audio pre-amp design

[Andy]

http://www.redcircuits.com/Page60.htm

What is the purpose of c1 and c2?

How does the feedback network work?

What's c7 for?

How does c6 & p2 work as a bass boost?

Does p1 form part of the input resistance or can i omit it if i dont
need to alter the gain?

sorry for all the q's... im better at digital than analogue

Thx in adv

Andy

 

[Jeff Liebermann]

On 2 Apr 2004 08:32:39 -0800, maccyd10@hotmail.com (Andy) wrote:

http://www.redcircuits.com/Page60.htm

What is the purpose of c1 and c2?

The R1/r2 voltage divider splits the power supply voltage in half to
supply bias to the + input of the op amp. C1 and C2 supply the
necessary filtering to keep any garbage on the power supply from
entering the + pin and getting amplifier. There are no frequency
response determining elements here.

How does the feedback network work?
What's c7 for?
How does c6 & p2 work as a bass boost?

Quite well. C7/R8 (22uf/560ohms) form approx a 20Hz RC network. It
reduces the feedback at frequencies above 20 Hz. C6/P2 (0.047uf/100K)
goes the opposite direction at 30 Hz. The result is a fairly flat
gain if both corner frequencies were identical. However, as P2 is
reduced in value, the 30Hz corner frequency goes up resulting in a
bass boost between 20 Hz and whatever c6/P2 have as a corner
frequency.

Does p1 form part of the input resistance or can i omit it if i dont
need to alter the gain?

P1 doesn't do anything other than set the gain. You can remove C4 and
P1 with no ill effects.

sorry for all the q's... im better at digital than analogue

You might wanna try one of the analog amplifier modeling programs
available.
http://www.penzar.com/links.htm

--
Jeff Liebermann 150 Felker St #D Santa Cruz CA 95060
(831)421-6491 pgr (831)336-2558 home
http://www.LearnByDestroying.com AE6KS
jeffl@comix.santa-cruz.ca.us jeffl@cruzio.com

 

[John Popelish]

Andy wrote:

http://www.redcircuits.com/Page60.htm

What is the purpose of c1 and c2?
R1 and R2 divide the instantaneous supply voltage in half. Even the

variations in the supply are divided. C1 and C2 form a low pass
filter to smooth the ripples in the supply and average the half supply
voltage down to so low a frequency that you can't hear those ripples.

How does the feedback network work?

C6 allows less feedback at low frequencies than it does at higher
frequencies, raising the gain at low frequencies (bass boost). The
pot and switch allow adjustment of the effect ot its elimination.

R8 divides the output signal down with R7, to raise the gain at all
audible frequencies.

What's c7 for?

It eliminates the gain boost of the R7 R8 divider at very low
(inaudible) frequencies), essentially opening up the path through R8
at DC, since there will be that divided supply voltage on the output
added to everything. This is a necessity when you use a single
supply, instead of a positive and negative supply.

How does c6 & p2 work as a bass boost?

See above.

Does p1 form part of the input resistance or can i omit it if i dont
need to alter the gain?

It does lower the input impedance at full volume. C5 is backwards,
regardless. The two inputs of the opamp must match for linear
operation, so there must be the same half supply voltage on the -
input as there is on the + input so there is the half supply voltage
applied to the right end of C5.

So you can just eliminate C4 and P1 and connect your signal directly
to C5.

--
John Popelish

 

[Ban]

Andy wrote:

http://www.redcircuits.com/Page60.htm

What is the purpose of c1 and c2?

Both caps are for decoupling the power supply. R3 and C2 can be omitted
without any change. C4 and R5 can be omitted as well, no purpose.

How does the feedback network work?

What's c7 for?

It is there because otherwise there would be an offset dependent on P2 at
the output and this would limit assymetrically the output voltage. R7 and R8
are there to make more gain without increasing R6. But if you do not want so
much gain (13dB minimal) you can omit R7, R8 and C7, then your gain is only
3dB.

How does c6 & p2 work as a bass boost?

for high frequencies C6 is conducting and keeps the gain at 13dB, low

frequencies have to go thru P2 and can be boosted another 15dB.

