This video is going to be about passive RC high pass filters. And it's going to
build upon a few of the concepts I covered earlier in my video about
low pass filters. So make sure you watch that one first!
So as a quick reminder if you put a resistor and a capacitor in this
configuration you get a low pass filter with a certain cutoff frequency.
This allows low frequency signals to pass through and higher frequency
signals get attenuated.
Well high-pass filters are almost exactly the same!
Just swap the position of the resistor in the capacitor.
Now the high-pass frequencies pass through unchanged and the
low frequencies get attenuated.
The reason why this works is that the voltage across a capacitor
cannot instantaneously change.
So if there's a sudden fast change on the input voltage
it will instantaneously appear on the output side of the capacitor.
So really high frequencies pass straight through the capacitor
as if it was just a piece of wire.
Also, series capacitors block DC.
After all, they're just parallel plates that never actually touch each other,
and in the DC steady-state
they're basically an open circuit
so you get zero volts on the output.
And in between these two extremes we get a frequency response curve like this.
It's very similar to the bode plot of a low pass filter - it's just flipped around.
At high frequencies, the signal passes straight through the filter unchanged.
As the signal approaches the cut-off frequency
it starts to decrease in amplitude.
At the exact cut-off frequency the signal's amplitude is reduced by
3 decibels or 29 percent.
Soon after that, the amplitude dies off at 20 decibels per decade
which means that for every time the input frequency gets reduced by a factor of ten,
the output frequency's amplitude will also get reduced by a factor of ten.
And the other cool similarity is that the equation for calculating the cut-off frequency
is exactly the same as the one for low-pass filters!
Now let's look at some examples.
First, let's have a little fun looking at signals that have been high pass filtered.
I'm going to design a high pass filter with a cut-off frequency of 30 kHz.
And feed in some signals from my function generator so that you can see
what the filter does to them.
Let's start designing the filter by choosing the resistor value.
I'm going to choose ten kilohms so I only put a light load on the function generator.
Next I use the rearranged cutoff equation that lets me
calculate the value of the capacitor I need.
It turns out that I need a 537 picofarad capacitor.
That's a weird non-standard value. So I am going to combine
some capacitors to get closer to the value I want.
This inaccuracy will affect the cut-off frequency by a few kilohertz
but it's close enough for the sake of demonstration.
So with 60 Hz on the input,the high pass filter is barely letting anything pass.
A 10 kilohertz about half the signal is being let through.
You can also see that there's a phase shift in the signal.
At 30 kHz the output is roughly 29 percent less than the input,
which is about what we expected because that's the cut-off frequency.
Finally I want to show you this example where I have a 3 volt sinusoid
laid on top of a nine volt DC offset on the input.
The high pass filter completely removes the DC offset
and you're left with the 3 volt sinusoid.
Another interesting thing to see is how high-pass filters affect square waves.
With this 600 Hz square wave you can see that only the sudden fast
edge transitions are passing through.
But the part of the square wave that's just a flat line gets blocked out.
If we crank the frequency up to 6kHz, a little more of the
original wave shape gets through.
At twice the cut-off frequency we get a waveform like this.
And at 1 MHz almost everything gets through.
Actually that's also because my function generator is pretty terrible at putting
out high-frequency square waves.
Alright time for another example.
In my video about op-amps I showed you a circuit that takes tiny signals from a
microphone and amplifies them so you can listen to very faint sounds.
To give you a quick review, over here we have 9 volts and a 5 kilohm resistor
which are used to power the microphone.
This gives us a 20 millivolt peak to peak signal laid
on top of a 9 volt DC bias.
Over here we have the amplifier configured for a gain of roughly 100.
The idea is that I want to take the 20 millivolts from the microphone
and amplify it to 2 volts so I can hear it on a pair of headphones.
Now in between the microphone and the amplifier, I have a high pass filter.
But what would happen if I didn't have that filter?
If I didn't have the high pass filter here,
the amplifier would get nine volts DC on the input and it would try to multiply
that by a hundred.
I obviously can't get 900 volts out of a 9 volt supply
and the amplifier's output would just end up maxed out to the +9 volt rail.
With a constant 9 volts DC output,
there's not going to be any audio and you'd seriously risk damaging your headphones.
So let's stop destroying our headphones and put the high pass filter back in.
The high pass filter removes the DC component of the signal coming
out of the microphone
so that the amplifier only amplifies the tiny 20 millivolt AC signal.
Using the cut-off frequency equation you can see that the cut-off frequency
is 15.9 Hz.
So any frequencies above that get passed through into the amplifier stage.
The result is a 2 volt AC signal on the output of the amplifier suitable
for driving headphones.
Finally do you see how I used a 100 kilohm resistor in the high pass filter?
Well there's a reason for that.
Having a high resistance value there will ensure that I don't overload
the microphone's unamplified output.
As with low-pass filters you should always pick the resistor value first,
then use the rearranged cutoff equation
to determine the capacitor you need to get the cut-off frequency that you want.
That's it for today and stay subscribed and I'll show you how to use LTSpice
to simulate complex filters.