Setting up a transistor

[Rock_on]

Here comes the noob again hahaha

But this is now more simple than before

Questions are :

1.) how can i set a transistor's operating voltage (ex. 10mA @ 5v)

2.) what does Vb do? (i know a transistor need it to turn it ON but what i mean is, what's the difference between 2v and 3v? Does higher Vb = more gain? Im sure it's no, but what does it do?)

3.) if im using voltage divider as a biasing technique, what should be my Rtop and Rbottom value? ( i know that same value = half the voltage, but what i mean here is how much current do i need? Example, we know that using 2 50K or 2 10k is still = to half the voltage, but the current is different.)

That's it
Please answer

[R.G.]

But this is now more simple than before

Questions are :

1.) how can i set a transistor's operating voltage (ex. 10mA @ 5v)

The simplest way is to use an emitter resistor. The emitter resistor always has a voltage across it of V = (emitter current) * (emitter resistance). You set the emitter voltage by taking note that the base voltage is always one diode-drop above (for NPN) the emitter, and setting the base voltage. The gain of the transistor action and the feedback of the emitter current be the right value to do this. But that only sets the emitter voltage and current.

The collector current is equal to the emitter current minus the base current. In a transistor with a gain more than 100, the error in ignoring the base current and saying that collector current equals emitter current is less than 1%, so you can assume the collector current is the same. Then to fix the collector voltage, you know the collector current is the same as the emitter current, so you pick a collector resistor which makes the collector voltage be where you want it.

2.) what does Vb do? (i know a transistor need it to turn it ON but what i mean is, what's the difference between 2v and 3v? Does higher Vb = more gain? Im sure it's no, but what does it do?)

It sets the base voltage, and the emitter voltage must follow along if the rest of the circuit will let it follow along. Gain is determined by other things.

3.) if im using voltage divider as a biasing technique, what should be my Rtop and Rbottom value? ( i know that same value = half the voltage, but what i mean here is how much current do i need? Example, we know that using 2 50K or 2 10k is still = to half the voltage, but the current is different.)

The voltage divider must provide the base current and still hold the base at the right voltage for biasing. In order to do this well, the current through the divider must be significantly larger than the base current. The base current is 1/HFE of the emitter current, and the emitter current is the base voltage minus one diode drop divided by the emitter resistor.

This presents a problem because it depends on HFE and you don't usually know exactly what HFE is. So you make a good guess at what HFE is - 100 or more is usually OK - and estimate the base current. Then the current in the divider needs to be much larger than the base current.

How much larger? Generally more than 10x larger than the base current


There is a huge background of things behind this very, very short explanation. Your question was approximately the topic of five junior-level EE lectures when I was in school, with homework and textbook assignments as well as prerequisite circuits courses, so I've boiled it down. A lot.

[Rock_on]

Ok so let's put on some specific values

Looking on the 2n3904's DC Current Gain section

At the test condition, with the Ic = 10mA, Vce = 1.0v, the min gain is 100 and max 300

I would like to have a gain of 100 in this example

So how do you solve for the values??

[R.G.]

Looking on the 2n3904's DC Current Gain section
At the test condition, with the Ic = 10mA, Vce = 1.0v, the min gain is 100 and max 300

I would like to have a gain of 100 in this example
So how do you solve for the values??

You can't.

First, the datasheet specifies current gain at 100-300. Current gain is not equal to voltage gain, which is what I think you want.

Second, it is possible to get large voltage gains, but not with simple four-resistor circuits. The reasons have to do with how you have to pick the resistors to get things to work at all.

One of the things that those lectures passed over was that the amount of voltage gain you can reliably get is limited by your needing to use feedback to stabilize things. Another was the internal operations of the transistor as expressed in the Shockley resistance.

What you can do is to get a voltage gain of 10-50 reliably, or a high gain where you take what the transistor's internal operations give you.

Here's the real problem. The voltage gain of the four-resistor NPN circuit closely approaches Rc/Re, where Rc is the resistor in the collector for *AC* purposes, not necessarily the DC value, and Re is the UNbypassed portion of the visible Re and the internal, you-can't-get-at-it emitter resistance.