Does p1 form part of the input resistance or can i omit it if i dont
need to alter the gain?

You can omit it, and turn C5 around, the pos. pole needs to go to R4. The
way it is drawn it will not work.

sorry for all the q's... im better at digital than analogue

Thx in adv

Andy

I need a digital guy to look over my design, can you help me there? I'll
mail you.

--
ciao Ban
Bordighera, Italy

 

[John Woodgate]

I read in sci.electronics.design that Jeff Liebermann
<jeffl@comix.santa-cruz.ca.us> wrote (in <qq5r601v12gif1i8cvgs9uu5jt737u
vfk4@4ax.com&gt about 'audio pre-amp design', on Fri, 2 Apr 2004:

On 2 Apr 2004 08:32:39 -0800, maccyd10@hotmail.com (Andy) wrote:

http://www.redcircuits.com/Page60.htm

What is the purpose of c1 and c2?

The R1/r2 voltage divider splits the power supply voltage in half to
supply bias to the + input of the op amp. C1 and C2 supply the
necessary filtering to keep any garbage on the power supply from
entering the + pin and getting amplifier. There are no frequency
response determining elements here.

Using two smallish caps and one extra resistor allows the d.c.
conditions to settle quicker than if C1 was much larger and C2 was
omitted.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Ian Bell]

Andy wrote:

http://www.redcircuits.com/Page60.htm

What is the purpose of c1 and c2?

C1 decouples the dc supply from the + input of the op amp and keeps and
reduces the amount of power supply noise reaching the opamp input. C2
provides further decoupling and also ensures the op amp + input is at ac
ground. R1 and R2 are there to allow the op amp, which is designed for sual
rail operation, to work from a single supply. This is a common but poor
design technique.

How does the feedback network work?

What's c7 for?

So that R8 is connected to ac ground but not zero volts.

How does c6 & p2 work as a bass boost?

At high frequencies C6 is effectively short sircuit so the feed back is
primarily determined by R6. At lower frequencies the impedance of C6
rises, the feedback is reduced and the gain increases. The degree to which
the gain increases is determined by P2.

Does p1 form part of the input resistance or can i omit it if i dont
need to alter the gain?

Yes it does form part of the input resistance and yes you can omit it if you
do not need to alter the gain.

HTH

Ian

 

[Animesh Maurya]

John Popelish <jpopelish@rica.net> wrote in message news:<406DA2E3.E0430EA0@rica.net>...

Andy wrote:

http://www.redcircuits.com/Page60.htm

What is the purpose of c1 and c2?
R1 and R2 divide the instantaneous supply voltage in half. Even the
variations in the supply are divided. C1 and C2 form a low pass
filter to smooth the ripples in the supply and average the half supply
voltage down to so low a frequency that you can't hear those ripples.

How does the feedback network work?

C6 allows less feedback at low frequencies than it does at higher
frequencies, raising the gain at low frequencies (bass boost). The
pot and switch allow adjustment of the effect ot its elimination.

R8 divides the output signal down with R7, to raise the gain at all
audible frequencies.

What's c7 for?

It eliminates the gain boost of the R7 R8 divider at very low
(inaudible) frequencies), essentially opening up the path through R8
at DC, since there will be that divided supply voltage on the output
added to everything. This is a necessity when you use a single
supply, instead of a positive and negative supply.

How does c6 & p2 work as a bass boost?

See above.

Does p1 form part of the input resistance or can i omit it if i dont
need to alter the gain?

It does lower the input impedance at full volume. C5 is backwards,
regardless. The two inputs of the opamp must match for linear
operation, so there must be the same half supply voltage on the -
input as there is on the + input so there is the half supply voltage
applied to the right end of C5.

So you can just eliminate C4 and P1 and connect your signal directly
to C5.

Hi John,

I am newbie to op-amps, with very little knowledge about practical
circuits.

Since the circuit is using a single +ve power supply, output cannot
swing in –ve direction, am I right.

Resistor R1 and R2 are splitting the supply into two half at the
non-inverting end, and the input signal rides on this dc level.

Finally a coupling capacitor is used at the output end to strip out
the dc signal and the amplified ac output is fed to the speaker.

