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How does this circuit work? Is it a pre-amp?

M

MRW

Jan 1, 1970
0
Hello!

Happy 4th of July to my American colleagues!

I have this circuit: http://bayimg.com/NaDlBAAbD

How does it work?

I initially thought that the first opamp is a microphone pre-amplifier
in an inverting configuration and the second one is used to invert the
signal back. But when I looked at it again, the second opamp is
configured as a voltage follower. Why is that?

I think that C6 and C7 are just used as a bypass capacitor for any
noise coming from VCC. R20 and C5 are just high pass filters that will
roll off any signal below 8Hz. R15 and R17 are used to biased the
input at half of VCC for single supply operation.

Is that the right assumption?

I also have a feeling that if the first opamp is indeed a pre-amp,
then it's gain is frequency dependent. But how do I find out the
equation for it? Is it just (-R13)/(reactance of C5) ?

Any other insights into this circuit?

Thanks!
 
A

Andrew Holme

Jan 1, 1970
0
MRW said:
Hello!

Happy 4th of July to my American colleagues!

I have this circuit: http://bayimg.com/NaDlBAAbD

How does it work?

I initially thought that the first opamp is a microphone pre-amplifier
in an inverting configuration and the second one is used to invert the
signal back. But when I looked at it again, the second opamp is
configured as a voltage follower. Why is that?

I think that C6 and C7 are just used as a bypass capacitor for any
noise coming from VCC. R20 and C5 are just high pass filters that will
roll off any signal below 8Hz. R15 and R17 are used to biased the
input at half of VCC for single supply operation.

Is that the right assumption?

I also have a feeling that if the first opamp is indeed a pre-amp,
then it's gain is frequency dependent. But how do I find out the
equation for it? Is it just (-R13)/(reactance of C5) ?

Any other insights into this circuit?

Thanks!

U6A is a current input amplifier, so the output voltage is -I_in * R13.

C5 is a DC-blocker not a filter; but it needs to be much larger than C7 to
ensure most of the signal current flows into the amplifier summing node.

The unnecessary voltage follower U7A has a capacitor from its output to
ground, which is a bit odd.

C6 and C7 are decoupling capacitors, as you spotted.
 
M

MRW

Jan 1, 1970
0
U6A is a current input amplifier, so the output voltage is -I_in * R13.


Thanks, Andrew! How would I be able to tell in the future if an opamp
is a current amplifier or not?
On this site, http://www.national.com/pf/LM/LM158.html , I don't spot
the indication right away.

C5 is a DC-blocker not a filter; but it needs to be much larger than C7 to
ensure most of the signal current flows into the amplifier summing node.

So the value of C5 is not determined by any frequency at all?

The unnecessary voltage follower U7A has a capacitor from its output to
ground, which is a bit odd.

Actually, this is the rest of the circuit: http://bayimg.com/BAdmjaAbD
(highlighted in the green box).
I'm still not sure what U7A does.. I figured it's independent from the
other part (highlighted in green).

Thanks!
 
M

MRW

Jan 1, 1970
0
U6A is a current input amplifier, so the output voltage is -I_in * R13.


By the way, is a current input amplifier synonymous to a current
feedback amplifier? If so, Opamps for Everyone has the following
statement: "In general, VFAs (voltage feedback amplifiers) are used
for precision and general purpose applications, while CFAs (current
feedback amplifiers) are restricted to high frequency applications
above 100 MHz."
 
E

Eeyore

Jan 1, 1970
0
MRW said:
By the way, is a current input amplifier synonymous to a current
feedback amplifier? If so, Opamps for Everyone has the following
statement: "In general, VFAs (voltage feedback amplifiers) are used
for precision and general purpose applications, while CFAs (current
feedback amplifiers) are restricted to high frequency applications
above 100 MHz."

Current feedback is something else entirely.

Graham
 
M

MRW

Jan 1, 1970
0
It's rubbish. Probably 'engineered' by a monkey on an off day.

Graham

Hi Graham,
I'm still learning about circuit configurations, so I really could not
tell.
Thanks!
 
M

MRW

Jan 1, 1970
0
It's shockingly incompetent..

Whoever put this lot together need to be shot.

Graham

Believe it or not, it was another classmate's instructor.. I just
copied the drawing on the computer because it was printed on paper. It
is supposed to be an FM transmitter circuit, and he asked me for help
in figuring it out....so I came here to get some answers.

The paper also says: "L2 is 112nH (this tunes to the middle of the FM
band, 98 MHz), with C13 at its center value 33pF. L2 is 5 turns of 22
swg enameled copper wire close-wound on a 5mm diameter former. You can
also have a fixed 33pF capacitor instead of the variable C13 and have
L2 as an adjustable molded coil."

.... btw, what's a "former"?
 
M

MRW

Jan 1, 1970
0
Current feedback is something else entirely.

Graham

Thanks, Graham! So how do I determine if the LM158 or any other ICs is
a current amplifier?
 
A

Andrew Holme

Jan 1, 1970
0
MRW said:
Thanks, Andrew! How would I be able to tell in the future if an opamp
is a current amplifier or not?
On this site, http://www.national.com/pf/LM/LM158.html , I don't spot
the indication right away.


The LM158 is not intrinsically a current amp. The circuit configuration in
which it is used is a current amp. Current flowing into the summing node
has nowhere else to go but through R13.

So the value of C5 is not determined by any frequency at all?


It's fairly non-critical. It needs to be large enough to NOT be a high pass
filter at audio frequencies.

Actually, this is the rest of the circuit: http://bayimg.com/BAdmjaAbD
(highlighted in the green box).
I'm still not sure what U7A does.. I figured it's independent from the
other part (highlighted in green).


