voltage gain of a transistor

[J M Fahey]

1) I analyzed voltage gain only, because it's a confusing point which often baffles even experienced people.
Yes, you will need to bias the base to whatever's necessary; in this case, to pass Ic/beta = 1mA/500 which will mean *around* 600mV Vbe, check the datasheet.
Simply biasing with a fixed voltage source (which can be a voltage divider) won't be quite stable (thermally) because needed voltage varies -2mV/ºC so in the real world some extra resistors are added to provide at least DC feedback, besides any needed AC feedback to tame those very high 384x into something usable.
But the basic design is as I said.
2) Enzo answered it: you can drive a base straight from a guitar pickup, as in the Dallas Rangemaster, the EH LPB1, etc; but if there is any chance of driving it from a high power low impedance signal source (such as plugging the speaker out or a headphone out with just 100 ohms in series) into a transisto base, you will need to have a series resistor, plus clamping diodes, etc.
Look at some Crate or Peavey schematics.
Note: an Op Amp input *is* a transistor base too !! (or a Fet gate, almost the same)

[lowell]

"Maybe you think Re is "0" because the emitter is grounded.
It's not so (this drives tube men crazy), you have an "equivalent", internal Re which depends on junction material and current passing through it.
For Silicon, Re(ohms) = 26/Ie (mA) so in this case, Re=26 ohms."

Does this apply when Re is bypassed w/ a largish Ce?

[lowell]

Also does this seem correct if using a BC547C? Fyi I only want a gain of 10 from this thing and I'm using it as the intial stage in a tube amp, therefore I have easy access to the higher B+. Also I'm using an Re to reduce gain... not sure if Re changes the idle current or not.

B+=60v
Rc=68k
Vc=30v
Ic=900ua
Re=1k
Zin=500*1k=500k
Ve=900mv

How do I bias this base to 1500mv? (900mv+600mv) If I use a voltage divider network and want even a 100k Zin a 1meg on top will only get my Vb down to 3v.

EDIT: I mean 6v for Vb.

[lowell]

I guess if I use a B+ of 20v then it makes sense... just like in your example JM.

B+=20v
Rc=10k
Vc=10v
Ic=1ma
Re=1k (to reduce gain)
Zin=500*1k=500k
Ve=1v
Vb=1.6v

This will give me a gain of 10 correct? I can then reduce the input to the next stage via a divider.

Voltage divider biasing network with a 1meg and 100k will get me down to a Vb of 2v.

[Wilder Amplification]

Are you voltage divider biasing this transistor's base to .7v? If not then how is the transistor biased ON?

There you go thinking in terms of voltage again.

When dealing with bipolar junction transistors -

CURRENT CURRENT CURRENT!!!

They are CURRENT controlled devices. The base will always have 0.7V RELATIVE TO THE EMITTER whether there is an emitter resistor or not. The emitter resistor DOES NOT set the gain. It controls base-emitter CURRENT, which controls the bias point.

You want a voltage controlled device that's similar to valves, go with an N-channel depletion mode MOSFET. Otherwise, realize that bipolar junction transistors are CURRENT CONTROLLED.

[jmaf]

You want a voltage controlled device that's similar to valves, go with an N-channel depletion mode MOSFET. Otherwise, realize that bipolar junction transistors are CURRENT CONTROLLED.

At the power levels he's using I'd guess almost any JFET would do.

[J M Fahey]

Dear Lowell, not at home now, and don't have schematic drawing software here, but tonight I'll post for you the schematic of a 10X voltage gain bipolar transistor stage.
Make that two: a +B derived supply one and a 9V battery powered one to fit in a small pedal box or inside your guitar.

[pdf64]

A silicon device (eg NPN or FET) can be connected into the cathode circuit of a triode to greatly increase the stage gain. I think that the tube then operates in grounded grid mode. See Vox 125
http://www.webphix.com/schematic%20h...s/v125lead.pdf
Doing it like that saves on a power supply for the FET; there's not much flexibility regards the FET current though. Pete.

[lowell]

There you go thinking in terms of voltage again.

When dealing with bipolar junction transistors -

CURRENT CURRENT CURRENT!!!

