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### Network # Common Emitter Amplifier with 2N3904

L

#### Lily Bepant

Jan 1, 1970
0
Hello,
I have been up to a question like this:
"Design a common emitter amplifier with an output impedance of 4.7Kohm
and a gain of -10 using 2N3904 powered by 12V DC power supply."

I got the pencil and paper started to design the thing. (I'm not an
electronics major)

I have jumped through many equations to find that: (I'm novice, so
don't assume that these are correct, please correct wherever possible)

Rc+Re=Vbb/Ic(max)

Ic(max).. I looked up 2N3904 manual and found that Ic(max) is 200 mA
for it. With 12V Vbb, it means Rc+Re= 60 Ohms. We want a gain of 10,
thus Rc/Re should be 10, which yields to the result that Re = 60/11 =
5.4 ohms Rc = 55 ohms (rounded these).

These values seem to be weird to me. What am I doing wrong?

Also, what would I take B (beta) of this thing? This is a design
question, therefore I understand that I should design something
independent of the B (beta) value. But how, when

R1+R2 = [(Vbb)]/[(10Ic)/B]

or not?
I'm interested in electronics but don't have enough knowledge on it.
Thanks for your guidance.

Best Regards,
Lily

A

#### Andrew DeWeerd

Jan 1, 1970
0
beta varies with the wind. Is there any reason you can't use an opamp for
your amplifier? Mucho easier.

J

#### John Popelish

Jan 1, 1970
0
Lily said:
Hello,
I have been up to a question like this:
"Design a common emitter amplifier with an output impedance of 4.7Kohm
and a gain of -10 using 2N3904 powered by 12V DC power supply."

I got the pencil and paper started to design the thing. (I'm not an
electronics major)

I have jumped through many equations to find that: (I'm novice, so
don't assume that these are correct, please correct wherever possible)

Rc+Re=Vbb/Ic(max)

Ic(max).. I looked up 2N3904 manual and found that Ic(max) is 200 mA
for it. With 12V Vbb, it means Rc+Re= 60 Ohms. We want a gain of 10,
thus Rc/Re should be 10, which yields to the result that Re = 60/11 =
5.4 ohms Rc = 55 ohms (rounded these).

These values seem to be weird to me. What am I doing wrong?

Also, what would I take B (beta) of this thing? This is a design
question, therefore I understand that I should design something
independent of the B (beta) value. But how, when

R1+R2 = [(Vbb)]/[(10Ic)/B]

or not?
I'm interested in electronics but don't have enough knowledge on it.
Thanks for your guidance.

A maximum rating is something you have to be sure your design does not
exceed. Staying well below the maximum current rating for the
transistor improves almost every aspect of operation, including power
consumed.

Some very rough rules of thumb:

1. The collector impedance of a common emitter amplifier is generally
high for low frequency operation, so you can approximate the output
impedance of the amplifier by assuming that it is made up mostly of
the collector load resistance.

2. The voltage gain cannot be higher than the collector resistor
divided by the emitter resistor. It will actually be a bit less than
this.

By 1, you should start with a collector resistance of about 4.7 k
ohms.

By 2, you should start with an emitter resistor of 470 or a bit less
(390, perhaps).

Note that these two choices will give an output impedance of about 4.7
k ohms, but a voltage gain of about 10 only when the amplifier runs
with no external load. If that is how you read the specification,
then you can continue with the rest of the design. If the spec
actually means that the voltage gain is to be 10 when the amplifier
has a 4.7 kohm external load connected then you have to design for an
unloaded gain of 20.

J

#### John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Lily Bepant <[email protected]>
I have jumped through many equations to find that: (I'm novice, so
don't assume that these are correct, please correct wherever possible)

Rc+Re=Vbb/Ic(max)

No, this is not correct. Vbb would be the voltage of an independent base
bias supply. And you do not design every stage to run at Ic(max)!
Ic(max).. I looked up 2N3904 manual and found that Ic(max) is 200 mA
for it. With 12V Vbb, it means Rc+Re= 60 Ohms. We want a gain of 10,
thus Rc/Re should be 10, which yields to the result that Re = 60/11 =
5.4 ohms Rc = 55 ohms (rounded these).

These values seem to be weird to me. What am I doing wrong?

