Amplification factor for common emitter amplifier

[amal banerjee]

Is it possible to calculate the amplification factor
for an amplifier,, e.g., a common emitter amplifier.
This is because the amplification depends very much
on the biasing condition, and so without a SPICE
simulation, it appears impossible to determine the
amplification factor. Am I right/wrong.

 

[Bill Sloman]

On Thursday, August 8, 2019 at 4:08:21 PM UTC+10, amal banerjee wrote:

Is it possible to calculate the amplification factor
for an amplifier,, e.g., a common emitter amplifier.
This is because the amplification depends very much
on the biasing condition, and so without a SPICE
simulation, it appears impossible to determine the
amplification factor. Am I right/wrong.

Any calculation depends on the model you adopt for the transistor.

Spice mostly depends on the Gummel-Poon model, which is pretyy good for normal bias conditions, but doesn't model reverse biassed conditions all that well.

Spice will run the VBIC model, which is a lot better, but the manufacturers don't publish VBIC parameters for their transistors, and treat the numbers as industrial secrets.

Win Hill is the kind of situation that would allow him measure all the VBIC parameters for a bread-and-butter transistor and publish them here, but it's not the kind of work he likes to do.

For hand calculation, the assumption that voltage between the base and the emitter controls a current from collector to emitter that is exponentially dependent on the base-emitter voltage works fairly well - the Ebers-Moll model - is frequently good enough.

This makes transconductance of a bipolar transistor at room temperature the collector current divied by the thermal voltage (26mV at room temperature), and the gain is then the product of that and the collector load impedance..

At high frequencies you have worry about the Miller capacitance from the collector the base, that cancels out some of the base current you are feeding in, and the Early - base-narrowing - effect means that some of the voltage applied to the collector leaks through and subtracts from the voltage you are trying to develop across the base-emitter junction, but you need to be aiming for a gain of a thousand or more before this gets substantial.

The short answer is that you can't work out the gain accurately by hand, but you can often get close enough for most practical purposes.

--
Bill Sloman, Sydney

 

[George Herold]

On Thursday, August 8, 2019 at 2:08:21 AM UTC-4, amal banerjee wrote:

Is it possible to calculate the amplification factor
for an amplifier,, e.g., a common emitter amplifier.
This is because the amplification depends very much
on the biasing condition, and so without a SPICE
simulation, it appears impossible to determine the
amplification factor. Am I right/wrong.

Do you have a copy of "Art of Electronics"? It's covered in there.

https://www.amazon.com/Art-Electronics-Paul-Horowitz/dp/0521809266

George H.

 

[legg]

On Wed, 7 Aug 2019 23:08:17 -0700 (PDT), amal banerjee
<dakupoto@gmail.com> wrote:

Is it possible to calculate the amplification factor
for an amplifier,, e.g., a common emitter amplifier.
This is because the amplification depends very much
on the biasing condition, and so without a SPICE
simulation, it appears impossible to determine the
amplification factor. Am I right/wrong.

First approximation is collector impedance / emitter impedance.

RL

 

[John Larkin]

On Wed, 7 Aug 2019 23:08:17 -0700 (PDT), amal banerjee
<dakupoto@gmail.com> wrote:

Is it possible to calculate the amplification factor
for an amplifier,, e.g., a common emitter amplifier.
This is because the amplification depends very much
on the biasing condition, and so without a SPICE
simulation, it appears impossible to determine the
amplification factor. Am I right/wrong.

One approximation is that the voltage gain is 40 times the DC voltage
drop in the collector resistor. That's a bit of algebra from
Gm=Ie/0.025. Ignoring stuff like Early voltage and Re. Close enough.










--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[bitrex]

On 8/8/19 2:08 AM, amal banerjee wrote:

Is it possible to calculate the amplification factor
for an amplifier,, e.g., a common emitter amplifier.
This is because the amplification depends very much
on the biasing condition, and so without a SPICE
simulation, it appears impossible to determine the
amplification factor. Am I right/wrong.

the small-signal amplification factor is easy to figure out, it's the
device trans-conductance at the operating point, times the load resistance.

For truly small signals (infinitesimal) this answer is exactly "right",
the Taylor series of the transistor/MOSFET curve drops to a single
linear term.

The larger-signal amplification factor you have to SPICE, or
approximate. As another user mentioned collector/drain impedance divided
by emitter/source impedance is often a good enough approximation in many
cases.

 

People did this years ago without PCs or even calculators in some cases. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET..

The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

 

[amal banerjee]

On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote:

People did this years ago without PCs or even calculators in some cases. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET.

