (higher Z) load.

I know I responded to you and you probably know all about this, but I think you forgot to mention a thing or two. Most of this is for the OP. What I said about what you said is to find out if you agree and perhaps start a much better argument. (there is an idea I aim to contact you with, but I gotta write it)

 

[Kevin Aylward]

"Steve Wilson" wrote in message
news:XnsAAAB3EB324456idtokenpost@69.16.179.22...

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

On Tue, 13 Aug 2019 20:14:05 GMT, Steve Wilson <no@spam.com> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

The transconductance Gm is 1/(.025/Ic) = 40 * Ic, and the voltage gain
is Gm * Rload. That magically collapses to Vgain = 40 times the DC
voltage drop in the collector resistor.

Transistors are actually pretty simple.

Here are two circuits with identical operating points. Your formula sets
the gain at 40 * 5 = 40.

You might ask someone to check that multiply for you.

Already posted the fix. Also included Legg's formula Vg = Zc/Ze, which you
ignored. Here's my post:

Steve Wilson <no@spam.com> wrote:

Here are two circuits with identical operating points. Your formula sets
the gain at 40 * 5 = 40.

40 * 5 = 200

Legg is right.

Vg = Zc / Ze

Circuit A has a gain of 0.535. Circuit B has a gain of 176.7.

Your equation is wrong.

George specified

(assuming no/small resistance from emitter to ground.)

Why did you snip that?

You have to include the impedance in the emitter.

176 is close to 200. A little Early effect will drop the gain some
from an ideal transistor. The 0.025 factor isn't exact either. And the
estimate is of course small-signal.

Your formula is useless and misleading.

Ahmmm....

You normally never run a transistor
with the emitter grounded. You need some way to stabilize the bias.

Sure, the bias needs stabalised...many designs use an emitter resister
bypassed with a big cap....taking the resister out of the game.

Some designs use a grounded emitter with a potential divider from collector
to base and ground, fixing the output voltage referred to as a Vbe
multiplier

Also,
there is usually some load on the collector, which you ignore. You could
also have transformer coupling or a resonant circuit in the collector,
which would produce a low dc resistance and consequently low voltage drop.
This would produce low gain in your formula.

Sure... What's your point?

Legg's formula includes the impedance on both the collector and emitter,
and includes loading due to the following stage and bypassing on the
emitter.

Legg's formula can also consider the DC case by omitting the load impedance
and emitter bypass. So you can get both the DC gain and high frequency
gain.

Your equation is only for DC gain, and ignores the resistance in the
emitter. It does not include the load impedance on the collector or
degeneration in the emitter. It is useless for real-world circuits.

It's a very well known, useful approximation for the maximum gain with a
resistive load. That is, the actual gain will always be less.

One can easily add in the effect of emitter resistance by computing Re' = re
+ Re,, re=1/(40.IC)

One can also include the effect of base resistance by adding rbb'/hafe to
Re'

With a current source load it becomes Va/vt as the maximum voltage, where Va
is the early voltage.

I have a tutorial on this here:

http://www.kevinaylward.co.uk/ee/bipolardesign1/bipolardesign1.xht

Pretty much all hand calculations are of limited value in the real world.

However, today, Spice is the only realistic way to design reliable multi
transistor circuits, well for asics anyway...

I'm designing fairly complex analog asics with 10k transistor system level
blocks, and I would say out of millions of simulations, I do hand
calculations as often as number as the fingers on that hand.

Its just not worth the agro to piss about when there are computers.

Even a diode resister circuit requires some effort to solve...

Id = Vt/R.W(is.R/vt.exp(Vs/Vt).... Where W is the Lambert W function...

http://www.kevinaylward.co.uk/ee/widlarlambert/widlarlambert.xht


-- Kevin Aylward
http://www.anasoft.co.uk - SuperSpice
http://www.kevinaylward.co.uk/ee/index.html

 

[George Herold]

On Wednesday, August 14, 2019 at 2:52:36 PM UTC-4, piglet wrote:

On 14/08/2019 4:24 pm, George Herold wrote:
On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

GH


If want it to be linear but emitter well and truly grounded one can
always de-suicide the bias by using feedback resistor to base from
collector. For max AC gain decouple it midway with a capacitor to ground.

piglet

piglet, :^) I've never done a suicide bias... mostly I've got
some opamp or digital output driving a transistor. (with neg.
feedback somewhere.)