They are CURRENT controlled devices. The base will always have 0.7V RELATIVE TO THE EMITTER whether there is an emitter resistor or not. The emitter resistor DOES NOT set the gain. It controls base-emitter CURRENT, which controls the bias point.

You want a voltage controlled device that's similar to valves, go with an N-channel depletion mode MOSFET. Otherwise, realize that bipolar junction transistors are CURRENT CONTROLLED.

Jon,
That's the thousandth time that I've read that explanation of how transistors work. Obviously I still don't quite get it. As JM has been saying, most explanations of transistors are bad ones. It IS TRUE though that the base/emitter junction (diode) needs to be forward biased to turn the damn things on... so no it's not ONLY current.

[lowell]

Dear Lowell, not at home now, and don't have schematic drawing software here, but tonight I'll post for you the schematic of a 10X voltage gain bipolar transistor stage.
Make that two: a +B derived supply one and a 9V battery powered one to fit in a small pedal box or inside your guitar.

What about what I posted. Will it work?

[J M Fahey]

OK, back home, here it is:
1) the schematic:

2) the Math.
Bring in a couple NSA approved Craig Super Computers
a) desired gain=10x , so Rc/Re=10
Choose Rc=10K so Re=1K (the small 26ohm@1mA Re_eq. can be lumped into it)
b) Ic=1mA (because I want it that way, capisce?)
Beta=500 (The "C" after "BC547" says so)
Ib=1mA/500=2uA
Ve=1mAx1K=1V
Vb=0.7V+1V=1.7V (approx.)
I choose a B+ of 24V, simply because I can use a 24V Zener to stabilize it.
Remember we are retrieving it from a +300V source, so I do not want surprises there.
Vc (idle) 10V below 24V=14V
Why 10V below 24V?; because 1mAx10K=10V. Let's keep it neat and easy.
I do this for a living, not for Nobel prizes, so simpler is better for me.
c) Biasing string.
I need Vb=1.7V.
I already said that Ib=2uA, so I choose to "waste" another 2uA through the Base to ground resistor.
I already post the closest standard values, please nobody complains about insignificant "errors" which are not so in the real world.
Base to ground voltage divider leg=1.7V/2uA=820K.
The +24 to base leg will pass 4uA (Ib+2uA) so it will be 22V/4uA=5M6.
d) Intrinsic input impedance= beta x 1K=500K
These will be in parallel with the 820K to ground (and the 5M6 to B+) so paralleling all:
Real input impedance ~300K. (not bad for a single humble ten cents bipolar transistor eh?)
e) Output impedance=10K
f) Max signal voltage before clipping: around 1/3 22V =~7V RMS
g) 300V->24V dropping resistor: I assume 1.5mA (I'm a cheap b*st*rd and hate waste), so Rdrop=280V/1.5mA =~180K.
h) Input capacitor: a cheap ceramic .01 (you can use a plastic one if you wish) gives me Fb around 50Hz into those 300K.
i) Output cap: a 1uF one will give me Fb=16 Hz.

So, in a nutshell: this is a simple bipolar transistor voltage amplifier, gain 10x, Zin=300K ; Zout=10K, Vout max=7V RMS.
Distortion? *Very* low (remember we are reducing the 384X possible to 10x through internal feedback)
Frequency response? = maybe not much into RF but definitely way over the top for audio. And much wider than any Op Amp.
This is a very practical no b*llsh*t design approach ; if you want the full Monty there are many good pages (and many bad ones too).
A good one, which says exactly the same using 40x more words:THE TRANSISTOR AS A VOLTAGE AMPLIFIER
Good luck and enjoy.
PD: too lazy to design the 9V version, turn that into a homework

[lowell]

JM you are awesome. Thanks so much for putting this in plain english and actually explaining I, V, R in all the necessary places.

When I use your voltage divider (yes voltage... I know sorry):

I get 820k/(820k+5.6m)=.125
.125*24v=~3v

I'm thinking the reason that my math is wrong is that I'm ignoring the CURRENT (Jon) through the base (which is also a resistance in parallel with the base-ground resistor) Is that right?