The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

I use the standard design rules and equations. The
starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed
absolute maximum values for Ic etc. The target Ic
value is checked to ensure that it would be less
than the absolute maximum value for Ic.

Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc.
Then the value of Vcc is easily calculated from
Vcc = 0.6Vcc + Vce
The values for Rc, and Re are calculated easily
from target Ic value and Vcc.

To calculate the base bias resistor values, the
maximum base current is easily calculated from
datasheet supplied value for minimum hfe, and
then the two base bias resistor values are calculated,
the second(grounded) using the calculated value
for the first(Vcc -> transistor base), and the
voltage divider formula. Of course, to calulate the
value for the first resistor, I use Vb = Ve + Vbe

I use emitter bypass resistor, whose value is
computed using Xc = 0.1Re(emitter resistance). I
also use input/output DC blocking capacitors
whose values are calculated easily as 1/frequency,
so that Xc = 1/(2*PI*freq*(1/freq)). Dc is blocked
but AX resistance is very low.

Please note that ALL the above calculations are
performed with a simple C language program, that
formats the results as a text SPICE netlist. Then
it can be used with any SPICE simulator, in my case
either HSpice or Ngspice. So, the PC is invaluable.

 

[John Larkin]

On Wed, 7 Aug 2019 23:08:17 -0700 (PDT), amal banerjee
<dakupoto@gmail.com> wrote:

Is it possible to calculate the amplification factor
for an amplifier,, e.g., a common emitter amplifier.
This is because the amplification depends very much
on the biasing condition, and so without a SPICE
simulation, it appears impossible to determine the
amplification factor. Am I right/wrong.

For a tube, amplification factor is Gm * Rp, transconductance times
the dynamic plate resistance. That corresponds to voltage gain with an
infinite plate resistor.

Transistors have nearly flat collector slopes, so equivalent mu is
huge, so huge it's not a practical thing to spec.


--

John Larkin Highland Technology, Inc trk

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[bitrex]

On 8/11/19 10:16 AM, John Larkin wrote:

On Wed, 7 Aug 2019 23:08:17 -0700 (PDT), amal banerjee
dakupoto@gmail.com> wrote:

Is it possible to calculate the amplification factor
for an amplifier,, e.g., a common emitter amplifier.
This is because the amplification depends very much
on the biasing condition, and so without a SPICE
simulation, it appears impossible to determine the
amplification factor. Am I right/wrong.

For a tube, amplification factor is Gm * Rp, transconductance times
the dynamic plate resistance. That corresponds to voltage gain with an
infinite plate resistor.

Transistors have nearly flat collector slopes, so equivalent mu is
huge, so huge it's not a practical thing to spec.

vacuum triodes have "internal negative feedback" vis a vis their dynamic
Rp, maybe part of the reason they're fancied so much.

You can make a nice wideband two-tube amp with an EF86 pentode and 6DJ8
or 12AU7 dual triode or something. pentode as common cathode amp, DC
coupled to one half of the triode. triode as cathode follower with the
cathode "load" the screen of the pentode - poor man's "ultra linear" for
local feedback around the first gain stage. can also bootstrap the
pentode's plate resistor for max gain. then AC couple into a split-load
triode, take output from cathode end, send plate output back to
pentode's cathode for the global feedback.

 

On Sunday, 11 August 2019 10:53:07 UTC+1, amal banerjee wrote:

On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote:
People did this years ago without PCs or even calculators in some cases.. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET.

The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

I use the standard design rules and equations. The
starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed
absolute maximum values for Ic etc. The target Ic
value is checked to ensure that it would be less
than the absolute maximum value for Ic.

Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc.
Then the value of Vcc is easily calculated from
Vcc = 0.6Vcc + Vce
The values for Rc, and Re are calculated easily
from target Ic value and Vcc.

To calculate the base bias resistor values, the
maximum base current is easily calculated from
datasheet supplied value for minimum hfe, and
then the two base bias resistor values are calculated,
the second(grounded) using the calculated value
for the first(Vcc -> transistor base), and the
voltage divider formula. Of course, to calulate the
value for the first resistor, I use Vb = Ve + Vbe

I use emitter bypass resistor, whose value is
computed using Xc = 0.1Re(emitter resistance). I
also use input/output DC blocking capacitors
whose values are calculated easily as 1/frequency,
so that Xc = 1/(2*PI*freq*(1/freq)). Dc is blocked
but AX resistance is very low.

Please note that ALL the above calculations are
performed with a simple C language program, that
formats the results as a text SPICE netlist. Then
it can be used with any SPICE simulator, in my case
either HSpice or Ngspice. So, the PC is invaluable.