But I cannot find any coupling capacitor at the out end?

Actually I am talking in reference to LM386 datasheet, they are
placing a capacitor in series with the speaker.

Datasheet is here http://www.national.com/ds/LM/LM386.pdf

You said that C7 is necessary, if a single supply is used, so is C7
acting as coupling cap, as I guess.

Finally I want to know, are cap C1 and C2 redundant once you stuck to
a regulated power supply.


Thanks a lot for your time

Best regards,
Animesh Maurya

 

[John Popelish]

Animesh Maurya wrote:

Hi John,

I am newbie to op-amps, with very little knowledge about practical
circuits.

This one illustrated many opamp concepts.

Since the circuit is using a single +ve power supply, output cannot
swing in –ve direction, am I right.

Right. That is why there is a divider that produces a bias voltage at
1/2 of the supply. All input and output signals swing around that
voltage, inside the coupling caps.

Resistor R1 and R2 are splitting the supply into two half at the
non-inverting end, and the input signal rides on this dc level.

After C5, yes.

Finally a coupling capacitor is used at the output end to strip out
the dc signal and the amplified ac output is fed to the speaker.

Yes.

But I cannot find any coupling capacitor at the out end?

Correct. That coupling capacitor has been neglected in this design.
It assumes that any downstream amplifier has a coupling capacitor, but
that is a foolish assumption. I would follow this amplifier with both
a coupling cap and a high resistor to ground to discharge any DC on
the output.

Actually I am talking in reference to LM386 datasheet, they are
placing a capacitor in series with the speaker.

It is a necessity in that application.

Datasheet is here http://www.national.com/ds/LM/LM386.pdf

You said that C7 is necessary, if a single supply is used, so is C7
acting as coupling cap, as I guess.

It is coupling the feedback divider to ground to raise the audio
frequency gain. At DC, it is effectively an open circuit, so there is
more total feedback at DC and thus, less gain.

Finally I want to know, are cap C1 and C2 redundant once you stuck to
a regulated power supply.

If the supply is perfectly steady, then they are redundant, as is R3.

--
John Popelish

 

[Ian Bell]

Animesh Maurya wrote:

John Popelish <jpopelish@rica.net> wrote in message
news:<406DA2E3.E0430EA0@rica.net>...
Andy wrote:

http://www.redcircuits.com/Page60.htm

What is the purpose of c1 and c2?
R1 and R2 divide the instantaneous supply voltage in half. Even the
variations in the supply are divided. C1 and C2 form a low pass
filter to smooth the ripples in the supply and average the half supply
voltage down to so low a frequency that you can't hear those ripples.

How does the feedback network work?

C6 allows less feedback at low frequencies than it does at higher
frequencies, raising the gain at low frequencies (bass boost). The
pot and switch allow adjustment of the effect ot its elimination.

R8 divides the output signal down with R7, to raise the gain at all
audible frequencies.

What's c7 for?

It eliminates the gain boost of the R7 R8 divider at very low
(inaudible) frequencies), essentially opening up the path through R8
at DC, since there will be that divided supply voltage on the output
added to everything. This is a necessity when you use a single
supply, instead of a positive and negative supply.

How does c6 & p2 work as a bass boost?

See above.

Does p1 form part of the input resistance or can i omit it if i dont
need to alter the gain?

It does lower the input impedance at full volume. C5 is backwards,
regardless. The two inputs of the opamp must match for linear
operation, so there must be the same half supply voltage on the -
input as there is on the + input so there is the half supply voltage
applied to the right end of C5.

So you can just eliminate C4 and P1 and connect your signal directly
to C5.

Hi John,

I am newbie to op-amps, with very little knowledge about practical
circuits.

Since the circuit is using a single +ve power supply, output cannot
swing in –ve direction, am I right.

As the circuit stands yes.

Resistor R1 and R2 are splitting the supply into two half at the
non-inverting end, and the input signal rides on this dc level.

And so does the output.

Finally a coupling capacitor is used at the output end to strip out
the dc signal and the amplified ac output is fed to the speaker.

But I cannot find any coupling capacitor at the out end?

Well spotted. The design as it stand has a large dc offset at the out put.
A series capacitor is needed to remove this. At the same time this will
allow the output to swing negative.

