Amplification factor for common emitter amplifier

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.  

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

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. 

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 

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

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.

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.

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

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