I usually design for much lower tr beta than the datasheet claims. Datasheets like to lie by omission, some si trs lose a lot of beta over time, it allows margin for some other minor issues & makes substitution effortless. In short it makes equipment more reliable.


NT

 

[George Herold]

On Sunday, August 11, 2019 at 5:53:07 AM UTC-4, amal banerjee wrote:

On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote:
People did this years ago without PCs or even calculators in some cases.. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET.

The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

I use the standard design rules and equations. The
starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed
absolute maximum values for Ic etc. The target Ic
value is checked to ensure that it would be less
than the absolute maximum value for Ic.

Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc.
Then the value of Vcc is easily calculated from
Vcc = 0.6Vcc + Vce
The values for Rc, and Re are calculated easily
from target Ic value and Vcc.

Hmm OK.. Amal, I think you are using the 'current gain' model of a transistor.
This is often just fine.
At the next 'level' there is the Ebers-Moll model... which looks a little
more deeply at the dynamics of the transistor junction.

You might want to look into that. AoE is a good book to start with.
If money is tight, start with the 2nd edition. ~$20-30 used.

George H.

To calculate the base bias resistor values, the
maximum base current is easily calculated from
datasheet supplied value for minimum hfe, and
then the two base bias resistor values are calculated,
the second(grounded) using the calculated value
for the first(Vcc -> transistor base), and the
voltage divider formula. Of course, to calulate the
value for the first resistor, I use Vb = Ve + Vbe

I use emitter bypass resistor, whose value is
computed using Xc = 0.1Re(emitter resistance). I
also use input/output DC blocking capacitors
whose values are calculated easily as 1/frequency,
so that Xc = 1/(2*PI*freq*(1/freq)). Dc is blocked
but AX resistance is very low.

Please note that ALL the above calculations are
performed with a simple C language program, that
formats the results as a text SPICE netlist. Then
it can be used with any SPICE simulator, in my case
either HSpice or Ngspice. So, the PC is invaluable.

 

[John Larkin]

On Mon, 12 Aug 2019 09:37:28 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Sunday, August 11, 2019 at 5:53:07 AM UTC-4, amal banerjee wrote:
On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote:
People did this years ago without PCs or even calculators in some cases. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET.

The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

I use the standard design rules and equations. The
starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed
absolute maximum values for Ic etc. The target Ic
value is checked to ensure that it would be less
than the absolute maximum value for Ic.

Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc.
Then the value of Vcc is easily calculated from
Vcc = 0.6Vcc + Vce
The values for Rc, and Re are calculated easily
from target Ic value and Vcc.

Hmm OK.. Amal, I think you are using the 'current gain' model of a transistor.
This is often just fine.

But not to compute voltage gain. VG is, first order, independent of
beta.


--

John Larkin Highland Technology, Inc trk

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[George Herold]

On Monday, August 12, 2019 at 1:33:01 PM UTC-4, John Larkin wrote:

On Mon, 12 Aug 2019 09:37:28 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Sunday, August 11, 2019 at 5:53:07 AM UTC-4, amal banerjee wrote:
On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote:
People did this years ago without PCs or even calculators in some cases. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET.

The collector resistor should usually be dropping about half the Vcc.. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

I use the standard design rules and equations. The
starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed
absolute maximum values for Ic etc. The target Ic
value is checked to ensure that it would be less
than the absolute maximum value for Ic.

Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc.
Then the value of Vcc is easily calculated from
Vcc = 0.6Vcc + Vce
The values for Rc, and Re are calculated easily
from target Ic value and Vcc.

Hmm OK.. Amal, I think you are using the 'current gain' model of a transistor.
This is often just fine.

But not to compute voltage gain. VG is, first order, independent of
beta.

OK. (I am but a tadpole when it comes to design/ understanding of transistors.)
But don't you need the 'intrinsic' emitter resistance (~25 mV/Ic)
to get the gain?
(assuming no/small resistance from emitter to ground.)

George H.

--

John Larkin Highland Technology, Inc trk

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[John Larkin]

On Mon, 12 Aug 2019 10:45:47 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Monday, August 12, 2019 at 1:33:01 PM UTC-4, John Larkin wrote:
On Mon, 12 Aug 2019 09:37:28 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Sunday, August 11, 2019 at 5:53:07 AM UTC-4, amal banerjee wrote:
On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote:
People did this years ago without PCs or even calculators in some cases. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET.

The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

I use the standard design rules and equations. The
starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed
absolute maximum values for Ic etc. The target Ic
value is checked to ensure that it would be less
than the absolute maximum value for Ic.

Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc.
Then the value of Vcc is easily calculated from
Vcc = 0.6Vcc + Vce
The values for Rc, and Re are calculated easily
from target Ic value and Vcc.