The circuit is an FM transmitter. U7A is an isolating amplifier. The 1nF
capacitors are there to prevent RF feedback causing instability.
 
M

MRW

Jan 1, 1970
0
The LM158 is not intrinsically a current amp. The circuit configuration in
which it is used is a current amp. Current flowing into the summing node
has nowhere else to go but through R13.



It's fairly non-critical. It needs to be large enough to NOT be a high pass
filter at audio frequencies.



The circuit is an FM transmitter. U7A is an isolating amplifier. The 1nF
capacitors are there to prevent RF feedback causing instability.


Thanks, Andrew! I understood most of what you said except for
(regarding C5):
It's fairly non-critical. It needs to be large enough to NOT be a high pass
filter at audio frequencies.

If C5 does not become a high pass at audio frequencies, then how is
the audio source getting thru past the opamp stages? I had the
impression that C5 would short as the frequency gets higher. Isn't
that what a DC block does?
 
A

Andrew Holme

Jan 1, 1970
0
Andrew Holme said:
The LM158 is not intrinsically a current amp. The circuit configuration
in which it is used is a current amp.

I should have said the configuration is a current INPUT amp, or
trans-resistance amp i.e. current input, voltage output.

It works like this:

The voltage at U6A pin 2 is held at VCC/2 by negative feedback. It never
changes. The capacitance microphone produces a small AC speech current
which flows through C5. Most of this current flows on through R13. Only a
vanishingly small fraction of the current enters pin 2.

The output voltage is therefore VCC/2 + I*R13
 
E

Eeyore

Jan 1, 1970
0
MRW said:
Eeyore wrote


Believe it or not, it was another classmate's instructor..

Nothing surprises me any more.

I just
copied the drawing on the computer because it was printed on paper. It
is supposed to be an FM transmitter circuit,

Uh ? It is ?

and he asked me for help
in figuring it out....so I came here to get some answers.
Hmm.


The paper also says: "L2 is 112nH (this tunes to the middle of the FM
band, 98 MHz), with C13 at its center value 33pF. L2 is 5 turns of 22
swg enameled copper wire close-wound on a 5mm diameter former. You can
also have a fixed 33pF capacitor instead of the variable C13 and have
L2 as an adjustable molded coil."

... btw, what's a "former"?

A former is something on which you can form a coil e.g.a piece of plastic
tubing.

Grahsm
 
E

Eeyore

Jan 1, 1970
0
MRW said:
Thanks, Graham! So how do I determine if the LM158 or any other ICs is
a current amplifier?

You're confusing 'current amplifier' with 'current feedback amplifier'. Not the
same thing at all.

Graham
 
A

Andrew Holme

Jan 1, 1970
0
MRW said:
Thanks, Andrew! I understood most of what you said except for
(regarding C5):

If C5 does not become a high pass at audio frequencies, then how is
the audio source getting thru past the opamp stages? I had the
impression that C5 would short as the frequency gets higher. Isn't
that what a DC block does?

Correct. I meant that C5 needs to be large so it doesn't attenuate the
lower audio frequencies. Strictly speaking, it is a high pass filter; but
to call it one suggests that only high frequencies pass through; when in
fact the cutoff frequency is so low that all audio frequencies can pass.
Only sub-audio and DC are blocked.
 
B

Bob Masta

Jan 1, 1970
0
The circuit is an FM transmitter. U7A is an isolating amplifier. The 1nF
capacitors are there to prevent RF feedback causing instability.

I'm totally mystified by this circuit. If it really is an FM
transmiter, then where is the carrier tuning? I have seen
FM transmitters that used a condenser (capacitor) mic
as the modulator for the tuning circuit, but that doesn't seem to
be the case here. And if it is a transmitter, wouldn't C4 be
trying to filter out the whole signal? (Assuming that's supposed
to be the output, since there is no antenna connection marked,)

BTW, if that really *is* supposed to be a condenser mic, good
luck finding one at a reasonable price: These are typically only
used as precision reference mics, at US$ 1000 and up. The
cheap little electret mics that we see everywhere use the same
underlying technology (sort of), but they have a permanently
charged backing plate (the electret) and, unfortunately for
tuning uses, an FET preamp that isolates the capacitance of
the mic from the output and turns it into a voltage change.

Best regards,



Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, Signal Generator
Science with your sound card!
 
L

leo123

Jan 1, 1970
0
I'm totally mystified by this circuit. If it really is an FM
transmiter, then where is the carrier tuning? I have seen
FM transmitters that used a condenser (capacitor) mic
as the modulator for the tuning circuit, but that doesn't seem to
be the case here. And if it is a transmitter, wouldn't C4 be
trying to filter out the whole signal? (Assuming that's supposed
to be the output, since there is no antenna connection marked,)

BTW, if that really *is* supposed to be a condenser mic, good
luck finding one at a reasonable price: These are typically only
used as precision reference mics, at US$ 1000 and up. The
cheap little electret mics that we see everywhere use the same
underlying technology (sort of), but they have a permanently
charged backing plate (the electret) and, unfortunately for
tuning uses, an FET preamp that isolates the capacitance of
the mic from the output and turns it into a voltage change.

Best regards,

Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, Signal Generator
Science with your sound card!

..........
 
A

Andrew Holme

Jan 1, 1970
0
Bob Masta said:
I'm totally mystified by this circuit. If it really is an FM
transmiter, then where is the carrier tuning? I have seen
FM transmitters that used a condenser (capacitor) mic
as the modulator for the tuning circuit, but that doesn't seem to
be the case here. And if it is a transmitter, wouldn't C4 be
trying to filter out the whole signal? (Assuming that's supposed
to be the output, since there is no antenna connection marked,)


Those bayimg URLs all look alike:
 
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