Using that wrong equation. Begin with Vcc - the *collector* supply
voltage = 12 V. Now, the terms of the question do not include anything
that allows you to calculate a required collector current. So you have
to choose one. With a 4.7 kohm collector load, and 11 V available once
you have dropped 1 V across the emitter resistor (assumed for the moment
it is decoupled by a parallel capacitor at signal frequencies) to get
reasonably stable d.c. operating conditions, you would get as much
output voltage swing as possible by dropping 5.5 V across the collector
resistor, which gives you a collector current of 5.5/4.7 = just under
1.2 mA. A standard value 820 ohm emitter resistor gives you 1 V on the
emitter then.
Also, what would I take B (beta) of this thing? This is a design
question, therefore I understand that I should design something
independent of the B (beta) value. But how, when

R1+R2 = [(Vbb)]/[(10Ic)/B]

or not?

Well, again, it's not Vbb but Vcc (unless you REALLY have a separate
base bias supply). What you do is to choose R1 and R2 using an average
value for beta and then see what happens to Ic with maximum and minimum
beta devices. In my book, beta min is 100 and beta max is 300 at Ic = 10
mA. I don't have a beta against Ic curve, so I'm going to assume that
the beta doesn't vary with Ic sufficiently to create a problem. The
average beta is 200.

Here, we have 12 V/(12 mA x 200)= 200 kohm. Very neat! Now there is 1 V
on the emitter and thus 1.7 V on the base. We can assume Vbe = 0.7 V
because it doesn't vary much between devices and Ic varies strongly with
it, making our d.c. stabilising circuit (emitter resistor and base
potential divider) a high-gain control loop.

So R1 + R2 = 200 kohm and R1 has 11 V across it while R2 has 1 V across
it. Strictly, it also has 0.5% less current in it, because that goes
into the base, but we won't be using high-precision resistors, so we can
neglect that. Thus R1/R2 = 11, and that's 2 equations with 2 unknowns,
R1 and R2. We get R2 = 200kohm/12 = 16.7 kohm and R1 = 183 kohm. We
would choose the standard values 180 kohm and 16 kohm.

I'll leave it to you to calculate what collector current you actually
get with 180 kohm, 16 kohm and 820 ohms, with beta values of 100, 200
and 300. For extra credit, what is Ic with a beta of 200 and the
transistor immersed in boiling water?

However, this stage has a gain of much higher than 10. The gain is
RcIe/26, with Ie in milliamps, giving a gain of 217. To get a gain of
just 10, we split the 820 ohm emitter resistor into Rc/10 = 470 ohms and
820-470 = 350 ohms (use standard value 360 ohms), and decouple (connect
a capacitor in parallel) the 360 ohms only. We could have just used the
470 ohms as the emitter resistor but that would have given only about
0.5 V on the emitter and if you do the calculations you will find that
Ic varies a lot more with beta than it does with Ve = 1 V.
I'm interested in electronics but don't have enough knowledge on it.
Thanks for your guidance.

Persevere!

I

#### Ian Bell

Jan 1, 1970
0
Lily said:
Hello,
I have been up to a question like this:
"Design a common emitter amplifier with an output impedance of 4.7Kohm
and a gain of -10 using 2N3904 powered by 12V DC power supply."

Quite possible but it is a pretty vague specification. Is it and ac
amplifier or a dc one? (I suspect it is ac). If it is ac over what
bandwidth should it have this gain and what is the impedance of the source
signal. if you have this info a basic design is quite straightforward.
I got the pencil and paper started to design the thing. (I'm not an
electronics major)

I have jumped through many equations to find that: (I'm novice, so
don't assume that these are correct, please correct wherever possible)

Rc+Re=Vbb/Ic(max)

Don't know where this came from. it's not a standard one AFAIK and Ic(max)
is not overly important for this design.
Ic(max).. I looked up 2N3904 manual and found that Ic(max) is 200 mA
for it. With 12V Vbb, it means Rc+Re= 60 Ohms. We want a gain of 10,
thus Rc/Re should be 10, which yields to the result that Re = 60/11 =
5.4 ohms Rc = 55 ohms (rounded these).

These values seem to be weird to me. What am I doing wrong?

First the IC(max) is a red herring.

