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 nor mal 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 mo del - 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 co llector 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, bu t you can often get close enough for most practical purposes.
-- Bill Sloman, Sydney
Do you have a copy of "Art of Electronics"? It's covered in there.
George H.
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
First approximation is collector impedance / emitter impedance.
RL
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 s o bias it near the apex of the gain curve.
You need an emitter resistor, almost everybody uses them. As long as the tr ansistor is operating the input resistance will be that emitter resistor ti mes the hfe of the transistor. Between that and the collector resistor, con sider it a ratio. Wherever you bias it, after looking at the spec sheet, th e 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 e mitter resistor much less, being a lower value of course. Now comes the bia s. You know where you want it so you can figure what the emitter voltage sh ould 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 g etting 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 tha t 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 al ways be fine. And it will make it thermally stable IF the base is biased fr om a voltage divider. If you just take a like 560K from the collector to th e 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 anyt hing. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Instead o f 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 y ou can make a transistor twice s good as it really is. These days the trans istors 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 mou se down and step away from the PC. Get some graph paper and a bunch of mech anical pencils. I recommend the kind that screws the lead out rather than t he kind where you push the edge. Or you can just buy some pencils and a pen cil 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 wonderfu l 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 th ey 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 1,45
0 mW and the rating of the transistor is 700mW, you have to find a differen t transistor.That's life.
On Saturday, August 10, 2019 at 8:16:01 PM UTC+5:30, snipped-for-privacy@gmail.com wrot e:
I just happen to have a little time before people start bugging me.
so bias it near the apex of the gain curve.
transistor is operating the input resistance will be that emitter resistor times the hfe of the transistor. Between that and the collector resistor, c onsider 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 S ET.
emitter resistor much less, being a lower value of course. Now comes the b ias. 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 vol ts above your intended emitter voltage.
e current effect on the emitter voltage, and in small signal it is actually getting hard to find transistors with less than 100 hfe.
e (or lower ) you want. then you go to the transistors and figure out one t hat is at the apex of its hfe curve at the current YOU WANT.
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.
ything. And there is more. Frequency response, Miller effect anyone ? Now y ou 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 bu t more bandwidth, and the same voltage gain pretty much.
he 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 tra nsistors are so good there isn't much need for it, but a few applications d o still exist.
ouse down and step away from the PC. Get some graph paper and a bunch of me chanical 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 p encil 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. Th en your design at least looks neat and that helps to understand it.
ful thing invented before we were born.
s any good someone does. Most manufacturers make them available, after all they are trying to sell them.
e 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 differ ent transistor.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.
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
ote:
. I just happen to have a little time before people start bugging me.
on so bias it near the apex of the gain curve.
e transistor is operating the input resistance will be that emitter resisto r 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.
he 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 voltag e should be. to the base you make a resistive divider providing about 0.6 v olts above your intended emitter voltage.
ase current effect on the emitter voltage, and in small signal it is actual ly getting hard to find transistors with less than 100 hfe.
nce (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.
t always be fine. And it will make it thermally stable IF the base is biase d from a voltage divider. If you just take a like 560K from the collector t o the base it will bias it, but it will not be stable thermally.
anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Inste ad 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.
the concept when I saw it but the math was very elusive. But using bootstr ap you can make a transistor twice s good as it really is. These days the t ransistors are so good there isn't much need for it, but a few applications do still exist.
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 th an 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 wha t 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.
erful thing invented before we were born.
t's any good someone does. Most manufacturers make them available, after al l they are trying to sell them.
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 diff erent transistor.c vs. Vce curves, and the listed
I usually design for much lower tr beta than the datasheet claims. Datashee ts like to lie by omission, some si trs lose a lot of beta over time, it al lows margin for some other minor issues & makes substitution effortless. In short it makes equipment more reliable.
