# Working principle of circuit

Discussion in 'The Projects Forum' started by prescott2006, Jan 23, 2010.

1. ### prescott2006 Thread Starter Active Member

Nov 8, 2008
72
1
i would like ask some questions about the circuit.this circuit is taken from thegimmick guitar tuner project.

1.what is the functions of each of the capacitor C1,C5.C6,C8 and the one in E3?below is my guessing:
C1:to filter out the dc component from the signal source.
C5&C6:to make sure the oscillator start up correctly.but what is the theory behind that?
C8 and the one inside E3:i don't know...

2.i know E2 is a multiple feedback filter,but why it uses this topology instead of others?

3.is the function of the potential divider below E1 is to provide a reference voltage to E1?then why it also connected to pin 5 of tl072 in E2?

4.the bridge rectifier at the bottom is b80c800,but i can't find this component.can i replace it with four 1n4007 diodes?

5.can somebody explain the working principle of the comparator?and why it is connected like that?(how to determine the resistor value and the resistor capacitor configuration as well.)

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2. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
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I agree with this assessment.
You are correct. The caps load the crystal so that it gets a kick on power up to get it started.
C8 is used to filter the V/2 reference provided by R2 and R3. The reference is needed because the opamp is being powered by a single supply
The choice of this active filter topology is at the designer's discretion. Many factors would have been involved. One influence would would be the designer's familiarity with the topology.
The divider is there to provide the reference for the opamp. The opamp is being operated as an inverting amp so the negative terminal requires a reference.
I suspect the bridge is provided to safegaurd against accidental reversal of the input voltage. Feel free to use four discrete 1n4007 diodes in its place.
One input to the comparator is seeing the signal and the other is seeing the signal delayed by the integrator. This is probably being done to prevent the output of the comparator from oscillating at the transitions of the input signal.

Last edited: Jan 23, 2010
3. ### prescott2006 Thread Starter Active Member

Nov 8, 2008
72
1
what does the capacitor filter?is it used to filter dc component?but if in that case we won't get V/2 anymore.i am pretty confused.kind to explain?

but why the voltage divider connected to the active filter at the same time?don't pin 5 should be grounded?

not very understand with this statement,can elaborate more?

another question,if i want to simulate the guitar signal by using signal generator,what voltage amplitude and frequency should be used?

4. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
The capacitor insures that the voltage at the junction of R2 and R3 will appear to be a low impedance to disturbances and therefore act as a stable DC reference voltage.
The input is connect to ground when the opamp is powered from a dual supply. For example, +5V and -5V. This opamp is being powered by 5v and ground so an artificial reference needs to be used. This artificial reference typically chosen to be at the voltage V/2. This gives the output of the opamp freedom to swing equally above and below the reference voltage.
Basically what is happening is that the input signal to the comparator is being compared to the delayed version of the input signal.
I would think any frequency in the low end of the audio range would be appropriate.

hgmjr

5. ### prescott2006 Thread Starter Active Member

Nov 8, 2008
72
1
concise and straight-to-the-point explanation.thanks very much.

off from the topic,what is the 'must-read' book to learn electronic.i don't want to dive to theory so much,what i want is practically useful.after googling,many people suggest that the art of electronic, make:electronic, getting started in electronic, what else do you recommend beside aac book? i would like to learn pic as well, but i am a totally beginner, so which book is recommended? i need a textbook instead of microchip tutorial and datasheet.

6. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218

Also if you can obtain any of Forrest Mims books, they are excellent introductions. Mr Mims is a member of the forum by the way.

hgmjr

7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,018
682
I think if you calculate the -3dB frequency of R7 and C10 (1Hz), you will find that the filter is there to extract the DC component of the signal as a reference. The input signal is compared with that DC level. It's basically another way of AC coupling. The designer could have used the Vcc/2 reference instead of the low pass filter, but the hysteresis resistor R9 would have coupled a tiny bit of the comparator output back the the Vcc/2 reference.

8. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
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I see your point ron h. It does seem that what you are saying makes sense. Such a scheme would mean that following long periods of input signal absense, it would take approximately 0.693 (natural log of 0.5) time constants or about 700 milliseconds to re-establish the comparator's reference voltage.

hgmjr

9. ### Ron H AAC Fanatic!

Apr 14, 2005
7,018
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No, I believe the DC voltage at the top of C10 will always be Vcc/2, due to the fact that the voltage divider R2/R3 biases the inputs of IC2A and IC2B. The input signal is AC coupled through C1.

10. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
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You are right. I was thinking digital rather than analog.

hgmjr

11. ### prescott2006 Thread Starter Active Member

Nov 8, 2008
72
1
i got one newbie question.how to connect electret mic to my circuit above?what value should i use for the resister and capacitor?i found the circuit in text file on web.
so basically i just connect the 'audio out' to 'guitar in' in my circuit?

• ###### Microphone.txt
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Last edited: Jan 25, 2010
12. ### prescott2006 Thread Starter Active Member

Nov 8, 2008
72
1
i am using 220 ohm as the current limiting resistor for the led,is it too small?what value is safe to prevent the led from burn out?