DC Blocking capacitor - How does it know where ground is?

[AnalogNewb]

DC Blocking capacitor -- How does it know where ground is?

Hi Everyone,

A DC blocking capacitor removes any DC offset from an input signal. I've been doing this for years without thinking about it.

My question is, how does it know what the local ground potential is?

My only thought on this is that the output side of the cap is just pushing and pulling electrons, and it adds to / subtracts from whatever the voltage on the input to the circuit already is.

With an inverting op-amp input, it's going into the "virtual ground" of the circuit so that sort of makes sense.

But now I am using a DC blocking cap to feed something that's not an op-amp: an ADM3491 differential line receiver.

So again, how does the cap know where ground is? I feel like the input impedance of of the circuit after the cap plays into this -- the ADM3491 has an input impedance of 9K.

Also, this is a manchester encoded digital signal that is intended to pass through AC coupled analog circuits.

Thanks.

 

[berkeman]

Hi Everyone,

A DC blocking capacitor removes any DC offset from an input signal. I've been doing this for years without thinking about it.

My question is, how does it know what the local ground potential is?

My only thought on this is that the output side of the cap is just pushing and pulling electrons, and it adds to / subtracts from whatever the voltage on the input to the circuit already is.

With an inverting op-amp input, it's going into the "virtual ground" of the circuit so that sort of makes sense.

But now I am using a DC blocking cap to feed something that's not an op-amp: an ADM3491 differential line receiver.

So again, how does the cap know where ground is? I feel like the input impedance of of the circuit after the cap plays into this -- the ADM3491 has an input impedance of 9K.

Also, this is a manchester encoded digital signal that is intended to pass through AC coupled analog circuits.

Thanks.

Welcome to the PF.

Never anthropomorphize DC blocking caps. They hate it when you do that.

A capacitor is a 2-terminal device, so there is no mechanism for any referencing to anything other than what is connected to the 2 terminals. And yes, the cap's effect depends on the impedance of the sourcing circuit, and the impedance of the circuit on the far side of the cap. It's all about complex impedance divisions.

If you can post a couple example circuits, we can do a better job of helping you get the right mental pictures to use...

 

[AnalogNewb]

Welcome to the PF.
Never anthropomorphize DC blocking caps. They hate it when you do that.

If you can post a couple example circuits...

Thanks and :-) .

I can't get to the computer with Altium on it right now, so let me try to boil it down a little.

I don't have a formal electronics education. I was always told that there was no such thing as absolute voltage, that all voltages are only relative to some other potential.

When people say that the DC voltage is removed by a blocking cap, I don't get how the cap "knows" where zero volts is. Zero (ground in my circuit) could be a normal wall outlet ground, ground to earth, or the plate circuit of an oscilloscope at +1000 volts.

In all of those cases, using a DC blocking circuit results in an AC waveform that is centered on whatever ground happens to be in that circuit.

I'm starting to think that there must really be an "absolute zero" voltage, and that would be when there are no surplus electrons available and no positive holes ready to accept electrons.

Put another way, the output of the capacitor can either supply electrons or suck them up, and in doing so adds or subtracts to the potential in the circuit that follows it. Let me see if I can do an ascii diagram

 -----------------                        -------------------
 | signal source |---------| |------------| ADM3491 Input   |
 |   Z = 600R    |                        |  Z = 12K        |
 -----------------         Cap            --------------------

The signal source has a nominal output impedance of 600 Ohms (although in the real world it is typically somewhat lower), and the ADM3491 has an input impedance of 12K.

I'm thinking that if you model the ADM3491 as a 12K resistor between input and ground, then the input naturally sits at zero volts when it is unconnected.

Connecting the cap to the input then adds or subtracts electrons to or from the zero volts that the input naturally sits at. Is that right?

I want to understand the reality of this type of circuit for two reasons:

1. The next step is devising a diode clip circuit to keep from blowing up the ADM3491.

2. I have another, unrelated application where I need first remove the DC offset on the input and then apply a DC offset of 2.5 volts to feed a single supply op-amp.

But first I would just really like to understand the DC blocking cap at its most basic.

Thanks.

 

[waht]

Just forget about grounding for a second. If there is a closed path somewhere between the terminals of a DC or AC voltage source, then current will flow.

A DC blocking capacitor just represents a short for AC only. And similarly, if there is a path taken from the AC source through the DC blocking cap and back to the AC source again, then current will flow.

 

[waht]

Consider the circuit:

http://ecelab.com/circuit-ac-amp-1a.jpg

The AC current originates at the voltage source V_in.

