BradBrigade wrote in message
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...
-Brad
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.
