# Transistor biasing

I know this question has been asked a million times, but I've read all
the answers and still don't understand. I've been studying this off
and on for years and can't seem to get it. I just give up for a few
months and come back.

I keep reading that in order to properly bias a transistor, you need to
first determine Ic then figure out Ib=Ic/Hfe. The problem is Hfe is
all over the place and all the datasheets I look at give the transistor
Hfe between 25 and 300. That doesn't help me. Maybe if it was between
50 and 60 I could average, but not 25 and 300. Plus, that data is for
a 10V supply, which I might not want, and give no data for other supply
voltages. I want Ic to be 100mA, but with no idea of what Hfe is, how
do I go from there and be confident that my design will work and not
explode in my face when I hook it up?

Ok, so Hfe varies with Ic. So the solution is to put a resistor before
the emitter and have negative feedback stabilize the gain. But how do
I determine the value of that resistor? And for that matter of the
potential divider on the base or the collector resistor. I search and
search and all I get are specific examples of biasing with no clue how
the values were determined. Do I really have to trial and error this
stuff? Is it even possible to plan a circuit on paper and not have it
behave wildly different when it is built?

One more thing. The negative feedback is supposed to keep the gain of
the amplifier constant regardless of the transistor inserted. But does
that mean regardless of the differences between individual transistors
of one type (like 2N222) or can I plug in ANY transistor of any type
and the gain should stay about the same?

Any help is appreciated. Maybe it'll jar my brain into finally
understanding...

2. ### Ralph MoweryGuest

Look here for an explination.

http://www.interq.or.jp/japan/se-inoue/e_dance26.htm

Forget about Hfe of the transistor. It mostly maters only in that you have
enough. The ratio of the collector and emitter resistors determin the DC
gain of the circuit. The AC gain can be improved by a capacitor across the
emiter resistor. Also do not try to calculate the gain to the smallest
fraction or many decimal places as most formulars are only close
aproximentations.

3. ### john jardineGuest

The general idea is to design for worst case conditions. If your minimum Hfe
looks likely to be a ball park 25, then design to this value (or preferably
less).
In practice the Hfe value is only of interest when setting the base bias
resistors and TBH it's usual to design most amplifiers without even looking
at a data sheet; just assume all small transistors have a Hfe of 100 and all
power transistors a Hfe of 10. Your Hfe estimates may be way off the mark
but the circuit will still work, (that's why the biasing resistors are
there). Only noticable change would be a small shift in the static collector
voltage.
The emitter resistor is simply set to a value that that will give as high a
emitter voltage as you can get away with. The higher the voltage the more
stable the biasing. A couple of volts or more is OK (say 22ohms at 100ma),
0.5V is pushing it a bit.
The potential divider at the base is set to pass around 4x to 10x times the
expected base current (your choice). The expected base current depends of
course on that Hfe but is now of little consequency as the base current
loading (ie due to Hfe differences) can vary over a massive range without
upsetting the base voltage (hence emitter voltage, hence collector voltage)
too much.
And yes, plug in pretty much any transistor and it'll work the same. Which
it must, seeing as you've just cornered the gain differences, so there's
pretty much nothing left but basic 'transistor action'. Which is the same
everywhere.
Easy eh!.
regards
john

I think, we need "beta" as high as possible so we can use only
resistors to determine the GAIN of the amplifier.

I'm confused with your explanation here, could you contrasted to using
single base bias resitor only, without using Emitter resistor as
negative feedback element.
-------------------------------------------------------------------
Using potential divider bias, hence Thevenin source and Therenin
resistor, I can't understand how that would stabilize Ic.

thank you to clear up my confusion

5. ### Jasen BettsGuest

it's easier in a spreadsheet, kirchoff's laws can be put in and you can see
the effects of different choices...

as long as circuit gain is significantly less than the Hfe, and the
replacement transistor can handle the conditions, (current voltage wetc)
yes.

Trying to use a transistor with a Hfe of 25 in a circuit with a gain
of 20 probabl;y wouldn't work too well...

Bye.
Jasen

6. ### Ralph MoweryGuest

I'm confused with your explanation here, could you contrasted to using
Unless you want to measuer each transisitor you use, you can not use the
single base bias resistor. Even then as the transisitor heats up the
circuit will often bceome unstable (thermal runaway) due to the changing of
the charistrics of the transistor. There are some circuits that try this
with a diode mounted on the same heatsink as the transistor to compensate
for the heating of the transistor.