NT
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.
ote:
. I just happen to have a little time before people start bugging me.
on so bias it near the apex of the gain curve.
e transistor is operating the input resistance will be that emitter resisto r 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.
he 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 voltag e should be. to the base you make a resistive divider providing about 0.6 v olts above your intended emitter voltage.
ase current effect on the emitter voltage, and in small signal it is actual ly getting hard to find transistors with less than 100 hfe.
nce (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.
t always be fine. And it will make it thermally stable IF the base is biase d from a voltage divider. If you just take a like 560K from the collector t o the base it will bias it, but it will not be stable thermally.
anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. Inste ad 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.
the concept when I saw it but the math was very elusive. But using bootstr ap you can make a transistor twice s good as it really is. These days the t ransistors are so good there isn't much need for it, but a few applications do still exist.
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 th an 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 wha t 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.
erful thing invented before we were born.
t's any good someone does. Most manufacturers make them available, after al l they are trying to sell them.
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 diff erent transistor.c vs. Vce curves, and the listed
Hmm OK.. Amal, I think you are using the 'current gain' model of a transist or. 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.
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
ses. I just happen to have a little time before people start bugging me.
ation so bias it near the apex of the gain curve.
the transistor is operating the input resistance will be that emitter resi stor times the hfe of the transistor. Between that and the collector resist or, consider it a ratio. Wherever you bias it, after looking at the spec sh eet, the emitter is going to have a certain voltage at the operating point YOU SET.
. 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 vol tage should be. to the base you make a resistive divider providing about 0.
6 volts above your intended emitter voltage.e base current effect on the emitter voltage, and in small signal it is act ually getting hard to find transistors with less than 100 hfe.
edance (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.
most always be fine. And it will make it thermally stable IF the base is bi ased from a voltage divider. If you just take a like 560K from the collecto r to the base it will bias it, but it will not be stable thermally.
ld anything. And there is more. Frequency response, Miller effect anyone ? Now you might want to lower the resistive divider that biases the thing. In stead of like 470K and 47K you might want 47K and 4.7K. Lower input impedan ce but more bandwidth, and the same voltage gain pretty much.
had the concept when I saw it but the math was very elusive. But using boot strap you can make a transistor twice s good as it really is. These days th e transistors are so good there isn't much need for it, but a few applicati ons do still exist.
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 an d 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 pape r. Then your design at least looks neat and that helps to understand it.
onderful thing invented before we were born.
that's any good someone does. Most manufacturers make them available, after all they are trying to sell them.
as 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 d ifferent transistor.
e Ic vs. Vce curves, and the listed
istor.
OK. (I am but a tadpole when it comes to design/ understanding of transist ors.) 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.
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.
John Larkin wrote:
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
40 * 5 = 200
Legg is right.
Vg = Zc / Ze
You might ask someone to check that multiply for you.
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.
cases. I just happen to have a little time before people start bugging me.
fication so bias it near the apex of the gain curve.
as the transistor is operating the input resistance will be that emitter r esistor times the hfe of the transistor. Between that and the collector res istor, 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 poi nt YOU SET.
Vcc. The emitter resistor much less, being a lower value of course. Now com es 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.
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.
impedance (or lower ) you want. then you go to the transistors and figure o ut one that is at the apex of its hfe curve at the current YOU WANT.
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 colle ctor to the base it will bias it, but it will not be stable thermally.
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 impe dance but more bandwidth, and the same voltage gain pretty much.
I had the concept when I saw it but the math was very elusive. But using b ootstrap 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 applic ations do still exist.
ut the mouse down and step away from the PC. Get some graph paper and a bun ch of mechanical pencils. I recommend the kind that screws the lead out rat her 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 p aper. Then your design at least looks neat and that helps to understand it.
A wonderful thing invented before we were born.
et that's any good someone does. Most manufacturers make them available, af ter all they are trying to sell them.
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.
the Ic vs. Vce curves, and the listed
ansistor.
istors.)
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.
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.
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