Removing DC from AC signal

Hi

I have seen and read a few times about removing a DC signal with a series capacitor and I have a few questions about it.

  1. How does it work, why a capacitor would remove the DC signal and not the AC signal?
  2. Usually they are electrolytic capacitors of around 100-200uF - i.e. polarized. But if the AC signal is oscillating between f.x. -5V and +5V, would that not blow the capacitor (when the voltage is minus)?
  3. If it removes the DC signal, would it remove the "DC offset", i.e. make a sine 0 to 5V moving to -2.5V to +2.5V?

Thank you!

Oh and:

  1. I unfortunately do not have an oscilloscope (only a multimeter) but I tried to simulate it using tinkercad https://www.tinkercad.com/ - and there I could see the DC voltage is only removed when there is a load after the capacitor. When there is a load this is what I see in the “simulated” oscilloscope (not sure how accurate it is).

ocilloscope.png

Capacitors in series "block" DC.* You are essentially making a [u]high-pass filter[/u], and since DC is "zero Hz" it get's filtered-out, leaving the "AC component".

The resistor is important. You need a DC current path to ground. EDIT - That should help with question 4. And also, if you're try to calculate the cutoff frequency for the filter you need to have a resistance or it would theoretically go down to zero-Hz and pass DC.

Usually they are electrolytic capacitors of around 100-200uF - i.e. polarized. But if the AC signal is oscillating between f.x. -5V and +5V, would that not blow the capacitor (when the voltage is minus)?

The DC component actually ends-up across the capacitor and the AC passes-through the capacitor.

If it removes the DC signal, would it remove the "DC offset", i.e. make a sine 0 to 5V moving to -2.5V to +2.5V?

Correct!

  • If you look at how a capacitor is made there is no direct connection between the pins so DC current can't flow and an Ohmmeter shows infinite resistance (after the capacitor charges up). Electrolytic capacitors are a little "different" and you might measure some leakage resistance.

Thanks man that is great help!

Just so I am sure I understand:

DVDdoug:
The DC component actually ends-up across the capacitor and the AC passes-through the capacitor.

So you're saying sending an AC sine wave oscillating from -5V to +5V will not blow a polarized capacitor?

And have I understood correctly that:

  • there needs to be a resistor for the DC offset to go - that's why I need a load (in my case a speaker) which acts like a resistor?
  • When there is no load and no resistor, the resistance is basically infinite, which means the cutoff frequency is 0, which means DC actually passes through?

The definition of a capacitor's behaviour in circuit is:

I = dV/dT

where I is current, dV/dT is the instantaneous rate of change of voltage in volts/second.

In the case of DC, dV/dT is by definition equal to zero.
Therefore I=0 (no current flows).

aarg:
The definition of a capacitor's behaviour in circuit is:

I = dV/dT

where I is current, dV/dT is the instantaneous rate of change of voltage in volts/second.

Alas you forgot the most important part, the capacitance.

I/C = dV/dt

Or put another way

I dt = C dV

Which is simply

dQ = C dV - charge changes with voltage, capacitance being the constant of proportionality.

With DC there is no change in voltage, so no change in charge, so no current. DC current cannot flow through a
capacitor, which is why it blocks DC.

The DC voltage across the capacitor is exactly as if the capacitor wasn't there and the ac voltage
decreased to nothing.

If there is no load after the capacitor there is nothing to define the DC voltage on the capacitor, so it will be unchanged from when you put the circuit together. In fact no current flows at all, ac or dc.

ToneArt:
Thanks man that is great help!

Just so I am sure I understand:
So you're saying sending an AC sine wave oscillating from -5V to +5V will not blow a polarized capacitor?

That 5V amplitude AC signal can exist on both sides of the capacitor without needing much voltage across the capacitor, in which case its not a problem. If the voltage across the capacitor is forced to the wrong polarity it will conduct and heat up.

The signal voltage (w.r.t. ground) and the voltage across the capacitor are, in general, different,
usually very different for a coupling-capacitor.

Any equivalent way of thinking about the High Pass Filter action is as a voltage divider. Except, the "resistance" of the capacitor is actually frequency-dependent reactance (i.e. it lies on the imaginary axis of the complex impedance plane):

So, at DC (w = 0), the transfer function is 0. It approaches unity as the frequency rises.

The 3dB point is when |jwRC| = 1 or w = 1/RC.

Thanks everybody it's a LOT clearer now :slight_smile:

Hi again - I have a bit more related to this if that's ok :slight_smile:

I noticed that quite often people add a small capacitor and a 10R resistor between the big capacitor and the ground - especially when outputting audio.

See for example here - on the right side after the last opamp the 0.1uF cap and 10R resistor:

3 questions:

  1. Why is there need for a small capacitor (0.1uF) there?
  2. Since the resistor is "after" the capacitor, does it have any effect?
  3. Like we spoke about the big capacitor only acts as an HPF if there is a resistor "discharging it", but having a load at the end like a speaker will act as the discharging resistor, so just the big capacitor on the output would be enough, correct?

Thank you!

PS: FYI the picture is from https://www.instructables.com/id/Arduino-Audio-Output/

ToneArt:
I noticed that quite often people add a small capacitor and a 10R resistor between the big capacitor and the ground - especially when outputting audio.

This has nothing to do with the original question (removing DC).

Zobel network.
Leo..

The 0.1uF and 10R at the output are to
load the chip to reduce its tendency
to oscillate.

Quote: "So you're saying sending an AC sine wave oscillating from -5V to +5V will not blow a polarized capacitor?".

That is not what was said at all. And the the polarized capacitor will act as a very low resistance short circuit to either -5v or +5v, depending on which way the polarized capacitor is placed in the circuit. Often sound like a gun shot!

Paul

That circuit shows the 922 opamp being used, first as a buffer and low pass filter, and secondly as a
power amplifier to drive the speaker.

Unfortunately this last part of the circuit is completely wrong and will not work.

You cannot parallel opamp sections like this, and secondly that opamp cannot drive 8 ohms, its limited
to 32 ohm loads or above.

With a real power amp the 10 ohm resistor and 0.1µF capacitor form a Zobel network which allows
the amp to be stable into inductive loads. An opamp will not need this, but then a normal opamp cannot
drive high current.

Thank you all for your replies.

Wawa:
This has nothing to do with the original question (removing DC).

Zobel network.

Ok I understand now thanks - and no it has nothing to do with the original question - but how could I know? :slight_smile: (I thought it was related to the capacitor)

@MarkT: the diagram shows a speaker, but I think the idea is more like "Line-level" type, so not a power amp :slight_smile:

And since it is not a power amp, I would guess the Zobel network is useless (since it will not be connected directly to a speaker), correct?

Paul_KD7HB:
Quote: "So you're saying sending an AC sine wave oscillating from -5V to +5V will not blow a polarized capacitor?".

That is not what was said at all. And the the polarized capacitor will act as a very low resistance short circuit to either -5v or +5v, depending on which way the polarized capacitor is placed in the circuit. Often sound like a gun shot!

I appreciate the help, but I am even more confused now: "the cap will act as a short circuit to either -5V or +5V" - what does that mean? Short circuit from the wave signal to the rails?
(and the gun shot, that's when the capacitor pops I imagine?)

There are much better anti-aliasing low pass filter configurations that you could use, considering the number of op amps available...