voltage gain of a transistor

[lowell]

Will someone please clear this up for me already?? I swear I've searched and searched for this answer online but can't find it. Obviously there's no simple answer. How does one calculate the voltage (not current) gain of a transistor? What values in the datasheet do I need and what equation will aid in designing a transistor amplifier? Yes Vc/Vb=gain but that doesn't tell me jack sh$t. Read me? I could figure out the gain of a transistor circuit that is already built, just not in the design stages. I believe Hfe is DC current gain, which really doesn't help me unless I want to use the transistor as a switch. I'm thinking from the little I can remember, the beta plays a part here. Please, someone clear this up for me once and for all.

[defaced]

It took me a sec to remember on what forum, but I knew RG explained it to me once. Maybe this will help: http://music-electronics-forum.com/t1244/#post172661

[Steve Conner]

Hi lowell, to a first approximation the voltage gain of the transistor itself is infinite. (It is actually limited by Early effect.)

So when used in a circuit with a resistor as the collector load, the voltage gain of the circuit is just the current gain of the transistor, times the resistance of the resistor.

If you load the collector with a current source instead of a resistor, you can get very high gain indeed, maybe 10,000.

But, if making a simple amplifier with a single transistor, it's not a good idea to use all of the current or voltage gain. The reason is that the gains are very non-linear, so the amp will generate lots of distortion, and not necessarily the pleasant tubey kind either.

And the transistor parameters vary so much with temperature and between parts, if you built 100 of the circuits they would all perform very differently, and would sound different in winter.

It's better to have a partially bypassed emitter (which limits and stabilizes the current gain) and a resistive collector load, and under those design constraints, the most voltage gain you can get is about 50 off a 9V battery supply.

PS: Beta, Hfe and hfe are technically different things, but I can never remember which one is which. (I think beta and hfe are both the small-signal current gain, and Hfe is the DC gain for large signals, or something.) When talking about transistor current gain in a hand-waving manner (which is all you should ever do as it's so unpredictable) either will do.

[lowell]

Steve,
Beautiful. Thank you for the response. I can't believe it's that simple!! So is this the correct equation then?

(Ie/Ib)*Rc
or
(Ic/Ib)*Rc
Because Ie=Ic

"It's better to have a partially bypassed emitter (which limits and stabilizes the current gain) and a resistive collector load, and under those design constraints, the most voltage gain you can get is about 50 off a 9V battery supply."

Is this gain of 50 clean gain or total gain? Finally, what if I were to use a transistor as the 1st preamp stage in a tube amp and I was using a 250v B+? What would your gain approximation be? I'd like to have a good overall handle on this.

[lowell]

Voltage Gain Av = (RL . Ie) / 25

I found this here: THE TRANSISTOR AS A VOLTAGE AMPLIFIER

So the load resistance divided by Ie divided by room temperature. I see from this that voltage gain is highly dependent on temperature. The warmer we get the less gain we get.

Also Ie can be approximated by: (Vin/Rb)*Hfe right?

[Steve Conner]

Hi lowell,

Yes, that equation will work, except it doesn't give voltage gain, because the input is a current Ib. To know voltage gain you need to calculate the transistor's gm ("transconductance"). With the partially bypassed emitter it's easy, gm is just 1/Re, and the voltage gain becomes:

Vout/Vin = Rc/Re (where Re is the unbypassed bit of the emitter resistance)

This assumes Re is large compared to the transistor's own non-linear emitter resistance, which is a function of the collector current. (approximately 26/Ie, answer in ohms if Ie in milliamps)

This is another limit on the gain: You can't have high gm with low current. If you want high gm you have to run the transistor hot, and that means you need to use a smaller Rc so that the extra current translates into a proper bias point, thus losing the voltage gain you were trying to make. 1mA idle current is a good starting point: the transconductance will be 1/26, which is 38mA/V.

For a guitar application you also need to know the input impedance which is beta*Re. That could be as low as 10k, which loads a guitar too heavily. (well, for tube purists: it's a vital part of the Tubescreamer and Fuzz Face sounds.)

This is why Peavey show Darlingtons in their Transtube patent, and why we prefer small MOSFETs like the LND150.

The 50 is clean gain. Distorted gain is always less, which is why people sometimes say "compression" when they mean distortion.

If you used 250V B+, you can use a bigger Rc. Hence the ratio Rc/Re can be bigger and so the gain. But it will start to become limited by Early effect, and the loading of the following stage. 100k like in a regular tube stage would be a good starting point, and you can maybe get 5 times the gain that a 12AX7 would have, because a BJT running at 1mA has much more transconductance than the 12AX7. 40 times more to be precise, so even after throwing a lot of it away by partial bypassing, you can still blow the tube into the next county.

[lowell]

Thanks Steve I'll try and absorb that in my sleep.

[Steve Conner]

Better still, build a couple of transistor amp circuits and plug a guitar into them, you'll soon get a feel for it! Because transistors vary so much, a feel is more useful than the math. Here's the basic circuit, it is the simplest possible one that works reliably. You can use it as a boost pedal and so on.
The Transistor Amplifier - Page 1

For a theorist or mathematician this can be depressing — there isn't one ‘right’ answer. For the rest of us it's good news as it means there are a wide range of values which are ‘OK’.

[lowell]

Steve,
I'm still trying to grasp all of this. I'm studying the J201 datasheet. http://www.fairchildsemi.com/ds/J2/J201.pdf I'm considering this for the input stage of my amp. I'm not looking for a lot of gain. Just a 3v p-p output... so a gain of about 7.5 if using humbuckers (400mv). I'm also considering starting w/ the Dumble ODS input circuit http://www.webphix.com/schematic%20h...eamps/jfet.pdf, subbing this jfet in for the NTE452.