Gain is roughly RC/Re. Output impedance is roughly Rc so make it the 4.7K
specified. So Re is 470 ohms.
Also, what would I take B (beta) of this thing? This is a design
question, therefore I understand that I should design something
independent of the B (beta) value. But how, when

R1+R2 = [(Vbb)]/[(10Ic)/B]

or not?

The gain being ruoghly Rc/Re comes from assuming beta is large and
conversely the negative feedback due to Re ensures the gain is relatively
independent of beta.
I'm interested in electronics but don't have enough knowledge on it.
Thanks for your guidance.

All you need to do now is to work out the bias circuit values to ensure dc
stability and give a sufficiently high input impedance.

HTH

Ian

F

#### Frank Bemelman

Jan 1, 1970
0
Lily Bepant said:
Hello,
I have been up to a question like this:
"Design a common emitter amplifier with an output impedance of 4.7Kohm
and a gain of -10 using 2N3904 powered by 12V DC power supply."

I got the pencil and paper started to design the thing. (I'm not an
electronics major)

I have jumped through many equations to find that: (I'm novice, so
don't assume that these are correct, please correct wherever possible)

Rc+Re=Vbb/Ic(max)

Ic(max).. I looked up 2N3904 manual and found that Ic(max) is 200 mA
for it. With 12V Vbb, it means Rc+Re= 60 Ohms. We want a gain of 10,
thus Rc/Re should be 10, which yields to the result that Re = 60/11 =
5.4 ohms Rc = 55 ohms (rounded these).

It's not mandatory to make Ic the maximum 200mA allowed. The wanted
output impedance is 4.7K, so the collector resistor should be 4.7Kohm.
The emitter resistor 4700/10 -> 470 ohms.

To give the amplifier maximum swing, the output needs to be at 6V
with no input signal. Ic = 6V/4700 -> 1.2mA
Voltage at the emitter is 1.2mA * 470ohm -> 0.6V
Vbe should be ~0.7V, so the voltage at the base needs to be 1.3 volt.
For this, you can use a voltage divider of 47K and 5K6.

Hope this helps.

--
Thanks, Frank.
(remove 'x' and 'invalid' when replying by email)

These values seem to be weird to me. What am I doing wrong?

Also, what would I take B (beta) of this thing? This is a design
question, therefore I understand that I should design something
independent of the B (beta) value. But how, when

R1+R2 = [(Vbb)]/[(10Ic)/B]

or not?
I'm interested in electronics but don't have enough knowledge on it.
Thanks for your guidance.

Best Regards,
Lily

P

#### Paul Burridge

Jan 1, 1970
0
It's not mandatory to make Ic the maximum 200mA allowed. The wanted
output impedance is 4.7K, so the collector resistor should be 4.7Kohm.
The emitter resistor 4700/10 -> 470 ohms.

Yeah, but no one seems to have mentioned that if the emitter resistor
is bypassed (as it may well be) then the the Vgain will be RC/RE+re
where re is the internal resistance of the emitter diode and typically
quite small - but worth remembering nonetheless. You're also assuming
the output is unloaded, which is a bigger assumption. If it's loaded,
the Vgain will be RC||RL/RE+re.
To give the amplifier maximum swing, the output needs to be at 6V
with no input signal. Ic = 6V/4700 -> 1.2mA
Voltage at the emitter is 1.2mA * 470ohm -> 0.6V
Vbe should be ~0.7V, so the voltage at the base needs to be 1.3 volt.
For this, you can use a voltage divider of 47K and 5K6.

A Ve of 0.6 seems very low. Should be at least a volt, IMHO.

F

#### Frank Bemelman

Jan 1, 1970
0
Paul Burridge said:
Yeah, but no one seems to have mentioned that if the emitter resistor
is bypassed (as it may well be) then the the Vgain will be RC/RE+re
where re is the internal resistance of the emitter diode and typically
quite small - but worth remembering nonetheless. You're also assuming
the output is unloaded, which is a bigger assumption. If it's loaded,
the Vgain will be RC||RL/RE+re.

I'm only considering/answering the topology the OP presents, which
uses just a emitter resistor and nothing else. Furthermore, the OP
wanted an output impedance of 4K7 with a gain of 10. For design purposes
that means the load is assumed as not present.
A Ve of 0.6 seems very low. Should be at least a volt, IMHO.

