# How is this high pass filter?

I am new to electronics. I am watching this video.

At 5:48, you can see the high pass filter: I understand the high pass filter, look at this: Then, I thought the first circuit should be like this: Do I misunderstand anything?

• Is that really an AM modulator? Isn't the audio signal just varying the dc bias of the "carrier"?
– user173271
Jul 19, 2021 at 11:23
• Perhaps yes ... How are the non-linearities of a TL072 ? Or perhaps on the decreasing "transfert function" at high frequencies ? To be see in deep ... Jul 19, 2021 at 13:10
• @James Something very strange ... If model of opamp considered as level 3, Boyle model ... opamp seems to be a multiplier ??? Jul 19, 2021 at 15:05
• @James could you write down the equations for the op-amp (on the right of the picture) to demonstrate what you say? I could not understand what you say
– Dat
Jul 21, 2021 at 10:05

The capacitor has a complex resistance which is defined by: $$X_{c}=\frac{1}{2\pi fC}$$ This means that high frequencies are "passed through" the capacitor (f is high, so Xc is low). The resistor with the value of 100k is there to set a minimum resistance. So in combination with the 680k resistor, a gain of 6.8 is set. That means that for higher frequencies, the gain is 6.8 and for lower frequencies this circuit will decrease its gain

• how do you calculate the 6.8 gain?
– Dat
Jul 20, 2021 at 3:01
• @Dat The TL072 amplifier is connected as an inverting amplifier (google op-amp inverting amplifier). This can be found by the feedback resistor with a value of 680k. The op-amp as inverting amplifier has a transfer function which is defined as Rf/Rin. In this case, these resistors are 680k/100k. This means an amplification of 6.8. Do you understand? Jul 20, 2021 at 10:28
• I think I could derive the result (6.8) with the pin 3 of the op-amp connected to ground, but pin 3 is connected to a lot of sources that made me confused
– Dat
Jul 20, 2021 at 16:04
• It's only logical that you get confused. It's not a simple circuit. On the non-inverting terminal, the 100k potentiometer in combination with the 680k connected to the 9V and the 100k in between them generates an offset. This can also be called a bias offset, as @james mentioned. This means that the output voltage of the op-amp also has an offset. The music from the mp3-player is super positioned onto this bias. All in all, the "sources" as you call it are there to generate this offset, but doesn't affect the amplification. The music will however change this DC bias continuously. Clear? Jul 20, 2021 at 20:29
• why do we need to generate an offset?
– Dat
Jul 21, 2021 at 4:02

Yes, you have a misunderstanding.

First and foremost, since there is a capacitor in series with the circuit, you can already determine that the capacitor will block any DC and pass only AC, so it is a high pass filter, even without looking further.

The second thing is that there is an op-amp in the circuit, and since that signal connects to an inverting op-amp circuit, effectively the op-amp inverting input can be thought of as a virtual ground - although in this circuit it's not ground because this is a single supply circuit, so the virtual half-voltage supply is set via the positive op-amp input. In a simpler example with dual power supplies it could be the virtual ground.

The modifications you present to the circuit would throw off the feedback and DC biasing, making the circuit not to work as intended.

• "single supply circuit, so the virtual half-voltage supply is set via the positive op-amp input". Where is "single supply circuit"? and where is the virtual half-voltage supply?
– Dat
Jul 20, 2021 at 3:16

Do I misunderstand anything?

You misunderstand that in this circuit, pin 2 of the op-amp (the inverting input) is a "virtual ground". This is made so by negative feedback. Negative feedback keeps pin 2 at the same voltage level as pin 3 (the non-inverting input). This means that current flow is needed through the 1 nF capacitor to produce an output voltage but, at 0 Hz, the impedance of a capacitor is infinite hence, there can be no current flow. Hence, it's a high-pass filter.

Because the input pin (2) is a virtual earth, the 100 kΩ resistor is ineffective as a filter component.

• Could you answer all this by writing down the equations?
– Dat
Jul 20, 2021 at 3:06
• What equation of what relationship are you looking for? Jul 20, 2021 at 7:37
• I don't understand the last sentence: "Because the input pin (2) is a virtual earth, the 100 kΩ resistor is ineffective as a filter component." Why is that?
– Dat
Jul 20, 2021 at 8:31
• Because the virtual earth is like zero ohms and 100 John's in parallel is ineffective. Jul 20, 2021 at 8:56
• I am more confused. What is John's ?
– Dat
Jul 20, 2021 at 15:49