If you have Rc=10K and Re = 1K, and the biasing allows this to work linearly, your gain is going to be very close to 10. If you split the emitter resistor into two 500 ohm resistors and use a great big capacitor to form an AC short around the bottom one, the gain will increase to very nearly 20. If you use the same BFC to bypass the entire emitter resistor, the gain does not go to infinity, but is instead limited by something called the Shockley resistance.

The Shockley resistance is an internal operation that acts like a resistor, but isn't. It's a quirk of how transistors amplify, and is only approximated. The Shockley resistance varies with emitter current, and acts like a resistor of 26mV/Ic. This effectively limits the gain you can get in complex ways.

And as you probably note, it varies with collector current.

So some results are:
You can get designed-for gains of 10-50, with the top end being very loose and unreliable.
You can get higher gains, but varying and ill-defined.


I can tell you how to do the normal four-banger setup. In fact, I've typed that in several times in this forum before. Can you find that?

[Kesh]

here's the way i learned. step by step

set up this



choose a reasonable value for Vc and Ic and Ve.

use ohm's law to calculate Rc from VP - Vc and Ic

work out Ib using the transistors typical gain/beta/hfe whatever you want to call it. Ib = Ic/hfe

work out Ie as Ib + Ic

calculate Re from Ie and Ve using ohm's law

calculate Vb as Ve + 0.6 (one diode drop)

pick a bias current I1 of about 10 x Ib

calculate R1 from Ib and VP-Vb using ohm's law.

calculate I2 as I1 - Ib

calculate R2 from I2 and Vb using ohm's law

voltage gain will be Rc/(Re + re) (re is the emitter's internal resistance). to get more gain bypass all or part of Re with a cap of high enough value.

[Rock_on]

here's the way i learned. step by step

set up this



choose a reasonable value for Vc and Ic and Ve.

use ohm's law to calculate Rc from VP - Vc and Ic

work out Ib using the transistors typical gain/beta/hfe whatever you want to call it. Ib = Ic/hfe

work out Ie as Ib + Ic

calculate Re from Ie and Ve using ohm's law

calculate Vb as Ve + 0.6 (one diode drop)

pick a bias current I1 of about 10 x Ib

calculate R1 from Ib and VP-Vb using ohm's law.

calculate I2 as I1 - Ib

calculate R2 from I2 and Vb using ohm's law

voltage gain will be Rc/(Re + re) (re is the emitter's internal resistance). to get more gain bypass all or part of Re with a cap of high enough value.

Thanks!! I think that's what im looking for since last time.

[PRR]

{sigh} If I'd known R.G. and Kesh had replied in another thread, I would not have typed all this.

> ex. 10mA @ 5v

Why would a small-audio stage ever run at 10mA? Whatever.

If you don't know any different:

* Use that "voltage divider bias" you mentioned.

* Set the Emitter voltage to 1V.

The rest follows from there.

10mA at 1V means a 100r emitter resistor.

5V up from that is 6V. Ass-ume a 9V supply? Then the collector resistor drops 9V-6V= 3V. At 10mA this is 300r collector resistor.

(And here's an issue. A 300r collector load can drive external loads as low as 1K. But small-audio loads are more often over 10K. So why use so much current? Define your load, pick Rc 2 to 10 times smaller, and work from there.)

The base voltage will be 0.6V up from the emitter voltage. 0.6V+1V= 1.6V.

If base current were zero we could pick "any" voltage divider that made 1.6V from 9V. However there is base current, and we don't know exactly, only roughly.

Base current is Ie/hFE. Whenever possible, use the highest hFE part which will stand the power. For small-audio, parts with hFE of 300-800 are readily available. Assume you get mostly worst-case parts. hFE=300.

Then base current is 10mA/300= 0.033mA.

We would like this 0.033mA to cause less than a tenth-volt sag in the voltage divider. 0.1V/0.033mA is 3K equivalent resistor. We also want a 1.6V/9V or 0.18 (AKA 9V/1.6V or 5.6) voltage ratio. 5.6 is not a small number, so the lower resistor can be slightly higher than 3K. 3.3K is a common part. The upper resistor is 5.6-1 or 4.6 times higher, 15.2K. Check: 3.3K||15.2K= 2.7K, we wanted a bit less than 3K for tolerable sag, this works.

An alternative way to find base bias resistors: take the square-root of your assumed hFE. Root-300 is 17. Set bias resistors to flow Ic/17. 10mA/17= 0.59mA. The lower resistor drops 1.6V at 0.59mA so must be 2.72K, upper drops 7.4V at 0.59mA so must be 12.5K.