Yep, zip, nada, nothing L

#### Lily Bepant

Jan 1, 1970
0
Hello again,
Using that wrong equation. Begin with Vcc - the *collector* supply
voltage = 12 V. Now, the terms of the question do not include anything
that allows you to calculate a required collector current. So you have
to choose one. With a 4.7 kohm collector load, and 11 V available once
you have dropped 1 V across the emitter resistor (assumed for the moment
it is decoupled by a parallel capacitor at signal frequencies) to get
reasonably stable d.c. operating conditions, you would get as much
output voltage swing as possible by dropping 5.5 V across the collector
resistor, which gives you a collector current of 5.5/4.7 = just under
1.2 mA. A standard value 820 ohm emitter resistor gives you 1 V on the
emitter then.

I didn't understand the maximum output voltage swing concept. I have
(tried to) built DC and AC circuits, tried some equations and loops
but couldn't come up with the values you guys have found. (I tried to
find the balance by calculating the intersection point of AC and DC
load lines to find the quiescent operating point. Won't the maximum
output voltage swing will be achieved where AC and DC lines
intersect?)
I think I'm missing something again here.

Thanks for the answers!
Lily

F

#### Fred Bloggs

Jan 1, 1970
0
Lily said:
Hello again,

I didn't understand the maximum output voltage swing concept. I have
(tried to) built DC and AC circuits, tried some equations and loops
but couldn't come up with the values you guys have found. (I tried to
find the balance by calculating the intersection point of AC and DC
load lines to find the quiescent operating point. Won't the maximum
output voltage swing will be achieved where AC and DC lines
intersect?)
I think I'm missing something again here.

Thanks for the answers!
Lily

This is a troll...

B

#### Bill Sloman

Jan 1, 1970
0
Hello,
I have been up to a question like this:
"Design a common emitter amplifier with an output impedance of 4.7Kohm
and a gain of -10 using 2N3904 powered by 12V DC power supply."

The output impedance of a 2N3904 transistor in a common emitter
configuration is of the order of 100k, so if you use a 4k7 load
resistor, the output impedance of the amplifier will be close to 4k7.

You want to use up some of that 12V biassing the transistor to get a
reasonably predictable collector current, so 1mA through 4k7, giving a
nominal (average) output voltage of 7.3V is probably okay.

At 1mA, the incremental emitter resistance of transistor is about 30R.

To get a gain of -10, you need an extra 440R between the emitter adn
0V. 430R is probably close enough.

1mA through 430R is 0.43V.

The Fairchild data sheet for the 2N3904

http://www.fairchildsemi.com/ds/2N/2N3904.pdf

gives the base-emitter "on" voltage as about 0.67V at 25 degrees
Celcius, so you need to bias the base at a nominal 1.1V to have 1mA of
collector current.

You can do this with a voltage divider. At 1mA and 25C the typical
current gain is about 230, minimum 70, so the base current could be as
high as 14uA, though 4.3uA would be typical.

So design the divider for an impedance of around 10k, and an unloaded
voltage of about 1.15V - 10k and 100k would be close enough.

If you want closer control over the bias voltage and the collector
current, you can bias the base further above the 0V rail, and use a
bigger resistor from the emitter to the 0V rail to define the nomimal
current. Thios obviously gives you less gain, so you by-pass this
resistor with a second, smaller resistor in series with a capacitor to
the 0V rail.

The capacitor blocks any DC current, but - for sufficiently high
signal frequencies - serves as an AC short circuit, so that you high
frequency gain is defiend by the parallel sum of the two resistors.

P

#### Paul Burridge

Jan 1, 1970
0
I didn't understand the maximum output voltage swing concept. I have
(tried to) built DC and AC circuits, tried some equations and loops
but couldn't come up with the values you guys have found. (I tried to
find the balance by calculating the intersection point of AC and DC
load lines to find the quiescent operating point. Won't the maximum
output voltage swing will be achieved where AC and DC lines
intersect?)
I think I'm missing something again here.

This was class A. wasn't it? You want to accommodate maximum output
voltage swing by 'setting' the collector voltage at midway between Vcc
and Vb. It's not critical with smaller signals but becomes
increasingly important as signals get larger (to avoid clipping).

L

#### Lily Bepant

Jan 1, 1970
0
Hello,
This is a troll...

What makes you think that I am trolling? I'm trying to gather some
information about the subject and still couldn't succeed.

