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  #1  
Old 07-06-2010, 04:30 PM
vindicate vindicate is offline
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Default Coupling

When coupling a circuit(like audio) with a capacitor, do you have to have your AC voltage go negative?

For instance if you had a a Sine wave that goes from +5v to 0V and coupled it with a capacitor, wouldn't the capacitor just fill to 5v and block all current?
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Old 07-06-2010, 04:43 PM
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Originally Posted by vindicate View Post
When coupling a circuit(like audio) with a capacitor, do you have to have your AC voltage go negative?
It depends. A capacitor blocks DC, but passes the effects of an AC signal. For instance, if you were going to connect your electret mic preamplifier to an audio amplifier that used a bipolar supply, you would likely want the signal to be centered around 0v.

If an AC signal is offset from 0v enough so that it does not traverse 0v, it is more correctly called rippled DC.
Quote:

For instance if you had a a Sine wave that goes from +5v to 0V and coupled it with a capacitor, wouldn't the capacitor just fill to 5v and block all current?
If the output side of the cap has a current path to ground via a resistor or high-value choke (inductor), then the average value of the output will be 0v - however, you will still see the AC signal.

Try it in your simulator. Use a high-value resistor (10k or greater) or a high-value inductor (say, 10mH) and see what your output looks like.
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Old 07-06-2010, 05:10 PM
vindicate vindicate is offline
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Ok, your right. If I have a sin from 0-5V the output after the coupling is 2.5 to -2.5.

It thought I tried that before but I guess not.

Again Sgt.Wookie you are the man. Thanks alot.
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Old 07-06-2010, 11:48 PM
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GetDeviceInfo GetDeviceInfo is offline
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all you really need is a change in signal, and this can take place at any opertaional voltage of the cap. The value of the cap and the rate of change will imply an impedance which will attenuate the signal.
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Old 07-07-2010, 02:34 PM
vindicate vindicate is offline
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Look at this example I made:



I'm trying to simulate a microphone as simply as possible. So the DC sawtooth generator is the mic. It's ranges from 500mv to 0mv.

On the right side of the capacitor(before the resistor divider) you get +250mv to -250mv. You add in the resistor divider and get 2.75 to 2.25v(roughly). That's all expected.

I would assume you would then pump that into an opamp to get amplification. But my question is, don't you want to go down to 0V not some arbitrary +V number? If I put what I have into a non-inverting amp with feedback, I would end up getting like 5.5V to 4.5V or something like that.

Am I completely wrong on this, or am I on the right track.
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Old 07-07-2010, 02:42 PM
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Originally Posted by vindicate View Post
Look at this example I made:



I'm trying to simulate a microphone as simply as possible. So the DC sawtooth generator is the mic. It's ranges from 500mv to 0mv.

On the right side of the capacitor(before the resistor divider) you get +250mv to -250mv. You add in the resistor divider and get 2.75 to 2.25v(roughly). That's all expected.
Your resistors are too low in value. Use 47k to ground and 33k to +V, or 100k to ground and 68k to +V. Otherwise you'll start getting distortion at lower frequencies.

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I would assume you would then pump that into an opamp to get amplification. But my question is, don't you want to go down to 0V not some arbitrary +V number? If I put what I have into a non-inverting amp with feedback, I would end up getting like 5.5V to 4.5V or something like that.

Am I completely wrong on this, or am I on the right track.
If you are using dual supply rails, then you would want the signal to be centered around 0v. However, you are likely going to be using a single supply, so you must bias the signal somewhere near Vcc/2.

Since you have already stated in another thread that you intend to use a TL081, you need to bias the input at Vcc/2 +1.5v to give you the best chance of avoiding clipping.

You will not be able to use a TL081 with a 5v supply, as it would only have a 0.5v output swing. You might as well use 9v in your simulations, because that's what you will be using.
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Old 07-07-2010, 05:14 PM
vindicate vindicate is offline
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I wasn't really trying to simulate the mic circuit that I'm going to build, I was trying to simulate a mic circuit in general just so I can get a better handle on the concepts.

In the simulator all they have is the "ideal" opamp so that's what I'm working with for now.

So you were saying that you want to bias the input at Vcc/2. In my example that would be 2.5V. So technically mu input is biased at VCC/2.

When you amplify that though, how do you make it so the "peaks" are amplified more, and not amplify the bias?

Can you amplify that from a 2.25 to 2.75(a vcc/2 bias) to a 5v to 0v(with a vcc/2 bias). The bias stays the same, just the min and max and amplified. Maybe it's not possible, to me that seems logical though.

Last edited by vindicate; 07-07-2010 at 06:05 PM.
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Old 07-09-2010, 04:06 AM
vindicate vindicate is offline
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Anyone?
..,
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Old 07-09-2010, 04:55 AM
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I wasn't really trying to simulate the mic circuit that I'm going to build, I was trying to simulate a mic circuit in general just so I can get a better handle on the concepts.
I see.

Quote:
In the simulator all they have is the "ideal" opamp so that's what I'm working with for now.
Why don't you download a more serious simulator? LTSpice is free, and very serious.

Quote:
So you were saying that you want to bias the input at Vcc/2. In my example that would be 2.5V. So technically mu input is biased at VCC/2.

When you amplify that though, how do you make it so the "peaks" are amplified more, and not amplify the bias?
You reference one input to Vcc/2.

One way to do that is to use a single resistive divider between Vcc and GND to establish the reference voltage, and then use much higher-value resistors to the cap and the opamp input. In this case, you could use 10k resistors in the voltage divider, and two 100k resistor to establish the bias on the opamp side of the cap, and the opamp itself.

Quote:
Can you amplify that from a 2.25 to 2.75(a vcc/2 bias) to a 5v to 0v(with a vcc/2 bias). The bias stays the same, just the min and max and amplified. Maybe it's not possible, to me that seems logical though.
You don't want to do that with audio. You want the average ref level and the input level to be very close to the same voltage.
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Old 07-09-2010, 05:15 AM
vindicate vindicate is offline
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Quote:
You reference one input to Vcc/2.

One way to do that is to use a single resistive divider between Vcc and GND to establish the reference voltage, and then use much higher-value resistors to the cap and the opamp input. In this case, you could use 10k resistors in the voltage divider, and two 100k resistor to establish the bias on the opamp side of the cap, and the opamp itself.
So you would use 2 resistor dividers? If thats what you mean I'm not quite sure where they would be placed in the circuit


As for LTSpice, I'm working on it. I haven't gotten used to it yet.
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