# BJT Amplifier Output Votlage Swing

Discussion in 'Electronic Basics' started by panfilero, Oct 2, 2012.

1. ### panfileroGuest

I'm having a hard time understanding the "negative swing" of a common-emitter amplifier. There is a website called thesignalpath.com where a BJT amplifier is described.

The set up is a BJT with a collector resistor, an emitter resistor, 2.5V and -2.5V supplies, and we're taking the output off the collector. The BJT has a Vce,sat = 0.2V According to the website

maximum Vout = Vdd
positive swing = VRc
minimum Vout = -2.3V + VRc
negative swing = Vce - 0.2V

I understand the first three things, but the last one, negative voltage swing has me confused... why is the negative voltage swing defined as Vce - Vce,sat?

here's a link to the source

much thanks!

2. ### Guest

What they're trying to say is that the negative swing is |VCC- - VCEsat|, or
the magnitude of the negative swing is the negative supply minus the
saturation voltage.

3. ### panfileroGuest

so you think it's just a typo? That he really meant Vcc - Vce,sat and not Vce - Vce,sat?

Yes.

5. ### Jon KirwanGuest

Shahriar must work for Rigol!

I think Dan pretty much nails it. But let me say it
differently. He's talking about the quiescent value for Vce.
He set Iq=2mA (quiescent current) in his design. This just
means the center operating point which, based on his
assumptions (Ie=Ic), also flows through Re and Rc. He
selected 1.6V as the drop for each and therefore the values
as 800 ohms each, too. This leaves 1.8V for the quiescent
Vce. (Vce will experience BIG changes in operation, but this
is the Vce value that happens when the signal is not
connected and the circuit is just sitting at its DC operating
point. That 1.8V is the Vce he is talking about.

The reason he subtracts 0.2V (for the saturation voltage) is
that Vce can't really get smaller than that without serious
distortion. (Actually, it's better to leave more like 0.8V
(so that the base-collector junction stays reverse biased,
but his power supply rails were VERY small and he simply
couldn't afford to waste it. Besides, it doesn't matter. He
only wants 1V swing peak to peak, anyway. So if he has 1.6V
on Re and 1.8V on Vce, then 0.5V downward would mean 1.3V on
Vce, which is WAY above the 0.8V I'm saying would be better.
So he never really drove that thing anywhere NEAR Vce=0.2V.
He was just using that figure for an absolute worst case
(that you would never really use in reality.)

Jon

6. ### panfileroGuest

Thanks felas, between yalls two explanations i have a much better idea what he was up to. much apprecaited.

7. ### George HeroldGuest

Say can I ask what is perhaps a silly question?
(I won't wait for permission :^)
Now, I don't do much design with transistors, (since opamps are so
much easier), but I thought that you should choose the collector
voltage to be ~1/2 the supply voltage for maximum swing of the
output. This comes from AoE... which IIRC also states that the
maximum gain from a common E amp is 1/2 the supply voltage divided by
the thermal voltage (25mV). (I know this ignores the CE saturation
voltage.)
So it seems that Shahriar could have choosen his operating point a bit
better and gotten closer to his gain of 100 in one stage. Say if you
want, even more gain can you run the collector even lower and give up
the maximum voltage swing?

George H.

8. ### Jon KirwanGuest

and noisier unless you pay dearly
That's another "rule of thumb," but it is not gospel. I've
gradually (that means I'm mentally slow) learned that there
are lots of considerations. Anyway, if you "work the
equations fully" and take into account all the important
things too then you will find that the 1/2 supply rule isn't
reality, either. It's decent, that's all. Centering the
maximum non-saturated collector swing involves more things
than that, though. And it's not always the goal, besides.

Since you bring of AofE, take a look at the student manual
for it. They actually walk you though a CE design starting on
page 115. You will see some competing considerations there
and no discussion at all about setting Iq. Which may also be
important. No rule of thumb is gospel. They just help a
little, is all. And it works a lot better when you have a
large magnitude supply rail pair than when you don't. This
guy was working with 5V.