P

#### Paul Burridge

Jan 1, 1970
0
What makes you think that I am trolling? I'm trying to gather some
information about the subject and still couldn't succeed.

I'm afraid there are just some folks here who take up perfectly good
space for no apparent reason, Lily. "Mr. Bloggs" is one of them.
Fortunately, most people here are very helpful and will do their best
to assist you in any way possible. Feel free to ask away as much as
you need to in order to gain a better insight into your problem with
this amp. It's what Usenet *should* be about.

J

#### John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Lily Bepant <[email protected]>
wrote (in said:
I didn't understand the maximum output voltage swing concept. I have
(tried to) built DC and AC circuits, tried some equations and loops
but couldn't come up with the values you guys have found. (I tried to
find the balance by calculating the intersection point of AC and DC
load lines to find the quiescent operating point. Won't the maximum
output voltage swing will be achieved where AC and DC lines
intersect?)
I think I'm missing something again here.

The way you specified your question, '4.7 kohm load', doesn't give
enough information to justify an assumption that this load is AC-
coupled. That's why I, and others, assumed it was the collector
resistor. If it's AC-coupled, then the data you provided is
insufficient to produce a solution without yet more assumptions.

F

#### Fred Bloggs

Jan 1, 1970
0
Paul said:
I'm afraid there are just some folks here who take up perfectly good
space for no apparent reason, Lily. "Mr. Bloggs" is one of them.
Fortunately, most people here are very helpful and will do their best
to assist you in any way possible. Feel free to ask away as much as
you need to in order to gain a better insight into your problem with
this amp. It's what Usenet *should* be about.

I have a solid record of providing REAL help to REAL people. The same
cannot be said of you- you are worthless trash who thinks SED is your
personal chat room. You have absolutely NOTHING to offer-and the other
contributors have already told you to take your elementary drivel to the
basics newsgroup. You never built anything in your life- you are a
dilettante and talker- too dumb to even run simulations and learn
electronics at the community college first semester level. You are a
total waste of time and it is people like you who are responsible for
USENET being abandoned by all but the trite-minded pedants. Like I said-
you are supeficialsuperficiala retiree idling away the last hours of
your waste of life- be sure and donate your organs or something useful.

F

#### Fred Bloggs

Jan 1, 1970
0
Frank said:
I'm only considering/answering the topology the OP presents, which
uses just a emitter resistor and nothing else. Furthermore, the OP
wanted an output impedance of 4K7 with a gain of 10. For design purposes
that means the load is assumed as not present.

Yep, zip, nada, nothing Isn't this typical- the damn jerk-off struggles to get a battery snap on
right- and now proceeds with technical critiques of working engineers.
He's a pathetic UK retiree maggot rotting in solitude in a slum
somewhere- you can see from his prose that he was probably a
slick-mouthed sales trash of some kind- and obviously a total zero when
it comes to anything constructive

F

#### Fred Bloggs

Jan 1, 1970
0
Lily said:
Hello,

What makes you think that I am trolling? I'm trying to gather some
information about the subject and still couldn't succeed.

Lemme see- you know about the various load lines, gains, and I/O
impedances, and reading data sheets- but you pretend you can't do a DC
bias? This makes no sense- you are a troll. Trolls are generally of low
intelligence, love to pseudo-intellectualize over definitions, and
usually deal with an elementary circuit that just will not work for some
reason.

J

#### John Larkin

Jan 1, 1970
0
Yeah, but no one seems to have mentioned that if the emitter resistor
is bypassed (as it may well be) then the the Vgain will be RC/RE+re
where re is the internal resistance of the emitter diode and typically
quite small - but worth remembering nonetheless.

Probably the reason nobody mentioned it is that it's not true.

John

F

#### Frank Bemelman

Jan 1, 1970
0
Fred Bloggs said:
Isn't this typical- the damn jerk-off struggles to get a battery snap on
right- and now proceeds with technical critiques of working engineers.
He's a pathetic UK retiree maggot rotting in solitude in a slum
somewhere- you can see from his prose that he was probably a
slick-mouthed sales trash of some kind- and obviously a total zero when
it comes to anything constructive

I was under the impression he was a 17 yr. old kid. Now, about battery
snaps and the like, don't you hate those battery compartiments where
all the AA(A) batteries point in the same direction v.s. the traditional
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