Collector to Base Resistor - Why?

[Lumpy]

Help me to understand the purpose of this
resistance, please.

Typical single Q NPN amplifier (guitar boost).

o--/\/\/--o +v
|
R? |
o--\/\/\--o--||-----o OUT
| /
| /C
| |/
IN o----||--o-----|B
|\
\E
\o--/\/\/--o
|
|
# GROUND

What is the R? doing in this circuit?

Taking it further, if I want to install back to
back clipping diodes at that C to B point, how
will that R? resistance interact?

Thanks -


Lumpy

 

[Eeyore]

Help me to understand the purpose of this
resistance, please.

Typical single Q NPN amplifier (guitar boost).

o--/\/\/--o +v
|
R? |
o--\/\/\--o--||-----o OUT
| /
| /C
| |/
IN o----||--o-----|B
|\
\E
\o--/\/\/--o
|
|
# GROUND

What is the R? doing in this circuit?

It's providing bias current for the transistor. Without it, it wouldn't be an
amplifier ! It's a failry crappy way of providing bias btw since the operating
point wil be heavily affected by beta/hfe.

Taking it further, if I want to install back to
back clipping diodes at that C to B point, how
will that R? resistance interact?

The resistor will be the last of your problems. That will also stop it being an
amplifier too. Nor do you need 'back to back' diodes either. Since when will the
collector ever be negative wrt to the base ?

I think you need to go learn some circuit basics.

Graham

 

[Lumpy]

I think you need to go learn some circuit basics.

That's why I came here. To gain knowledge from
those willing to help a novice.

I'd be happy to exclude you from that pool of
knowledge if you insist on being such an asshole.


Lumpy

 

[Bob Eld]

Help me to understand the purpose of this
resistance, please.

Typical single Q NPN amplifier (guitar boost).

o--/\/\/--o +v
|
R? |
o--\/\/\--o--||-----o OUT
| /
| /C
| |/
IN o----||--o-----|B
|\
\E
\o--/\/\/--o
|
|
# GROUND

What is the R? doing in this circuit?

Taking it further, if I want to install back to
back clipping diodes at that C to B point, how
will that R? resistance interact?

Thanks -


Lumpy

The resistor, R provides base current as has been mentioned but it also
provides negative feedback and helps stabilize the operating point of the
transistor. The operating point is establised by the Beta of the transistor
and the voltage drop across R. If the collector voltage gets too high, the
transistor turns on harder reducing the base drive which lowers the
collector voltage. Likewise, if the collector voltage is too low, the
collector rises turning the transistor on harder. This is a stabilizing
action. The local negative feedback also reduces distortion. Think of "R" as
the feedback resistor in an op-amp circuit where the transistor is a "poor
man's" op-amp. This circuit was very common years ago in simple transistor
amplifiers like in radios, etc. Bob

 

[Eeyore]

That's why I came here. To gain knowledge from
those willing to help a novice.

I think you ought to read a decent book on the subject first. A half-decent
working knowledge of simple circuitry is a reasonable prerequisite before asking
question as basic as this one !

Graham

 

[James Beck]

I think you need to go learn some circuit basics.

Graham

This was posted to SEB.
The B is for BASIC.

 

[Peter M. Groen]

That's why I came here. To gain knowledge from
those willing to help a novice.

I'd be happy to exclude you from that pool of
knowledge if you insist on being such an asshole.


Lumpy

"Lumpy"....

When someone advices you to go out and learn something about basics, it
isn't meant to piss you off. Your reaction is absolutely out of proportion.
The only thing you have assured is no help in the future and that is a bad
way to start your knowledge build-up. The question you asked is part of one
of the first lessons in Transistor technology.

Go out, buy a book about basic electronics (Any), read it, try it, read it
again and design your own. After that.. Maybe, someone is willing to help
you again. (Ever heard of kill-files? They work like a black hole).

One pointer for your next question: It is bad practice to build your
question like : "I have a problem, someone has the answer, Gimme, gimme,
gimme and if you don't you're an asshole."

Try it in the following format the next time:
- Tell us what you want to achieve.
- What you read already about the subject.
- Your approach so far. << Very, very important >>
- Where are you stuck..

Don't forget that people on Usenet (Some excluded) are here to share
knowledge to those who are willing to put some effort in the subject
themselves. And they are all volunteers. Putting a question on the internet
doesn't give you the right to demand an answer. You'll just have to wait if
someone is willing to spend some of his or hers *spare* time in your
problem. If you're here expecting direct answers to your problem, you're in
the wrong place. You'll only find pointers here to (maybe) a solution.
There's always an effort required from your part.

For now your only achievement was insulting someone who was willing to help
you point you in the right direction and the only problem you have at the
moment is your attitude.

Kind regards,

Peter

 

[Peter M. Groen]

This was posted to SEB.
The B is for BASIC.

That sure doesn't mean this is an educational institute, right? People here
are allowed to expect some effort from the one who's asking questions.
After all, I went to 8 years of study and almost 25 years of experimenting
to earn my Electronics Engineering Degree.

Kind regards,

Peter

 

[Eeyore]

This was posted to SEB.
The B is for BASIC.

Sure but not not s.e.k ( sci.electronics.kiddies )

Graham

 

[John Popelish]

Help me to understand the purpose of this
resistance, please.