Hmm OK.. Amal, I think you are using the 'current gain' model of a transistor.
This is often just fine.

But not to compute voltage gain. VG is, first order, independent of
beta.
OK. (I am but a tadpole when it comes to design/ understanding of transistors.)
But don't you need the 'intrinsic' emitter resistance (~25 mV/Ic)
to get the gain?
(assuming no/small resistance from emitter to ground.)

Exactly. But that's independent of beta.

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.

 

[Steve Wilson]

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.

Version 4
SHEET 1 1288 680
WIRE 560 -208 288 -208
WIRE 1104 -208 560 -208
WIRE 288 -192 288 -208
WIRE 560 -144 560 -208
WIRE 1104 -144 1104 -208
WIRE 288 -96 288 -112
WIRE 560 -48 560 -64
WIRE 608 -48 560 -48
WIRE 656 -48 608 -48
WIRE 1104 -48 1104 -64
WIRE 1152 -48 1104 -48
WIRE 1200 -48 1152 -48
WIRE 560 -32 560 -48
WIRE 1104 -32 1104 -48
WIRE 320 16 288 16
WIRE 496 16 320 16
WIRE 1008 16 992 16
WIRE 1040 16 1008 16
WIRE 288 32 288 16
WIRE 560 96 560 64
WIRE 624 96 560 96
WIRE 656 96 624 96
WIRE 1104 96 1104 64
WIRE 1168 96 1104 96
WIRE 1248 96 1168 96
WIRE 560 112 560 96
WIRE 1104 112 1104 96
WIRE 1248 112 1248 96
WIRE 288 128 288 112
WIRE 1248 192 1248 176
WIRE 560 224 560 192
WIRE 560 224 288 224
WIRE 1104 224 1104 192
WIRE 1104 224 560 224
WIRE 288 240 288 224
WIRE 288 336 288 320
FLAG 320 16 Vin
FLAG 288 128 0
FLAG 608 -48 Q1C
FLAG 624 96 Q1E
FLAG 288 -96 0
FLAG 288 336 0
FLAG 1152 -48 Q2C
FLAG 1168 96 Q2E
FLAG 1008 16 Vin
FLAG 1248 192 0
SYMBOL npn 496 -32 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL voltage 288 16 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 0.005 1k)
SYMBOL res 544 -160 R0
SYMATTR InstName R1
SYMATTR Value 1k
SYMBOL res 544 96 R0
SYMATTR InstName R2
SYMATTR Value 1.856k
SYMBOL voltage 288 -208 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value 10
SYMBOL voltage 288 224 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V3
SYMATTR Value -10
SYMBOL npn 1040 -32 R0
SYMATTR InstName Q2
SYMATTR Value 2N3904
SYMBOL res 1088 -160 R0
SYMATTR InstName R3
SYMATTR Value 1k
SYMBOL res 1088 96 R0
SYMATTR InstName R4
SYMATTR Value 1.856k
SYMBOL cap 1232 112 R0
SYMATTR InstName C1
SYMATTR Value 1
TEXT 720 -352 Left 2 !.tran 0 10m 0 1u
TEXT 720 -384 Left 2 ;'Voltage Gain
TEXT 464 288 Left 2 ;Vg = 5.35e-3 / 10e-3 = 0.535
TEXT 1016 288 Left 2 ;Vg = 1.767/ 10e-3 = 176.7
TEXT 552 256 Left 2 ;A
TEXT 1096 248 Left 2 ;B

 

[Steve Wilson]

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

 

[John Larkin]

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.

 

[George Herold]

On Tuesday, August 13, 2019 at 2:42:54 PM UTC-4, John Larkin wrote:

On Mon, 12 Aug 2019 10:45:47 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Monday, August 12, 2019 at 1:33:01 PM UTC-4, John Larkin wrote:
On Mon, 12 Aug 2019 09:37:28 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Sunday, August 11, 2019 at 5:53:07 AM UTC-4, amal banerjee wrote:
On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote:
People did this years ago without PCs or even calculators in some cases. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET.

The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

I use the standard design rules and equations. The
starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed
absolute maximum values for Ic etc. The target Ic
value is checked to ensure that it would be less
than the absolute maximum value for Ic.

Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc.
Then the value of Vcc is easily calculated from
Vcc = 0.6Vcc + Vce
The values for Rc, and Re are calculated easily
from target Ic value and Vcc.

Hmm OK.. Amal, I think you are using the 'current gain' model of a transistor.
This is often just fine.