The formular in the text book is for a "perfect" transisitor where the
transistors parameters are known and do not change due to heat and other
factors.

There are 3 common circuits that can be used to bias a NPN transistor. The
single resistor from V+ to the base, The one that adds an emitter resistor,
the one that adds a resistor from the base to the negative voltage. Each
one is more stable than the previous. The single resistor to the base is
way too unstable for mass production.

7. ### John PopelishGuest

The point of view that makes base biasing sensible is to understand
that junction transistors are turned on by voltage. That is, the base
to emitter junction must be forward biased by some particular voltage
to turn on some particular collector current. But while this bias
voltage is doing its job, the base emitter junction is also leaking
diode current. So there is an incidental ratio of collector current
to base current (that varies with conditions and with different
devices) and that ratio is called beta.

The resistor divider method is a way to produce the appropriate base
voltage, while also making a voltage source that has a low enough
effective (Thevenin) series resistance to provide the needed base
current without much distorting that voltage. If the divider passes 4
to 10 times as much current as the worst case (lowest beta) base
requires, then its voltage is sufficiently stiff to hold the bias
voltage acceptably stable for devices of higher beta, as well.

The emitter resistor is a negative feedback mechanism that provides an
other input voltage (remember that the input is the voltage between
the base and emitter, so both those terminals are inputs) related to
collector current. The difference between the voltage produced by the
emitter resistor and the base divider voltage is the actual input bias
voltage for the transistor.

9. ### john jardineGuest

Yes.
Last month I bought 200 off (cheap) BC557C transistors and in an idle moment
measured the Hfe of one of them. It came out at a rather pleasing 450.
Curious, I started to measure the others in the packet. Surprised to find
all of them had values of 450 +/-5% !!. (still though design 'em in as if
they have an Hfe of 100).
The single resistor bias method is theoretical, only to be found in
textbooks and is worthless in practice. (sorry

John P' knows the 'current/voltage' dichotomy and his provided explanation
would also be mine.
regards
john

:The resistor divider method is a way to produce the appropriate base
:voltage, while also making a voltage source that has a low enough
:effective (Thevenin) series resistance to provide the needed base
:current without much distorting that voltage. If the divider passes 4

:to 10 times as much current as the worst case (lowest beta) base
:requires, then its voltage is sufficiently stiff to hold the bias
:voltage acceptably stable for devices of higher beta, as well.

for same type of transistor, would the one with higher beta draw more
Ib compared to one with lower beta.

:The emitter resistor is a negative feedback mechanism that provides an

ther input voltage (remember that the input is the voltage between
:the base and emitter, so both those terminals are inputs) related to
:collector current. The difference between the voltage produced by the

:emitter resistor and the base divider voltage is the actual input bias

:voltage for the transistor.

wow....thats a clever idea to regards those terminals as inputs. I
could never have thought that.

Based on your explanation I tried to simplify so my lame brain can
digest it. I think, the variation on voltage across, BE junction, due
to R-emitter, provide the negative feedback.

11. ### bgGuest

The gain of an amplifier using feedback is given by --- A' = A / 1+BA
A' = Gain with feedback
A = Gain without feedback (open loop gain)
B = fraction of output fed back to the input (feedback factor)

Here are a few examples -
If A = 50 and B=1% Then A' = 50/ 1+.05 = 47.6
If A = 100 and B =1% Then A' = 100 / 1+1 = 50
If A = 200 and B = 1% Then A' = 200/ 1+ 2 = 66.7

Note that with 1% of feedback, the gain with feedback varied by about 40%
while the open loop gain varied by 400%. If this had been a transistor with
beta varying from 50 to 200, by using feedback we could design a stage with
a current gain that would fall somewhere between 47 and 66.

Now look what happens if we increase the feedback to 50%
If A = 50 and B=50% Then A' = 50/ 1+ 25 = 1.92
If A = 100 and B =50% Then A' = 100 / 1+ 50 = 1.96
If A = 200 and B = 50% Then A' = 200/ 1+100 = 1.98

For all practicle purposes, with 50% feedback, the stage gain is 2 while
the open loop gain varies by 400%. Again, if this was a transistor
amplifier, any transistor in the lot would make an accurate current gain of
2

The downside to negative feedback is that we waste alot of gain but the
upside is that we can make the gain very predictable. The reduction in gain
is a direct indication of the effectiveness of the feedback. .