[J M Fahey]

You're welcome.
Yes, he current through the upper leg of the biasing voltage divider is 4uA, twice the actual base current.
Most design examples are "teacher's/classroom examples" so they apply the basic math correctly, fine, but play it real safe so "any" transistor works, so they assume beta=50 ; current through the voltage divider 10x the actual base current (which was already estimeted as 10x a modern day example) and so on.
The classroom example works, but its performance is poor in the real world compared to what *can* easily be made with modern yet cheap and abbundant parts.
One example I saw used 1K as the lower biasing resistor , almost useless for most guitars.
Old Peaveys and Acoustics were fine examples of bipolar transistor design.
In fact, since transistors are so cheap , usually two direct coupled ones are used in many gain stages, providing much higher performance, hard to beat even using modern op amps.

[lowell]

"c) Biasing string.
I need Vb=1.7V.
I already said that Ib=2uA, so I choose to "waste" another 2uA through the Base to ground resistor.
I already post the closest standard values, please nobody complains about insignificant "errors" which are not so in the real world.
Base to ground voltage divider leg=1.7V/2uA=820K.
The +24 to base leg will pass 4uA (Ib+2uA) so it will be 22V/4uA=5M6." -JM

JM how did you know ahead of time that you were gonna "waste" 2ua through the base-ground resistor? Also is the conventional current going into the base or out of it?

[Wilder Amplification]

Jon,
That's the thousandth time that I've read that explanation of how transistors work. Obviously I still don't quite get it.

OK...and rather than investigate that explanation and question what it means, you'd much rather not get it and continue thinking in terms of voltage. How do you expect to ever "get it" if you keep doing that?

JM how did you know ahead of time that you were gonna "waste" 2ua through the base-ground resistor? Also is the conventional current going into the base or out of it?

He used this really super cool equation called "Ohm's Law".

OK...first some basics.

Picture water flowing through a hose. Water is your "current" while the size of the hose poses a resistance to this water flow. Resistance to water flow creates pressure..., which represents "voltage". Without resistance to water flow, you have no pressure to speak of.

Now...picture electrical current. Your circuit has a resistance. Resistance to electrical current flow creates voltage across this resistance. To support this, if you have no resistance (i.e. a short), your voltage goes to nothing while current goes to maximum.

This voltage value that gets generated across the resistance to current flow is proportionate to the value of said resistance and the value of said current flow. It is expressed mathematically as -

E = I x R where -

E = Electromotive Force in "Volts"
I = Current in "Amps"
R = Resistance in "Ohms"

Simple Ohm's Law.

Now...forget about the resistors in a transistor circuit for a second. Let's assume we have a bipolar junction transistor with a current gain of 100. This means that the collector current will be 100x the base current. This also means that if the transistor experiences a change in base current, it will also experience a change in collector current and that change in collector current will be 100x the change in base current. This all assumes that the transistor is not fully saturated.

If our theoretical transistor was being operated as a switch sinking current for a high current load, it would take at least 1/100th of the maximum rated load current to turn the transistor all the way on (full saturation).

Now...just like a diode that maintains a 0.7V drop across it no matter how much current is flowing through it, the base-emitter junction will always maintain a 0.7V drop across it regardless of current flowing through it. Resistors are used in the base-emitter circuit to control how much current flows through the base-emitter junction, and the base-emitter current controls how much current can flow through the collector-emitter junction. This is why they are CURRENT controlled devices. The voltage drop across the base-emitter junction remains the same at all times while the current flowing through the junction changes with input signal. And changing the current through the base-emitter junction changes the current through the collector-emitter junction by the current gain factor.

Up to the point of saturation, collector current = Transistor Current Gain x Base current while emitter current is the sum of the collector current and the base current.

With our current gain of 100 let's say we have 1mA of base current. This means that 100mA of collector current will flow. If we have 2mA of base current, then 200mA of collector current will be allowed to flow.

Of course, this assumes that the collector load resistor is sized to allow that much collector current to flow at the given source voltage. If it isn't, then collector current will not increase with any further increase in base current. You have just hit "saturation".

Make sense?