But not to compute voltage gain. VG is, first order, independent of
beta.
OK. (I am but a tadpole when it comes to design/ understanding of transistors.)
But don't you need the 'intrinsic' emitter resistance (~25 mV/Ic)
to get the gain?
(assuming no/small resistance from emitter to ground.)

Exactly. But that's independent of beta.

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.

Well we are in violent agreement, my only point was that
you need to Ebers-Moll or some other model to figure
out what the intrinsic resistance is.
(and no one wants to run a transistor gain so high that
the intrinsic resistance matters.)

George H.

 

[Phil Hobbs]

On 8/13/19 7:42 PM, George Herold wrote:

On Tuesday, August 13, 2019 at 2:42:54 PM UTC-4, John Larkin wrote:
On Mon, 12 Aug 2019 10:45:47 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Monday, August 12, 2019 at 1:33:01 PM UTC-4, John Larkin wrote:
On Mon, 12 Aug 2019 09:37:28 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Sunday, August 11, 2019 at 5:53:07 AM UTC-4, amal banerjee wrote:
On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, jurb...@gmail.com wrote:
People did this years ago without PCs or even calculators in some cases. I just happen to have a little time before people start bugging me.

Transistors have a gain curve. Generally people want linear amplification so bias it near the apex of the gain curve.

You need an emitter resistor, almost everybody uses them. As long as the transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, consider it a ratio. Wherever you bias it, after looking at the spec sheet, the emitter is going to have a certain voltage at the operating point YOU SET.

The collector resistor should usually be dropping about half the Vcc. The emitter resistor much less, being a lower value of course. Now comes the bias. You know where you want it so you can figure what the emitter voltage should be. to the base you make a resistive divider providing about 0.6 volts above your intended emitter voltage.

If you got a transistor with a gain of 100 it doesn't much matter the base current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.

Anyway, you choose the collector resistor to give you the output impedance (or lower ) you want. then you go to the transistors and figure out one that is at the apex of its hfe curve at the current YOU WANT.

An emitter resistor around a tenth of the collector resistor will almost always be fine. And it will make it thermally stable IF the base is biased from a voltage divider. If you just take a like 560K from the collector to the base it will bias it, but it will not be stable thermally.

That is as simplistic as it can really get if you really want to build anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead of like 470K and 47K you might want 47K and 4.7K. Lower input impedance but more bandwidth, and the same voltage gain pretty much.

And then there's bootstrap. Took me some time to figure that out. I had the concept when I saw it but the math was very elusive. But using bootstrap you can make a transistor twice s good as it really is. These days the transistors are so good there isn't much need for it, but a few applications do still exist.

You want to design kid ? (I can call you that because I am old) Put the mouse down and step away from the PC. Get some graph paper and a bunch of mechanical pencils. I recommend the kind that screws the lead out rather than the kind where you push the edge. Or you can just buy some pencils and a pencil sharpener. Get one of those old programming flow charts, that's what I use. It has lines all over it and is easy to align to the graph paper. Then your design at least looks neat and that helps to understand it.

So you can know the characteristics of the device, a transistor. A wonderful thing invented before we were born.

And you can have all the info, if Digikey doesn't have a spec sheet that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.

If you need a certain lowness of output impedance and you want to bias the transistor at the apex of hfe and find that volts times amps add up to 1,450 mW and the rating of the transistor is 700mW, you have to find a different transistor.

That's life.

I use the standard design rules and equations. The
starting point is of course the transistor datasheet - specifically the Ic vs. Vce curves, and the listed
absolute maximum values for Ic etc. The target Ic
value is checked to ensure that it would be less
than the absolute maximum value for Ic.

Vc is selected at ~ 0.5Vcc, and the Ve ~ 0.1Vcc.
Then the value of Vcc is easily calculated from
Vcc = 0.6Vcc + Vce
The values for Rc, and Re are calculated easily
from target Ic value and Vcc.

Hmm OK.. Amal, I think you are using the 'current gain' model of a transistor.
This is often just fine.

But not to compute voltage gain. VG is, first order, independent of
beta.
OK. (I am but a tadpole when it comes to design/ understanding of transistors.)
But don't you need the 'intrinsic' emitter resistance (~25 mV/Ic)
to get the gain?
(assuming no/small resistance from emitter to ground.)

Exactly. But that's independent of beta.

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.

Well we are in violent agreement, my only point was that
you need to Ebers-Moll or some other model to figure
out what the intrinsic resistance is.
(and no one wants to run a transistor gain so high that
the intrinsic resistance matters.)

Generally not in a vanilla CE amplifier, right--the emitter degeneration
provides local feedback, which linearizes the stage. But you can also
apply local resistive feedback from the collector to the base, as in
your bog-standard bipolar MMIC.

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