Current gain -
It is typical to use enough feedback so that the stage current gain drops to
1/4 or less of the gain without feedback. All we need to know is the minimum
beta to expect. For example, if you were using a transistor type, where
beta varied from 100 to 200, obviously you would never design a stage
having a gain of 150 unless you hand picked the transistors, but you could
design a stage with a current gain of 20 and expect all of the transistors
to work.

In a common emitter transistor amplifier -
Rl is the collector load resistor
Re is the emitter resistor
Ra is the resistor from Vcc to the base
Rb is the resistor from base to ground

The ratio Rb/Re is the stage current gain. This ratio is kind of equivalent
to B. That is, as Re gets larger, or Rb gets smaller, we are using more
negative feedback and the stage current gain is reduced. If we had a
transistor with a minimum beta of 100 , it would be typical to set Rb/Re to
about 25. A higher ratio is acceptable if you can tolerate the gain
variations, or a lesser ratio, if really tight control of the gain is
needed. It would also be possible to use a feedback resistor from the
collector to the base or global feedback form another stage to stabilize
this stage but that is a different story. For now we will look at biasing
when the feedback is due to Re.

A typical transistor will have a maximum beta at a certain collector
current, or a fairly constant beta over a range of collector current. This
range of collector current is where the transistor was designed to operate.
When Vbe is at .5 volts, the transistor collector current is at its lowest
range. When Vbe is at .6 volts the collector current is just about right,
and when Vbe is at .7 volts, the transistor is conducting heavily or near
saturation. We don't know the exact value of IC for a given Vbe but we can
get an idea of where IC is, over it's usefull range. I always shoot for a
Vbe of .6 volts and then tweak as neccessary. If we have an emitter resistor
of 1K and we want to operate at a collector current of 1ma, then the voltage
across the emitter resistor is Ie x Re = 1 volt Note that Ie is Ib+Ic, but
if Ib is a small part of Ie which usually is only about 1% of Ie, then for
all practicle purposes Ie = Ic. The base voltage is .6 volts higher than the
voltage across the emitter resistor or 1.6 volts. This base voltage appears
across Rb. Ra then has to drop Vcc - Erb at a current of Irb. This gets you
close to the value of Ra. Do not expect it to be exact. the reason being,
is that Vbe increases 60mv for a tenfold increase in Ib. It is an
exponential relationship and temperature dependent. Therefore the value of
Vbe is somewhere between .5 to .7 volts over the usefull range of Ic. If you
want an equation for Ra, try this approximation -

Ra = E/I = (Vcc - Erb) / Irb
And Erb = Emitter voltage + .6 volts
Emitter voltage = Ic * Er

Up to this point, you should have a grasp on the ratio Rb/Re to set current
gain, and how to get a ballpark value for Ra which supplies the forward bias
base current, but we have said nothing about the value of Vcc or the
voltage gain. Providing we use enough feedback to reduce the stage current
gain, the voltage gain will be Rl/Re. Our AC input voltage is divided up
across the base emitter junction and Re, and the output voltage is across
Rl. In terms of the AC signal voltage , voltage gain = E out/Ein = Collector
Ac voltage / Base to Ground AC voltage. If the signal developed across the
base emitter junction is small compared to the siganl developed across Re
(which it will be with enough feedback from Re), then the Ac voltage gain is
Rl/Re.