(Different numbers for the two processes because of different simplifying assumptions.)

From inspection, the voltage gain is Rc/Re or 3. This is small gain. It can be increased with an emitter bypass R and C.

The audio input impedance is the base bias divider equivalent, parallel with base input impedance Ze/hFE, where Ze is the equivalent Re with any bypass. If we bypassed the 100r with 42 ohms (to get Ze of 30r and voltage gain near 10), the base input impedance is 9K. Bias resistance is near 3K. Total input impedance is near 2.2K.

[kingswayguitar]

great thread
if i have a dc bias network to the base of

V+(11.7V)....47k....base(1.398V)...10k...ground,

it appears as though i have 0.219mA through the 47k and 0.14mA through the 10k.
so 0.08mA must be pouring into the base, correct?
i have an emitter resistor of 100R sitting at 0.820V and clean gain.
the collector is 780R/5.3V.

[Thecomedian]

Ok so let's put on some specific values

Looking on the 2n3904's DC Current Gain section

At the test condition, with the Ic = 10mA, Vce = 1.0v, the min gain is 100 and max 300

I would like to have a gain of 100 in this example

So how do you solve for the values??

As R.G. said, current gain != voltage gain. It's one of those things that I don't think I've ever heard any book really cover. It doesn't matter how much you push the transistor, you can't get the hFE number to be Voltage. Since Voltage gain is so fractional to current, this is why people love high gain transistors, and why the industry bothers with making them at all. It would be too easy to get the voltage gain you wanted if voltage increased at a 1;1 ratio with current.

what you want to know about is CCVS's, or Current Controlled Voltage Source.

http://en.wikipedia.org/wiki/Dependent_source

Here's an 'okay' place to start. Try to find some transresistance models to work with if you're into mathing it out.

[PRR]

> one of those things that I don't think I've ever heard any book really cover.

Why should they have to say that?

If you don't know that Voltage != Current, you shouldn't be reading chapter 13, you should re-re-read Chapter 1 until you have the basics digested.

If _I_ was running the school, you'd have to show you could do the calculation that kingswayguitar just posted, in a jiffy, after a 20 mile hike, before you were allowed to say "transistor".


BTW, it's perfectly possible to have Gv==Gi even in cascade; and to get very close with just one transistor per stage.


> this is why people love high gain transistors, and why the industry bothers with making them at all

The fact that hFE < infinity is a "flaw". Consider: vacuum-tubes and FETs have no "base current", yet we use them (sometimes beCAUse they have low-low-low input current). Circuit design is somewhat simpler if base current didn't exist. Circuit performance can usually be better if hFE is known to be very high.

But "high" is relative. I was trained on "assume hFE>50", which is often high-enough. This was shortly after the time when popular-price transistors could have hFE more like 5; yet we did stuff with them.

Question: what's better than one transistor hFE=500? Two transistors hFE=50. With proper connection you have effective current gain approaching 50*50= 2,500. And there was a period when hFE=500 was 50 cents, hFE=50 was 10 cents. Low transistor count is not necessarily a virtue.

[Thecomedian]

because it took me a while to realize that while playing with spice. I'm quite sure im not the only one who believed that voltage gains were underwhelming when getting 400 gain currents.

Don't more transistors cause more noise, though?

[mistahead]

because it took me a while to realize that while playing with spice. I'm quite sure im not the only one who believed that voltage gains were underwhelming when getting 400 gain currents.

Don't more transistors cause more noise, though?

Yes... and I'll assume we can't tolerate it...

Consider if the cost to implement a bandpass filter to remove the noise is twenty cents worth of passives, now again if it thirty cents worth of passives...

[R.G.]

The algorithm for biasing a single transistor in stabilized-bias setup is pretty straightforward.

However (you knew that was coming, right?) the process of picking a collector current and voltage is not necessarily, and the process of making the circuit work with the signal source you are stuck with and the load that's going on the collector is not.

You can't pick your signal sources, in most cases. They are thrust upon you, and you must cope with them how you may. And the load you must drive is really part of the collector resistor in an AC sense, so it directly lowers voltage gain.

