# BJT Amplifier Output Votlage Swing

Started by October 2, 2012
```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!
```
```On Tue, 2 Oct 2012 08:02:21 -0700 (PDT), panfilero <panfilero@gmail.com>
wrote:

>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
>

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.
```
```On Tuesday, October 2, 2012 11:19:15 AM UTC-5, k...@att.bizzzzzzzzzzzz wrote:
> On Tue, 2 Oct 2012 08:02:21 -0700 (PDT)
>
> wrote:
>
>
>
> >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
>
> >
>
>
>
>
> 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.

so you think it's just a typo? That he really meant Vcc - Vce,sat and not Vce - Vce,sat?
```
```On Tue, 2 Oct 2012 12:54:16 -0700 (PDT), panfilero <panfilero@gmail.com>
wrote:

>On Tuesday, October 2, 2012 11:19:15 AM UTC-5, k...@att.bizzzzzzzzzzzz wrote:
>> On Tue, 2 Oct 2012 08:02:21 -0700 (PDT)
>>
>> wrote:
>>
>>
>>
>> >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
>>
>> >
>>
>>
>>
>>
>> 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.
>
>so you think it's just a typo? That he really meant Vcc - Vce,sat and not Vce - Vce,sat?

Yes.
```
```On 10/2/2012 8:02 AM, panfilero wrote:
> 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!

Probably a big part of the reason for your confusion is that he does not
explain where the equation comes from and it is not true until he adds
the emitter capacitor a little later in the presentation. Without that
capacitor it is not true.

The circuit shown is an inverting amplifier.  I.e. when the input (base)
voltage goes higher, the output voltage goes lower and vice versa. Thus we
get the maximum output voltage with a low input and the lowest output
voltage with a high input voltage.

The maximum output voltage occurs when the transistor is turned off and
there is no current flow through Rc so Vrc goes to zero. This occurs with
a low input voltage. The output voltage is Vcc.  The output voltage at the
bias point was Vcc - VRC. The swing in the output voltage was VRC. (Where
Vcc is the positive supply voltage = 2.5 volts, and VRC is the bias voltage
across the collector resistor.)

The minimum output voltage occurs when we have driven the transistor into
saturation. This occurs with a high input voltage which will give us high
base, emitter, and collector currents. At saturation the voltage across
the transistor is Vcesat. The increase in the collector current increases
the voltage across the collector resistor. Without the emitter capacitor,
it will also increase the  emitter voltage. With the emitter capacitor
the voltage at the emitter is held constant (for AC signals). The output
voltage will be Ve plus Vce (where Ve is the emitter voltage and Vce is
the voltage from the collector to the emitter).  As we increase the collector
current Vce can swing from its bias value down to Vcesat.  If the emitter
voltage is held constant then all of this voltage swing be at the output so
the output swing will be VCE - Vcestat (where VCE is the bias collector to
emitter voltage). This is the result that he gives.  If the emitter capacitor
is not present then part of the transistor voltage swing will go to the
emitter resistor and the swing in the output voltage will be smaller.

Dan

```
```On Tue, 2 Oct 2012 08:02:21 -0700 (PDT), panfilero
<panfilero@gmail.com> wrote:

>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 =3D 0.2V  According to the website
>
>maximum Vout =3D Vdd
>positive swing =3D VRc
>minimum Vout =3D -2.3V + VRc
>negative swing =3D 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
>
>http://thesignalpath.com/blogs/2012/09/23/tutorial-on-the-theory-design-=
>
>much thanks!

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=3D2mA (quiescent current) in his design. This just
means the center operating point which, based on his
assumptions (Ie=3DIc), 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=3D0.2V.
He was just using that figure for an absolute worst case
(that you would never really use in reality.)

Jon
```
```On Tuesday, October 2, 2012 10:02:21 AM UTC-5, panfilero wrote:
> 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!

Thanks felas, between yalls two explanations i have a much better idea what he was up to. much apprecaited.
```
```On Oct 3, 4:13=A0am, Jon Kirwan <j...@infinitefactors.org> wrote:
> On Tue, 2 Oct 2012 08:02:21 -0700 (PDT), panfilero
>
>
>
>
>
> <panfil...@gmail.com> wrote:
> >I'm having a hard time understanding the "negative swing" of a common-em=
itter amplifier. =A0There 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. =A0The =
BJT has a Vce,sat =3D 0.2V =A0According to the website
>
> >maximum Vout =3D Vdd
> >positive swing =3D VRc
> >minimum Vout =3D -2.3V + VRc
> >negative swing =3D 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
>
> >http://thesignalpath.com/blogs/2012/09/23/tutorial-on-the-theory-desi...
>
> >much thanks!
>
> 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=3D2mA (quiescent current) in his design. This just
> means the center operating point which, based on his
> assumptions (Ie=3DIc), 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=3D0.2V.
> He was just using that figure for an absolute worst case
> (that you would never really use in reality.)
>
> Jon- Hide quoted text -
>
> - Show quoted text -

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.
```
```On Wed, 3 Oct 2012 09:09:01 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

>On Oct 3, 4:13=A0am, Jon Kirwan <j...@infinitefactors.org> wrote:
>> On Tue, 2 Oct 2012 08:02:21 -0700 (PDT), panfilero
>><snip>
>> 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=3D2mA (quiescent current) in his design. This just
>> means the center operating point which, based on his
>> assumptions (Ie=3DIc), 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=3D0.2V.
>> He was just using that figure for an absolute worst case
>> (that you would never really use in reality.)
>
>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),

and noisier unless you pay dearly

>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.)

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.

>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?

Yes, he doesn't say it but he makes that point indirectly
when he talks about the max gain of A=3D-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
```
```On Wed, 03 Oct 2012 15:53:32 -0700, I wrote:

>On Wed, 3 Oct 2012 09:09:01 -0700 (PDT), George Herold
><gherold@teachspin.com> wrote:
>
>>On Oct 3, 4:13=A0am, Jon Kirwan <j...@infinitefactors.org> wrote:
>>> On Tue, 2 Oct 2012 08:02:21 -0700 (PDT), panfilero
>>><snip>
>>> 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=3D2mA (quiescent current) in his design. This just
>>> means the center operating point which, based on his
>>> assumptions (Ie=3DIc), 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=3D0.2V.
>>> He was just using that figure for an absolute worst case
>>> (that you would never really use in reality.)
>>
>>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),
>
>and noisier unless you pay dearly
>
>>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.)
>
>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.
>
>>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?
>
>Yes, he doesn't say it but he makes that point indirectly
>when he talks about the max gain of A=3D-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

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
```