Some, but not all, considerations: The DC operating point
should, for temperature stability, have the emitter resistor
with as much voltage drop as you can afford to have. This is
because of kT/q in the BJT emitter (it's the cause of little
re which depends on Ie and thereby also on Iq). That voltage
(around 26mV at room temp) is HIGHLY temperature dependent.
So dwarfing it with a drop across Re helps make it
irrelevant. AofE's student manual recommends at least 1V
there to make the T-dependent 26mV not so important. The
author of that video didn't say any of this, but the 1.6V he
assigned is not only reasonable it's also a good idea for
temperature stability. This, of course, "steals away" some of
the range then available between Rc and Vce. Also, you should
allow for a continuously reverse biased BC junction, if
possible, as well. So that sets up a minimum Vce of 0.8V-1.0V
that you should NOT put into your calculations of the "center
point" for the collector. Again this steals away some of what
you plug into your calculations. If you have a 15V power
supply or more you won't care. Just center Vc and be done
with it. But if you are stuck designing something for a pair
of 1.5V batteries, then you start thinking more about the
details and struggling just a little differently when
balancing your priorities. There is no bright line rule
anywhere. Your brain cannot ever be fully disengaged and
there is always more to learn, too, I think. No matter what
you think you know, there is something else out there that
you haven't yet experienced and your "rules" will then get
you in trouble if your brain isn't turned fully on.

So assume I have 5V and use AofE's rule of 1V for Re. And
then apply my own 1V for Vce-min. This leaves 3V total (5V
minus 2V) for collector "swing." So I take the 1V for Re, add
1V for Vce-min, then add 1.5V for half of the 3V swing and
wind up with 3.5V for quiescent Vc, right? That's not 2.5V.
It's 3.5V. But it maximizes the swing under my rules of not
allowing BC to forward bias and allowing AofE's rule for
temperature stability.

I don't use the 1/2 V-supply rule unless I'm teaching someone
who knows nothing about BJTs and wants to just get started.
Yes, he doesn't say it but he makes that point indirectly
when he talks about the max gain of A=-64 depending upon
40*Iq*Rc, or 40 times the quiescent voltage drop of Rc.

But as I've said above, you have other considerations as well
that compete with just throwing more voltage drop at Rc.

Jon

9. ### Jon KirwanGuest

Damn, forgot to mention: I'm just an ignorant hobbyist and
not a professional. I've had ZERO formal electronics training
-- and I mean ZERO when I say it. I'm "book-learned" with
very very modest experience behind it. Nothing I've said
should be taken over what a trained professional says about
any of this.

Jon

10. ### Jon KirwanGuest

By the way, take note that Shahriar designed in 1.6+1.8 or
3.4V for the quiescent collector voltage. Compare that with
my "3.5V" above. Note the similarity? Yet due to different
explanations. Mine considerations point out the "whys" of
getting there, so that you can do your own thinking in
different cases and objectives. His do not. That's the only
complaint I'd have the video. But he was keeping this very
simple (and I think he'd already considered other things
before coming up with his "1/3rd for each" rule for talking
purposes.)

Jon

11. ### George HeroldGuest

Big Grin! Thanks Jon. You're much better at discrete design than I
am.
(Which isn't saying much.)
After spouting off I was reading a bit in AoE.
(I use to have the lab manual, but it went to a new home
a few years ago, hopefully guiding the next generation.)

I totally agree about not wanting to have the gain based on the (25mV)
thermal voltage. But to do that you can't bypass the entire emitter
resistor. (Maybe that is the next lesson?) I do like having big
supply voltages, then wasting a volt here or there doesn't cost you
much.

George H.

12. ### Jon KirwanGuest

The capacitor (soon) acquires a voltage equal to the DC
operating point across Re. All it does is short out the AC
part. But the DC operating point (and therefore Iq) at the
tip of the emitter holds solid. (Shahriar, shortly after
about 6 minutes into the video, shows you the Ie equation he
easily derived, but without showing you how.)

The point of Re is to maintain the biasing point so that the
expected useful collector peak to peak remains valid. And it
does. The capacitor doesn't impact that, since it is almost
entirely determined at DC (the AC does wiggle things very
slightly.) So the biasing point, and Iq, remain stable over
temperature.

Temperature impacts kT/q, which impacts A due to the fact
that the value of "40" in the AC gain of 40*Iq*Rc is just
q/kT and is inversely dependent on T. But exact gain usually
isn't the goal in a single BJY stage -- just a rough design
guide for validating assumptions (The room allowed for
collector swing, for example.) If exact gain overall is
more work on considering sources of time and temp drifts and
their impacts -- or else you stick the circuit into a more
temp stable place like the body cavity of a living human.)

You can improve on the temperature dependence of AC gain,
while keeping Re's DC operating point, by inserting another
resistor (Rx) in series with the emitter bypass capacitor.
Then the gain stops being 40*Iq*Rc and looks a lot more like
Rc/Rx.

And you can then improve the "stiffness" of the base (greatly