Can someone explain how its possible to get a negative voltage by AC coupling an audio signal in a circuit that is powered by a DC battery? In other words, how can the signal drop below ground potential of the battery (i.e., become negative)?

 

Because the coupling cap is actually an open in the circuit, you're not measuring a voltage relative to the batter you're using, but you are measuring the drop of voltage from a median to a lower voltage as the audio signal changes.

IE: Before the coupling, you may see the voltage go from 3v up to 5v, then down to 1v, then back up to 3v. After the coupling, the voltage will start as a reading of 0v, go up to 2v, then down to -2v, and then back to neutral of 0v.

That's the reasoning in a nutshell, it's more complicated in practice, as it deals with the change of voltage and what way the current flows through the overall circuit, but that should be a good start for you.

~Tim S

 

It normally requires some Diodes but the Base Emitter of a transistor will do.

Look at it this way.

At one signal level a capacitor gets charged to some voltage, and then the signal changes and the top end of the capacitor drops quicker than the voltage across the capacitor does. That capacitor maybe was from near ground to 5 Volts. The top end drops to near ground and now the other end might be -5 Volts. I say near ground because this type of action normally only happens when there is something between the capacitor and actual ground, like a diode or that transistor base emitter.

These types of bouncing capacitors are used in charge pumps and voltage doubler circuits to create circuit supplies like a negative rail for a low current double sided supply, or the boost voltage needed above the rail to gate drive a top side NMOS switch.

 

http://www.allaboutcircuits.com/vol_3/chpt_3/8.html

 

Ok, so it basically has to do with the time constant of the capacitor compared to the frequency of the applied signal? The example I'm working with has an audio signal being AC coupled by a series capacitor, then applied to a peak detector circuit consisting of two diodes, one of them with the anode to ground and cathode to the signal side, the other in series following the AC coupling capacitor and first diode (basically a clamping diode and a half wave rectifier). How is it possible for the clamping diode to become forward biased? Perhaps this was already answered, but it would help to understand my specific application.

 

I think I get it now. The coupling capacitor provides a phase shift, which creates a condition where the signal side of the cap becomes more negative than the coupled side, which causes current to flow in the opposite direction (which is the negative voltage I am looking at). Except I'm still confused as to how this voltage can be more negative than the ground reference. Again, the circuit is powered by battery and the applied audio signal shares a ground with the circuit, which is why I'm having difficulty understanding how the signal can drop below ground (become negative).

 

...and the signal at the clamping diode is shifted upwards? This really surprised me. I thought the signal would be half the amplitude, not the same amplitude (+/- 1V signal applied to clamping diode results in 0-2V signal).

Update...did a little reading, capacitor charges on negative half cycle which then adds to the signal during the positive half cycle, thereby doubling the voltage and making it appear as though the signal shifted upwards, when in fact the amplitude doubled.

 

Well the way I learned it is that you split the circuit into calculating the DC voltage and the AC voltage.

The AC in could say, be a microphone or a line in. This would provide the signal source, and since it is AC, it may range between say, -1v and +1v . Let's say for this example, it's a signal who's AC characteristics are that it goes down a volt, then up two volts, then back down a volt.

Let's also say you're using a 9v battery and within the circuit, you're holding the output at 5v.

After sending the signal through our circuit ( There's lots of different types of there ) the output BEFORE the coupling would read from 4v up to 6v then back down to 5v. This voltage is the 5v we set our circuit at, then at different points along our signal, it goes down to 4v ( which is 5v -1v of AC signal ) and up to 6v ( 5v + 1v of AC signal ) then down to 5v ( +0v AC signal ) .

After the coupling, we eliminate the DC component of voltage because of the capacitor ( a common coupling component ) , which restricts direct currant, and only 'allows' through changing currant. At rest, it would measure 0v because there is no change in voltage. As the circuit lowers to 4v, the voltage on the output of the cap would read -1v because it's reading the lowing voltage of the circuit from what the cap was previously charged to. Then when the circuit reads 6v, the cap would read +1v because it's reading the difference from what it was lowered to, to what the voltage it trying to 'reach'.

I would love to explain more about how the capacitor's characteristics with charging and discharging relate to how you read the different voltages, but I'm really not up to that level of detail at this time. Long day of work. But I hope this helped! If you can understand it with just that, I'm glad to hear I helped you, otherwise feel free to ask some questions!

~Tim S

 

I think some simple loop equations help explain this...

Say you have the situation of a DC voltage source + a capacitor + a resistor in series.

The capacitor will charge all the way up to the DC voltage and the voltage across the resistor becomes zero:

Vcap + Vresistor = Vsrc = Vdc,
Vcap = Vdc

This is a DC voltage which the capacitor holds onto for quite some time, based on the time constant.

Now say the voltage source has an AC signal;
Vsrc = Vdc + Vac

The capacitor is still Vcap = Vdc if the frequency is high enough (not enough time to charge or discharge), and that loop equation is still true:

Vresistor + Vcap = Vsrc
Vresistor + Vcap = Vdc + Vac

But since Vcap = Vdc, you get:

Vresistor = Vac

If the signal is too low frequency then the capacitor will change its voltage significantly and this falls apart.

 

I understand for the most part. Only thing I'm struggling with is that the AC/audio signal I am referring to is first being amplified by a transistor in a circuit powered by a 6V battery. Thus the transistor output swings as high as 6V and no lower than 0V. Then the signal is AC coupled by a series capacitor and input to a clamping circuit. The signal at this point (cathode of clamping diode) drops as low as -0.7V referenced to circuit ground. My question is how can I read a negative voltage across a circuit element when 0V is the lowest potential in the circuit?

 

I understand for the most part. Only thing I'm struggling with is that the AC/audio signal I am referring to is first being amplified by a transistor in a circuit powered by a 6V battery. Thus the transistor output swings as high as 6V and no lower than 0V. Then the signal is AC coupled by a series capacitor and input to a clamping circuit. The signal at this point (cathode of clamping diode) drops as low as -0.7V referenced to circuit ground. My question is how can I read a negative voltage across a circuit element when 0V is the lowest potential in the circuit? Is it simply because the capacitor is holding a charge from the previous positive cycle and therefore on the negative cycle, the diode is forward biased? In other words, my negative voltage reading is relative.

It is about the capacitor holding a charge still, yes, and it is also about the reference.

Basically, with no audio signal the capacitor charges up to say 3V.

When the transistor's swing their output to 6V the output voltage is (with Vsrc being transistor output):
Vsrc = Vcap + Vresistor
6 = 3 + Vresistor,
Vresistor = 3V

When the transistors swing their output to 0V, you get:
0 = 3 + Vresistor
Vresistor = -3V

Switching elements are what let this happen, since you charge the capacitor to some value then suddenly change the reference.