So I drew a resistor from collector to base, do I still keep the R
from the power rail to base? (or just get rid of that...
as long as the load has a resistive part, current is going to flow.)
And then split that resistor in the middle with a cap to ground...
'cause that's going to be too much bias (on average).

I'm much better with pictures, words confuse me at times.

George h.

 

[piglet]

On 15/08/2019 1:27 am, George Herold wrote:

On Wednesday, August 14, 2019 at 2:52:36 PM UTC-4, piglet wrote:
On 14/08/2019 4:24 pm, George Herold wrote:
On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

GH


If want it to be linear but emitter well and truly grounded one can
always de-suicide the bias by using feedback resistor to base from
collector. For max AC gain decouple it midway with a capacitor to ground.

piglet

piglet, :^) I've never done a suicide bias... mostly I've got
some opamp or digital output driving a transistor. (with neg.
feedback somewhere.)

So I drew a resistor from collector to base, do I still keep the R
from the power rail to base? (or just get rid of that...
as long as the load has a resistive part, current is going to flow.)
And then split that resistor in the middle with a cap to ground...
'cause that's going to be too much bias (on average).

I'm much better with pictures, words confuse me at times.

George h.

Here is a picture of simple self bias. It is negative feedback, broken
at AC in the lower sketch.

<https://www.dropbox.com/s/ze0vo98nut8gwqc/_SelfbiasedQ.jpg?raw=1>

piglet

 

[Jasen Betts]

On 2019-08-14, George Herold <gherold@teachspin.com> wrote:

On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

It's been a term of art here (in sci.electronics) for as long as I can remeber.

Basically the the base is connected to some sort of current source, and the
expression for emitter current is dominted by the beta.

--+-+--------+-----+-+-------+-- VCC
| | | | | R
| | R | R +-+
| R +-+ R | R |
R | R | | c | c
| c | c +b +b
+b +b | e | e
e e R | R |
| | | | | |
----+--------+-----+-+-----+-+-- GND
Q1 Q2 Q3 Q4

Q1, Q2, Q3 suicide bias, Q4 maybe not

--
When I tried casting out nines I made a hash of it.

 

[George Herold]

On Thursday, August 15, 2019 at 6:31:22 AM UTC-4, Jasen Betts wrote:

On 2019-08-14, George Herold <gherold@teachspin.com> wrote:
On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

It's been a term of art here (in sci.electronics) for as long as I can remeber.

Basically the the base is connected to some sort of current source, and the
expression for emitter current is dominted by the beta.

--+-+--------+-----+-+-------+-- VCC
| | | | | R
| | R | R +-+
| R +-+ R | R |
R | R | | c | c
| c | c +b +b
+b +b | e | e
e e R | R |
| | | | | |
----+--------+-----+-+-----+-+-- GND
Q1 Q2 Q3 Q4

Q1, Q2, Q3 suicide bias, Q4 maybe not

--
When I tried casting out nines I made a hash of it.

Thanks, Isn't Q3 the 'first circuit' one learns to bias a common emitter?
Well, it was the first circuit I learned.

GH

 

[George Herold]

On Thursday, August 15, 2019 at 4:49:11 AM UTC-4, piglet wrote:

On 15/08/2019 1:27 am, George Herold wrote:
On Wednesday, August 14, 2019 at 2:52:36 PM UTC-4, piglet wrote:
On 14/08/2019 4:24 pm, George Herold wrote:
On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

GH


If want it to be linear but emitter well and truly grounded one can
always de-suicide the bias by using feedback resistor to base from
collector. For max AC gain decouple it midway with a capacitor to ground.

piglet

piglet, :^) I've never done a suicide bias... mostly I've got
some opamp or digital output driving a transistor. (with neg.
feedback somewhere.)