Can you identify which paths AC current would take and return back to the original source?Hint: there could be three paths

 

[AnalogNewb]

Just forget about grounding for a second. If there is a closed path somewhere between the terminals of a DC or AC voltage source, then current will flow.

A DC blocking capacitor just represents a short for AC only. And similarly, if there is a path taken from the AC source through the DC blocking cap and back to the AC source again, then current will flow.

Thanks for trying to answer my question but I fully understand that. I've been using DC blocking caps in circuits for 20 years. I'm trying to understand the physics involved not just the rule of thumb.

 

[vk6kro]

Consider the following circuit:[PLAIN]http://dl.dropbox.com/u/4222062/capacitor.PNG

When power is first applied, the capacitor is uncharged and so there will be 6 volts across each 100 K resistor and 60 µA flowing in them. This will charge up the capacitor and eventually, after 100 seconds or so, there will be 12 volts across the capacitor and no voltage across the resistors. The charging current decreases as the capacitor charges up as there is less and less voltage across the resistors.

Now, briefly close the switch. The capacitor cannot discharge rapidly becaise it has to do so through the lower 100 K resistor.
The + side of the capacitor has been grounded, so the other side is 12 volts less positive than ground. So, it goes to minus 12 volts.
If you kept doing this, the voltage across the lower resistor would alternate between zero volts and minus 12 volts.

This is an extreme example, but it shows how DC blocking works. It all depends on the capacitor being fully charged and not having enough time to discharge.

If you varied the input voltage (across the switch if it was open circuit) from 13 volts to 11 volts, the voltage across the lower resistor would vary from +1 volt to -1 volt because the capacitor keeps its 12 volt charge and this subtracts from the input voltage.

 

[berkeman]

I don't have a formal electronics education. I was always told that there was no such thing as absolute voltage, that all voltages are only relative to some other potential.

When people say that the DC voltage is removed by a blocking cap, I don't get how the cap "knows" where zero volts is. Zero (ground in my circuit) could be a normal wall outlet ground, ground to earth, or the plate circuit of an oscilloscope at +1000 volts.

In addtion to the other good info posted already, I'll just say "correct" to the part about all voltages being relative. And blocking DC does not meaning that the cap ground references the signal, unless the downstream components are connected to ground. You often use a DC blocking cap to convert a ground-referenced signal to some other reference voltage, like when you have an audio signal from a microphone going into a circuit powered by a single 9V battery.

EDIT -- this may have already been said above, but think of a capacitor as an AC short, and a DC open. It's effect in the circuit between those extremes has to do with its value in uF, and the impedances of the upstream and downstream circuits.

 

[AnalogNewb]

Thank you all for attempting to answer my question.

Sadly I seem to be unable to communicate well enough to ask it correctly, and I'm also having trouble following and/or seeing the relevance of some of the answers presented. That's my failing I'm sure.

Thanks again.

 

[vk6kro]

It will take a little more effort on your part.

Capacitors do not "know" where ground is. They don't need to. That was the first part of your question.

If you want to understand DC blocking, work through post 7 above and then we can talk about what you don't understand. It is easy stuff really, and you will need it to work out the clamping circuits you need.

 

[AnalogNewb]

Capacitors do not "know" where ground is.

I was speaking metaphorically. Of course I know that they can't know what ground is, having only two pins and not being connected to it. You guys must think I'm a complete moron.

The question is how to they SEEM to know where ground is when in reality they cannot?

It will take a little more effort on your part.

I put a LOT of effort into writing my posts but nobody bothered to read them, you guys just skimmed and posted the first thing that popped into your head.

I also spent a good half hour going through post #7. I think I get it, I don't have any questions about it. But it doesn't address the question, that I can see.

Look, I know I'm not being easy to work with or giving you much satisfaction. I'm very grateful that you tried to help and I appreciate that you are here answering questions.

I'm too tired to continue this tonight but tomorrow I am going to go over the whole thread again. Frankly I think I've figured out the answer to the question myself, but it involves the concept of "absolute voltage" and apparently that's not allowed.

I hope you guys will give me one more chance tomorrow after I've had a chance to sleep on it and reconsider my wording.

Thanks again.

 

[vk6kro]

But first I would just really like to understand the DC blocking cap at its most basic.

That is what you asked.

Capacitors are either charged or they are not. If they are, they have a voltage across them which may not be removed easily if there is no easy low resistance path for them to discharge.

Absolute voltages do not have any relevance in these circuits. Only relative voltages matter.
Capacitors work just fine even in a space station.

[PLAIN]http://dl.dropbox.com/u/4222062/capacitor2.PNG

In the above circuit, assume the capacitor is already charged to 6 volts, so the right side of it is always at 6 V less positive than the left side. It is just like a battery, if you like.