Typical single Q NPN amplifier (guitar boost).

o--/\/\/--o +v
|
R? |
o--\/\/\--o--||-----o OUT
| /
| /C
| |/
IN o----||--o-----|B
|\
\E
\o--/\/\/--o
|
|
# GROUND

What is the R? doing in this circuit?

It provides a bias current to get the transistor conducting a little,
so that it can respond to the AC signal arriving through the base
capacitor. Since it delivers a current roughly proportional to the
collector voltage, the bias point is somewhat stabilized for different
transistor gains (if a high gain transistor over reacts to the bias
current by turning on too much, the low output collector voltage
lowers the bias, reducing the effect of higher current gain).

Unfortunately for high impedance guitar source, the negative feedback
does not only apply to DC bias current, but to the signal frequencies
as well. The net effect is that the feedback tries to hold the input
node at a more fixed voltage, lowering the input impedance of the
amplifier. Guitar signals work best into high (relative to the guitar
pickup) impedance amplifiers.

Taking it further, if I want to install back to
back clipping diodes at that C to B point, how
will that R? resistance interact?

The feedback resistor will limit the gain to some maximum value when
neither diode is conducting. You also need to think about the DC
voltage across this resistor, and what that will do to the diodes.
For instance, you may need to add a capacitor in series with the diode
pair, so that only AC signals are altered by those diodes, not the DC
bias point.

 

[Michael A. Terrell]

This was posted to SEB.
The B is for BASIC.

Ignore him. He's an asshole.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[Michael A. Terrell]

Sure but not not s.e.k ( sci.electronics.kiddies )

Graham

You're right. We didn't see any need to create a newsgroup just for
you and the other ignorant trolls.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[James Beck]

That sure doesn't mean this is an educational institute, right? People here
are allowed to expect some effort from the one who's asking questions.
After all, I went to 8 years of study and almost 25 years of experimenting
to earn my Electronics Engineering Degree.

Kind regards,

Peter

I guess maybe I don't understand what USENET is for then.

Jim

 

[Richard Seriani, Sr.]

After all, I went to 8 years of study and almost 25 years of experimenting
to earn my Electronics Engineering Degree.

Kind regards,

Peter

Darn, that must have been one heck of a tough school. That is 29 years
longer than most four-year degrees take to complete

Richard

 

[Lumpy]

...Think of "R" as the
feedback resistor in an op-amp circuit where the transistor is a
"poor man's" op-amp. This circuit was very common years ago in simple
transistor amplifiers like in radios, etc. Bob

Thanks very much, Bob. The op-amp FB analogy
helps very much.


Lumpy

 

[Eeyore]

Ignore him. He's an asshole.

Pot, kettle, black.

Graham

 

[Lumpy]

That sure doesn't mean this is an educational institute, right?

Whatever your personal fantasy is about what
this NG is, it's a public, unmoderated forum
with the words electronics and basics in the
title. So if your nose is so far up your own
ass the the purpose of YOUR being here is to
look down on people asking basic electronic
questions, that's your own problem.

Congrats on that 8 year electronic degree.
Isn't there a newsgroup for people of
your self perceived importance that
doesn't lower you to basic status?


Lumpy

 

[Lumpy]

The feedback resistor will limit the gain to some maximum value when
neither diode is conducting. You also need to think about the DC
voltage across this resistor, and what that will do to the diodes.
For instance, you may need to add a capacitor in series with the diode
pair, so that only AC signals are altered by those diodes, not the DC
bias point.

Thanks John.

Here's a real world example -
http://digitalcartography.com/eir/onetransSoftAndHardClip.gif

Disregarding the diodes at the output for now,
I'm concentrading on the D1 D2 pair. No cap.
It clips the audio signal, but way too hard.
So I'm assuming, perhaps incorrectly, that
what I want to do is put some R in series
with the diode pair. I don't have a scope,
but the sound I hear at the output is that
of a really hard clipped square wave.

Thanks -


Lumpy

 

[John Popelish]

Thanks John.

Here's a real world example -
http://digitalcartography.com/eir/onetransSoftAndHardClip.gif

Disregarding the diodes at the output for now,
I'm concentrading on the D1 D2 pair. No cap.
It clips the audio signal, but way too hard.
So I'm assuming, perhaps incorrectly, that
what I want to do is put some R in series
with the diode pair. I don't have a scope,
but the sound I hear at the output is that
of a really hard clipped square wave.

Without a series DC blocking capacitor, one of the diodes provides
enough average bias current to pull the collector all the way down to
1 diode drop above the base voltage at no signal. and the other one
is not used at all, because the collector to base junction would do
its job.

Try it with a fairly large blocking capacitor (a few microfarads, with
the positive end toward the collector, if electrolytic) and also a
resistor in series with the diodes, to soften the clipping. Something
around R/4 to R/10 might work better. Then adjust the value of R to
get an average collector voltage a little above half of the supply
voltage, so the diodes control the clipping, not the transistor
saturating.

 

[Stanislaw Flatto]

Darn, that must have been one heck of a tough school. That is 29 years
longer than most four-year degrees take to complete

Richard

Hi Sr.!
Your primary school math is perfect;^).
But how many years it took, after the degree, to admit that on some
subject you know nothing, and approach someone more experienced to ask?
"Engineer knows everything about nothing, merchant knows nothing about
everything."

Have fun

Stanislaw
Slack user from Ulladulla