 

[Animesh Maurya]

John Popelish <jpopelish@rica.net> wrote in message news:<406FA02F.2BD88B52@rica.net>...

It assumes that any downstream amplifier has a coupling capacitor, but
that is a foolish assumption. I would follow this amplifier with both
a coupling cap and a high resistor to ground to discharge any DC on
the output.

Thanks, John and Ian for the helpful reply.

John, maybe due to poor understanding or lack of knowledge,
I am unable to grasp your above statement.

Can you please put it in some more detailed way.

Even I don't understand what is a downstream amplifier.

Thanks again

Best regards,
Animesh Maurya

 

[John Popelish]

Animesh Maurya wrote:

John Popelish <jpopelish@rica.net> wrote in message news:<406FA02F.2BD88B52@rica.net>...
It assumes that any downstream amplifier has a coupling capacitor, but
that is a foolish assumption. I would follow this amplifier with both
a coupling cap and a high resistor to ground to discharge any DC on
the output.

Thanks, John and Ian for the helpful reply.

John, maybe due to poor understanding or lack of knowledge,
I am unable to grasp your above statement.

Can you please put it in some more detailed way.

Even I don't understand what is a downstream amplifier.

Thanks again

Best regards,
Animesh Maurya

The name, 'preamp' implies that this amplifier feeds another one.

I it were permanently connected to this next stage, and that next
stage had an input coupling capacitor, adding a second one ot this
output would be redundant. But if this preamp provided one of several
choices of input to the power amplifier and those choices could be
switched during operation (as by a multi pole selector switch), then
there would be a very loud pop when this output was selected as the
next stage input cap charges up to this bias voltage. And another pop
when the switch selected any other signal that had a different DC
bias.

You could cure all this popping by adding a series capacitor for each
input, with a high resistance bleed resistor to ground after each one,
so that all signals would enter the selector switch with zero average
voltage.

Is that any clearer?

--
John Popelish

 

[Ian Bell]

Animesh Maurya wrote:

John Popelish <jpopelish@rica.net> wrote in message
news:<406FA02F.2BD88B52@rica.net>...
It assumes that any downstream amplifier has a coupling capacitor, but
that is a foolish assumption. I would follow this amplifier with both
a coupling cap and a high resistor to ground to discharge any DC on
the output.

Thanks, John and Ian for the helpful reply.

John, maybe due to poor understanding or lack of knowledge,
I am unable to grasp your above statement.

Can you please put it in some more detailed way.

Even I don't understand what is a downstream amplifier.

It just means the amplifier that receives the output from this preamp.

Ian

 

[Animesh Maurya]

John Popelish <jpopelish@rica.net> wrote in message news:<40715C21.4E1D9EF0@rica.net>...

Animesh Maurya wrote:

John Popelish <jpopelish@rica.net> wrote in message news:<406FA02F.2BD88B52@rica.net>...
It assumes that any downstream amplifier has a coupling capacitor, but
that is a foolish assumption. I would follow this amplifier with both
a coupling cap and a high resistor to ground to discharge any DC on
the output.

Thanks, John and Ian for the helpful reply.

John, maybe due to poor understanding or lack of knowledge,
I am unable to grasp your above statement.

Can you please put it in some more detailed way.

Even I don't understand what is a downstream amplifier.

Thanks again

Best regards,
Animesh Maurya

The name, 'preamp' implies that this amplifier feeds another one.

I it were permanently connected to this next stage, and that next
stage had an input coupling capacitor, adding a second one ot this
output would be redundant. But if this preamp provided one of several
choices of input to the power amplifier and those choices could be
switched during operation (as by a multi pole selector switch), then
there would be a very loud pop when this output was selected as the
next stage input cap charges up to this bias voltage. And another pop
when the switch selected any other signal that had a different DC
bias.

You could cure all this popping by adding a series capacitor for each
input, with a high resistance bleed resistor to ground after each one,
so that all signals would enter the selector switch with zero average
voltage.

Is that any clearer?

Thanks thanks a lot, I got that point. You are so clear, I really appreciate it.

Best regards,
Animesh Maurya