Now rotate the pot. At the top, the voltage is at + 12 V so the output must be +6 volts.
12 - 6 = +6

In the middle, it is at + 6 V so the output must be at 0 volts.
6 - 6 = 0

At the bottom, it is at Zero Volts so the output must be at - 6 Volts.
0 - 6 = -6

Do this rapidly and you have a waveform theat goes from + 12 V to Zero volts causing the output to swing from + 6 volts to - 6 volts. In other words, the DC component of the signal has been removed.

This is how a blocking capacitor works.

 

[sophiecentaur]

Re the above circuit.
Once you connect it up, the capacitor will start to discharge through the series resistance (between 'out' and the wiper of the pot) until there are zero volts across it.
You can, briefly, obtain a non-zero voltage across it if you move the pot up or down but, the volts across the C will return to zero (exponentially - according to the RC time constant).
Oscillating the pot position fast enough will produce an AC voltage at the output. The faster the oscillation, the higher the value of AC volts - tending to 6V pk-pk about zero.

 

[vk6kro]

If the pot is in the centre position, the capacitor will charge to 6 volts which it reaches in about 60 seconds.
It has a time constant of about 10-15 seconds, so you can turn the pot quite slowly without appreciably affecting the voltage across the capacitor.

If you turn the pot from end to end, the output will be 12 volts peak to peak, centred around zero volts.

Incidentally, the position of the pot wiper willl determine if the capacitor charges higher or lower than 6 volts. If it is above the centre point, it will slowly charge higher than 6 volts and if it is below the centre point it will slowly discharge to less than 6 volts.
It will not fully discharge around the 100 K resistor and the bottom part of the pot because the pot has a voltage across it and so does the wiper unless it is at the bottom of its travel.

The whole point of this is that the movement of the pot is a slow motion version of an input signal. Next we say that you move the pot so fast that the capacitor doesn't change charge at all and the output is a faithful version of the input but now centred around 0 volts instead of 6 volts.

 

[nchavan]

Hello I am new to Physics FOrums.
lets try to answer the question "how capacitor knows where is the ground"?

We will consider the construction of the capacitor, capacitor consist of two metal plates and in between the plates the dielectric material is present. If you apply on type of charge to one end of the capacitor it induces opposite type of charge on the other part. (Read theory of capacitor on any physics site or the basic electronics books.)

So in case of DC blocking the signal is shifted to fix level either positive or negative so this fix voltage creates the opposite charge on the other side of the capacitor, so when ac signal with some offset passes through capacitor, the other terminal of the capacitor acts as a adder two different voltage sources mating at one point, so the opposite charge develop because of fix offset voltage and the shift in the ac signal gets nullify so you get output to near to zero.

I think this is how it is blocking the dc, and pulling the signal to the ground.

I hope I have convey a right answer to you.

 

[sophiecentaur]

I'm afraid that the initial question is not a proper question - if it were, it wouldn't need to use the word "know" in it. It sounds innocent and reasonable enough but it really has no answer.
The fact is that the DC conditions on one side or the other of a capacitor cannot communicate themselves across it. Once things have settled down (true DFC conditions have established) the 'DC Earth' on one side can be 1000V different from the 'DC Earth' on the other and there would be nothing to indicate it if you couldn't put a meter across it, or if the Capacitor didn't fail (not a true capacitor any more).
So you can't expect anything back from this thread but a lot of confusion, I think.

 

[chaoseverlasting]

When you apply a voltage across a capacitor, you bias it. For example, if you apply +9v at one end and then you connect the other end to -9v, you have told the capacitor that -9v voltage is the ground voltage. This is because you're forcing the movement of current in one direction with the voltage drop.

Of course, this explanation works for dc only.

 

[sophiecentaur]

Bias is just a word and you can't tell a capacitor anything - it hasn't a brain.

A charge will flow into or out of a capacitor, around any resistive path in the circuit until Q=CV, then it will stop (after a theoretically infinite time but, in practice, fairly quickly). After that time, no one can have a clue, looking at just one side, what the Potential difference V across the capacitor is.
Why should the capacitor care what the "ground voltage" is? It has just two terminals and all it does is follow the above rule.

If you are talking just about DC circuits (after things have settled down) then you can ignore any capacitors in the circuit for any calculations you may want to do (i.e. you could disconnect them all and it would make no difference).

My point is that anthropomorphising is not a good way of getting an understanding of what is 'really going on". It can only serve to confuse.

btw, in the first circuit in the thread, there is absolutely no way of knowing what DC potential will exist across the capacitor because no dc current will flow as there is, apparently, no DC path from left side to right side. What you have is a series high pass filter which has zero response at DC so no DC info can pass.