How does the op amp operate in this circuit.........and I know it needs a power supply "Vcc" to work,do normal batteries work for that ? ops:

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I\'m a newbie that needs help =)

 

There are a couple simple rules that you can use to solve op amp problems.

V- = voltage at - terminal
V+ =voltage at + terminal
Vo = Voltage at output
A = open loop gain (set it to something big like 1 million for most opamps)


open loop gain equation: Vo=A(V+ - V-) use to find output voltage when there is no feedback between Output and -.

Virtual Ground rule: If there is a feedback connection between Vo and V- then V+ and V- will be equal.

Remember that the opamp cant output voltages bigger than its power supply.

To answer your question more directly it looks like the opamp provides gain but limits (or "clips") the output at the turn on voltage of the feedback diodes. So you will get an amplified waveform as long as the output is below ~.7V. If it is above .7v the signal is limited at .7V.

You will need 2 batteries for this circuit 1 to provide the + supply and 1 to provide the - supply.

Hope this helps
Brent

 

Your circuit is of a non-inverting amplifier. When the amp is working in linear mode (ie. when the output is not being switched hard to the +ve or -ve rails) the output will force the -input to closely follow changes at the +input. What happens at the output depends on the resistor/capacitor feedback network R9/C7 and R10/C6.

At low frequencies, eg. 50Hz,the reactances of C6 and C7 are very high, so the only effective feedback component is R9. That allows all the output voltage to be fed to the -input and the amp will have unity gain.

As the frequency is increased, the reactance of C6 will decrease. The voltage fed back to the -input will be partially shunted to ground through R10 and C6, and so the output needs to have a greater amplitude to get enough equalising voltage to feed back to the -input. So as the frequency increases the gain will start to increase at the rate of 6dB per octave with the 3dB point at 3.18KHz (where the reactance of C6 equals R10).

As the frequency is further increased, the reactance of C7 comes into play - shunting R8 and effectively increasing the feedback. The output will start to reduce at 6dB /octave with the 3dB point at 33.86KHz. At frequencies above that C7 becomes progressively more effective and the gain will again reduce to unity.

The circuit is therefore a bandpass filter with a bandwidth from 3.18KHz to 33.86KHz.

The 'soft clipping' diodes will stop the output exceeding the input by 0.7V- for either polarity.

The 'hard clipping' diodes serve no purpose unless the input is DC referenced to ground in some way - but the circuit does not indicate this.

 

Thanks for the answers...I will read them well and revise some stuff then come back with more questions :P

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I\'m a newbie that needs help =)

 

Pebe,I've just simulated the circuit, and actually the output signal was clipped but by the hard clipping diodes not the soft clipping ones-which had no effect and I had same output when i removed them-....

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I\'m a newbie that needs help =)

 

You have to apply the right input to see the effect of the soft clipping. Take out the hard clipping and try a 1kHz, +/- 0.3v pp square wave. Try a +/- 0.3v sine wave at 100Hz, 1kHz, and 10kHz.

 

how did you choose the values of frequency?

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I\'m a newbie that needs help =)

 

I seem to remember with germanium diodes ccts like these were
log-ish compressors? feedback thru the broader knee region?

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Take a look at Pebe's reply. The soft clipper has the most effect at frequencies within the passband. Square waves have harmonics that go way beyond their fundamental frequency, so you will see some effect on any low-frequency square wave that has significant energy in the passband.