In many cases, you can't get a high enough input impedance without either going to funny biasing or an input buffer. The algorithmic answers to the input bias divider string usually produce an input resistance that's not high enough for guitar. This is one reason for wanting very high HFE - it lets you bias the transistor with high value resistors. A close competitor is that high-HFE transistors tend to make the assumptions and approximations that you made in setting the circuit up work out really well.

Noise is a whole other can of worms.

The right number of stages for lowest noise is always one. It's only ever not one if you simply can't do the overall job that way. It is one of those fortunate things left to us by the deity who set things up that high gain bipolars also tend to be quieter. More stages give more noise, although probably not grossly more if you're careful.

And to correct one misconception, you cannot filter noise out of your signal once it's in there. It's like those cosmetic commercials where they say "reduces the APPEARANCE of fine lines". You can remove some of the prominent annoyance of noise by filtering, but any signal that happens to be at a similar frequency to the noise you're getting rid of is removed too.

Paul's right about not skipping the dull chapters in the textbook. You're not going to win a marathon if you never learned to tie your shoes. Skipping to noise considerations and filtering is a few steps removed from the important stuff of learning to bias transistors in the many and several ways it can be done, and learning to reason about why one would be used over another.

[R.G.]

It occurs to me to ask a silly basics question that seems to be completely lacking in most people studying effects.

Who was Thevenin, who was Norton, and why were their circuits equivalent?

[artifus]

who was maxwell?

c'mon rg - you had a subtle dig at puretube for such shenanigans not so long ago...

[R.G.]

who was maxwell?

He invented coffee.   

c'mon rg - you had a subtle dig at puretube for such shenanigans not so long ago...

I did, but  my version of this (1) is a footnote to a long explanation which preceded it, (2)  it's clear that it's for "extra credit" study, and (3) it's not the only way I interact here.

[artifus]

fair dues - i thought maxwell was just the guy with the silver hammer anyway.  

*leaves apple on desk on way out of class*

[mistahead]

R.G. said:
"And to correct one misconception, you cannot filter noise out of your signal once it's in there. It's like those cosmetic commercials where they say "reduces the APPEARANCE of fine lines". You can remove some of the prominent annoyance of noise by filtering, but any signal that happens to be at a similar frequency to the noise you're getting rid of is removed too."

Thanks for this clarification - I always tend to end up thinking of transient noise which is quite different to the signal, from network geek background a lot of my signal doesn't overlap background noise freqs. as the above suggests (thankfully) but in music I understand that any noise in the audible range is definately going to be in a part of the signal I need to keep (analogue vs. digital also).

And to clarify - I'm reading those early chapters, very slowly lol.

[PRR]

> Don't more transistors cause more noise,..?

Everything hisses. This is normally small. If we apply an amplifier, the hiss gets bigger.

You can make dust look like boulders if you use enough magnification.

So the noise level is primarily about GAIN.

The number of transistors does NOT have to matter. Some very-low-noise amplifiers are built with chips containing 37 transistors.

In a moderately rational design, the first stage causes 90+% of amplifier hiss, and the other stages need not be designed for low-hiss. And in most well-developed audio systems, the *passive transducer* (microphone or other pickup) contributes more hiss than that first stage.


> a silly basics question

Perhaps not so completely "basics". W. Marshall published an extended essay in IEEE Trans:

On the Application of Thevenin and Norton Equivalent Circuits and Signal Flow Graphs to the Small-Signal Analysis of Active Circuits
W. Marshall Leach, Jr., Senior Member, IEEE
IEEE Transactions On Circuits And Systems- Vol 43, No 11, November 1996
The expressions for the small-signal gain, input resistance, and output resistance of active circuits can often be written by inspection if the small-signal Thevenin and Norton equivalent circuits seen looking into each terminal of the active devices are known. These circuits can also be used to simplify the noise analysis of active devices. In the analysis of circuits with feedback, simultaneous equations must be solved. Mason's signal flow graph is a convenient tool for obtaining the solution. 

[Thecomedian]

wouldn't it be good to darlington or "fuzz face" (which is different from darlington vis. collector drives base instead of emitter drives base) a pair of transistors, using the first one as low noise so that any noise amplification provided by Q2 is extremely small, while gain is still big?

Paul's right about not skipping the dull chapters in the textbook.

As far as that goes for me, I've been over the beginning chapters online and in hand me down books until my eyes bleed.