Few things: I cannot find any maximum Id or Vd rating. ?? I see the Vdg and Vge ratings. I guess maybe those are sufficient enough to fill in the blanks? Vdg is 40v so we can assume (since the gate will be 0v) that Vd cannot exceed 40vDC. Also the source cannot exceed +40vDC again because the gate will be 0vDC (this wouldn't work anyway but good to suss out specifics). The datasheet also says (I think) that the jfet is ON at 1ma idle current. Do I need to set my Vd (ha ha VD) to be around 20vDC so as not to exceed the 40v even during operation? If so lets assume a Vd of 20vDC. This also translates to 20mw of power. (just fyi) That means if I use a 10k Rd: 10k*1ma=10v across Rd at idle. Therefore my B+ must be brought down to 30vDC? Maybe I'm way off here. I'm sure the Rs plays a part here as it does in biasing a tube. The larger that Rs is, the more "off" the jfet is... right?

Ofcourse I'm willing to just build this and see, but I want to learn this stuff too.

[jmaf]

Steve,
I'm still trying to grasp all of this. I'm studying the J201 datasheet. http://www.fairchildsemi.com/ds/J2/J201.pdf I'm considering this for the input stage of my amp. I'm not looking for a lot of gain. Just a 3v p-p output... so a gain of about 7.5 if using humbuckers (400mv). I'm also considering starting w/ the Dumble ODS input circuit http://www.webphix.com/schematic%20h...eamps/jfet.pdf, subbing this jfet in for the NTE452.

Few things: I cannot find any maximum Id or Vd rating. ?? I see the Vdg and Vge ratings. I guess maybe those are sufficient enough to fill in the blanks? Vdg is 40v so we can assume (since the gate will be 0v) that Vd cannot exceed 40vDC. Also the source cannot exceed +40vDC again because the gate will be 0vDC (this wouldn't work anyway but good to suss out specifics). The datasheet also says (I think) that the jfet is ON at 1ma idle current. Do I need to set my Vd (ha ha VD) to be around 20vDC so as not to exceed the 40v even during operation? If so lets assume a Vd of 20vDC. This also translates to 20mw of power. (just fyi) That means if I use a 10k Rd: 10k*1ma=10v across Rd at idle. Therefore my B+ must be brought down to 30vDC? Maybe I'm way off here. I'm sure the Rs plays a part here as it does in biasing a tube. The larger that Rs is, the more "off" the jfet is... right?

Of course I'm willing to just build this and see, but I want to learn this stuff too.

I've only done some experiments with the j201 so I'm by no means an expert. jfets are always(?) depletion mode so you'll work them in the negative gate region, similarly to a class a preamp tube. The gate will likely be negative with respect to the terminal you choose as source so they listed that as the biggest potential you can develop inside the device, from drain to gate.

Since drain and source are whichever leg you choose in the J201 that is not the gate, they listed the gate as your reference because it's the only fixed terminal, that's probably why they don't list Vds instead like you said I think you oughta "fill in the blanks".

They provide 0.650 Watts maximum dissipation, so you can probably assemble the current from that. Just my 2 cents..

[Steve Conner]

RG has an article on biasing FETs that should answer your questions. Don't try to run them at more than Idss. It varies between parts up to a factor of 3, so you have to select them or adjust each circuit individually. And the source and drain really are interchangeable.

[Wilder Amplification]

Lowell...try changing your perspective and remembering that bipolar transistors are CURRENT controlled devices. They care not about voltage. The supply voltage against the collector load resistor value sets the max collector current that can flow, while the current gain of the transistor sets how much base current is required to drive the transistor fully on to allow the max current to flow. The base resistor combination determines how much voltage will be required at the base to arrive at that required base current. Therefore, the base resistor combination will have LOTS of control on the voltage gain.

Perhaps you should read up on just exactly how transistors work?

[J M Fahey]

They care not about voltage.

Sorry to disagree.
It's at best a simplification, poorly chosen (in my opinion) to illustrate Tubes vs. Transistors differences.
It causes more confusion than what it "explains".
Transistors *do* have voltage gain; in fact a very high voltage gain, unless reduced by feedback.
Suppose you have a very common BC547C (meaning Beta=500), with 10K load resistor, +B 20V, biased so it has 10V on its collector (for symmetrical clipping/maximum signal possible), so it's passing 1mA idle.
Emitter grounded, as not to add any external feedback into the equation.
Av=Gain=Rl/Re , Rl being 10K as stated before.
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.
Av=Gain=10000/26=384 (I already said it was high)
Intrinsic input impedance= Beta x Re=26x500=13000 ohms. Not bad.
So, in a nutshell, if you have a generator which can drive 13000 ohms and excite said transistor's base with , say, 10mVAC, you'll have 3840mVAC or 3,84VAC on its collector.
Granted, you won't have the 220K or 1M input impedance you can easily get with most tubes, but that's beyond the point.
And you can trade Voltage gain for input impedance by adding external emitter-to-ground resistors.
For example, adding 180 ohms will rise the total Re to around 200 ohms, provide 100K input impedance, and voltage gain around 50, similar to a classic 12AX7 stage.
Not bad at all.
EDIT: please take notice that I am not depending on resistors in series with the base either to provide some input impedance or to transform a "current driven" device into a "voltage driven" one by making said current pass through said resistor.
I am applying signal straight to the transistor base.

[lowell]

JM this helps. I have one question if you could please. Are you voltage divider biasing this transistor's base to .7v? If not then how is the transistor biased ON?

Actually another q too. I read that it's important to have a series base resistor to limit current and protect the transistor. ?

[Enzo]

Well, that depends upon where the wire to the base is coming from.