Ideally we can vary Vcc without significantly changing Ic provided that Ra
is tied to a supply other than Vcc, because if we vary Vcc, then we vary Ib
which will vary Ic. The value of Ra is probably best calculated after a
value is chosen for Vcc. So far we can set current gain with the ratio Rb/Re
and voltage gain with the ratio Rl/Re. To select Vcc we need to consider the
Q point collector voltage and the transistor power dissapation. If we had Ic
with no input signal = 1ma and Rl = 10K and Re = 1K then we drop 10 volts
across Rl and 1 volt across Re. Lets assume Vcc = 24 volts. When the
transistor is not conducting (open circuit), Vc = 24 volts. When the
transistor is fully conducting (short circuit) then Vc = 2.2 volts.
Visualize this as a simple voltage divider. When the transistor is an open
switch, the collector voltage is Vcc and when the transistor is a closed
switch we have a voltage divider of Rl in series with Re. The transistor
will not be a perfect switch with 0 ohms but this is close enough for an
estimate. Therefore our output signal can vary from 2.2 volts to 24 volts.
We can have a peak to peak output signal of 24-2 or 22 volts. The largest
voltage swing we can have at the output would occur when the collector Q
point voltage sits in the middle of this swing or 2 +11 volts = 13 volts.
This is class A operation. In a typical amplifier, we would not want to
swing the out put signal this close to its limits because it would distort.
If you
needed a 22 volt peak to peak signal you would have to raise Vcc to provide
headroom. But this example might be perfectly ok for say a 20 volt or less
peak to peak swing. As the value of Vcc goes up, the transistor power
dissapation also goes up which causes the transistor to heat up or smoke or
both.

Hopefully this gives you a better understanding of the DC bias. Set the
current gain Rb/Re, so it is small compared to beta. Set the voltage gain
Rl/Re to any desired value, but remember that a high voltage gain means Rl
has to be large compared to Re and this means that Vcc also has to be
higher. The collector Q point voltage puts a limit on just how far the
output signal can swing. The largest swing for class A operation is when the
collector Q point voltage is midway between Vcc and Ve when the transistor
is shorted. Typically, voltage gains max out at 20 to 30 in practicle
amplifiers. Also when the voltage gain is high, the bandwidth is reduced and
the output Z is raised, but these are compramises you don't need to deal
with just yet.

For low frequency or DC operation, the input resistance is Ra and Rb in
parallel and the output resistance is Rl.

http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

---------------------------------------------------------------------------------------------------------------------
Kevin,

I would strongly suggest you write a book about that. It sound more
autoritative.

It doesn't have to be thick but its enough to emphasize that transistor
is NOT a current controlled device.

I believe all of us here will support you.

best wishes

13. ### Kevin AylwardGuest

I have played with the idea of doing a book on and off over the years. I
might have a go after I finish the SS update I am working on.

I think that there is a place for something in-between Wins Art of
Electronics and the high brow, pretentios academic texts.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

Thank you to everyone that replied. I think my biggest problem was
that I didn't realize that the transistor Hfe was not the same as the
amplifier gain. That cleared a lot of confusion. Also, despite
reading it over and over, it just never really sunk in that a diode
drops about .7V regardless of the current through it. I was really
mixed up with that when trying to set the base current. I've been able
to design a few amplifiers on paper now and have them perform as I had
calculated (or very nearly so). I feel much better now.

Thanks again!

15. ### John PopelishGuest

If there is no emitter resistor, both bases look like a diode to the
emitter voltage, so, the base current would be independent of beta.
What would change would be the collector current. If there is an
emitter resistor, also, then as the collector current tends to rise,
so does the emitter voltage, so the higher beta unit would draw less
base current and the collector currents would be more similar (though
the high beta unit would still have a somewhat higher collector
current than the lower beta unit).
Right.

Except that there are two kinds of negative feedback possible. One is
based on the collector current (That is mostly what the emitter
resistor reacts to. Of course, the smaller base current also passes
through it.) The other kind is based on the collector voltage. You
can add the second kind by deriving the base voltage divider from the
collector voltage, instead of from the supply voltage. That way, if
the collector current increases, it lowers the voltage applied to the
base divider and reduces the collector current increase.

You can use both types of negative feedback in the same circuit or one
or the other. Current negative feedback raises both the input and
output impedance, while voltage feedback lowers both. So you can
combine them in various combinations to lower the gain (and stabilize
it for beta variations) and also get the input or output impedance
where you want it.

16. ### Jasen BettsGuest

transistors from a single batch are likely to be well matched
my (non authporitive) reference lists BC557 has Hfe 110-330, I think
A,B,and C variants of a transistor model usually offer higher Hfe,
I use it here, well a single bias resistor from the collector to the base,
it works great to amplify a surplus 600 Ohm dynamic microphone connected
to a soundcard input that (I think) expects an electret microphone.

admittedly I did experiment with different base resistors until I found one
that worked well.

+-[4M7]--- C -------------->
22uF | to soundcard
|| | Q1
--->>---||---+---- B BC547B +--------->
mic ||+ |
600R |
--->>------------------ E ----+

Bye.
Jasen