So I drew a resistor from collector to base, do I still keep the R
from the power rail to base? (or just get rid of that...
as long as the load has a resistive part, current is going to flow.)
And then split that resistor in the middle with a cap to ground...
'cause that's going to be too much bias (on average).

I'm much better with pictures, words confuse me at times.

George h.



Here is a picture of simple self bias. It is negative feedback, broken
at AC in the lower sketch.

https://www.dropbox.com/s/ze0vo98nut8gwqc/_SelfbiasedQ.jpg?raw=1

Thanks.
GH

piglet

 

On Thu, 15 Aug 2019 06:35:17 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Thursday, August 15, 2019 at 6:31:22 AM UTC-4, Jasen Betts wrote:
On 2019-08-14, George Herold <gherold@teachspin.com> wrote:
On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

It's been a term of art here (in sci.electronics) for as long as I can remeber.

Basically the the base is connected to some sort of current source, and the
expression for emitter current is dominted by the beta.

--+-+--------+-----+-+-------+-- VCC
| | | | | R
| | R | R +-+
| R +-+ R | R |
R | R | | c | c
| c | c +b +b
+b +b | e | e
e e R | R |
| | | | | |
----+--------+-----+-+-----+-+-- GND
Q1 Q2 Q3 Q4

Q1, Q2, Q3 suicide bias, Q4 maybe not

It's not. There's feedback, essentially a vbe-multiplier.

Q2 isn't suicidal because there's feedback there too. The transistor
can't saturate no matter how high the beta.

--
When I tried casting out nines I made a hash of it.

Thanks, Isn't Q3 the 'first circuit' one learns to bias a common emitter?
Well, it was the first circuit I learned.

GH

Usually an emitter resistor would be added. Otherwise it's still
suicide bias.

One nice option is to use the circuit of Q4, but return the lower
resistor to a negative supply. That defines the operating point better
over temperature.

 

[Phil Hobbs]

On 8/14/19 11:39 AM, jlarkin@highlandsniptechnology.com wrote:

On Wed, 14 Aug 2019 08:24:28 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Wednesday, August 14, 2019 at 7:46:32 AM UTC-4, tabb...@gmail.com wrote:
On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

What's suicide bias? I went searching for biploar transistor and suicide bias... but got a lot of mental health sites. (maybe lithium doping would help. :^)

GH

A resistor from Vcc to the base, with the emitter grounded. That's
basically a constant base bias current, so Ic is proportional to beta.

Early transistor circuits tended to do that. As do some cheap products
still.


I've done it in some special cases. One can now buy beta-graded
transistors, like BCX70J, that make the idea less silly.

This is cool:

https://www.dropbox.com/s/tqw2qe9mi4t8zmz/Suicide_Slicer.JPG?raw=1

I'm doing that as a CML-to-TTL converter.

That's only suicide bias if the transistor doesn't saturate, or if the
AC gets turned off. Otherwise you have a nice switched-cap resistor
there, working through the low base Z of the saturated transistor.

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

 

[Phil Hobbs]

On 8/14/19 10:25 AM, jlarkin@highlandsniptechnology.com wrote:

On Wed, 14 Aug 2019 04:46:28 -0700 (PDT), tabbypurr@gmail.com wrote:

On Wednesday, 14 August 2019 11:09:55 UTC+1, Steve Wilson wrote:

Your formula is useless and misleading. You normally never run a transistor
with the emitter grounded.

odd thing to say

You need some way to stabilize the bias. Also,

There's more than one way to do that. And despite the myths, suicide bias IS used in commercial circuits.


NT

The same simple concepts can be appied to a differential amplifier
biased by a current sink.

Suicide bias has poor control over Ic hence poor control over the
voltage drop in the collector resistor, hence less predictable voltage
gain.

For max gain and lowest noise in a single transistor amp, ground the
emitter.

Improves stability too. A bit of CB feedback stabilizes the operating
point.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com