Hi,

Is it possible to have a variable HPF that is passive? I've been breadboarding a simple RC filter and putting a 100k pot to ground. It works when the pot is zero, but as I roll it up the sound goes away. I suspect the output impedance is changing and therefore doesn't work with my system: it's an electric 9v mic (guitar) into an amp. For reference, I'd like the HPF to be roughly 40-1000 hz, which I can calculate the cutoff.

I've also been working toward this circuit, which is an electret mic preamp. If I pot R3, which is before the opamp, would that work? Then the opamp would maintain the output impedence?

 

Is it possible to have a variable HPF that is passive? I've been breadboarding a simple RC filter and putting a 100k pot to ground. It works when the pot is zero, but as I roll it up the sound goes away. I suspect the output impedance is changing and therefore doesn't work with my system: it's an electric 9v mic (guitar) into an amp. For reference, I'd like the HPF to be roughly 40-1000 hz, which I can calculate the cutoff.

It would be helpful if you posted a schematic so we can see what component values you're using.

If it looks like this:

You're changing the frequencies passed when you change the resistance.

 

It would be helpful if you posted a schematic so we can see what component values you're using.

If it looks like this:
View attachment 282703
You're changing the frequencies passed when you change the resistance.

Right, that's exactly what I did. So the junction between C1 and R1 was the output, across ground.
C1 = 22nf
R1 = 2k2 + 100k pot.
So with the pot at 0, the cutoff is 3288 hz.
With pot at 100k the cutoff is 71 hz.

But I only got sound when the pot was zero or close to it. I think because the output impedance was changing so much?

 

Also, what's a simple program that will allow me to draw circuits like DL324 did above?

 

But I only got sound when the pot was zero or close to it. I think because the output impedance was changing so much?

What is the input source and what are you using to measure the output?

Also, what's a simple program that will allow me to draw circuits like DL324 did above?

I use an old version of Eagle. I print to PDF so I don't get the distracting, useless colors:

 

The input source is an electret mic, basically this circuit.



Only I added the variable HPF at the end before the preamp (guitar amp). So C1 was 22nf and then R1 to ground.
I didn't measure the output, just using my ears.

The above circuit works as is... and the HPF resistor is the input impedance of the the preamp, 1M ohm.

 

OK, my first schematic in Eagle. Thanks for the tip. Anyway, this didn't work when I rotated the pot.

 

OK, my first schematic in Eagle. Thanks for the tip.

Nice! Don't forget to show all of the connections. I assumed you had the pot connected as a rheostat.

Anyway, this didn't work when I rotated the pot.

What kind of amplitude do you get from the microphone and how much are you amplifying the output?

 

Full disclosure, I'm a mechanical engineer and haven't done this type of circuit analysis and breadboarding since college 30 years ago. I've built/modified guitar amps and pedals, but never dug into the actual circuit design. So feel free to tell me what I'm doing wrong.



Here's my breadboard. I just stuck the pot in the board. I don't necessarily understand the difference between a potentiometer and a rheostat. Although a quick search on the internet suggests pots are installed in parallel with a voltage source and a rheostat is in series with the load. I'm not sure... it's possible that I'm using the pot wrong.

I don't know what kind of amplitude I get from the mic, I generally have to gain it up a lot after this circuit. That's why I think I need the circuit in post #1 that is a preamp. I just want to add a variable HPF to it, I think.

The circuit in post #6 does work, but it's a fixed HPF.

 

Here's my breadboard. I just stuck the pot in the board.

I can't see how the pot is connected. Is the thing at the top of the photo the mic or a power connector?

I don't necessarily understand the difference between a potentiometer and a rheostat.

A rheostat can be a pot connected so that only two terminals are used.

 

Try changing the capacitor to 2.2µF, the resistor to 22kΩ, and the pot to 1 megohm.

 

Ooooh. I only connected 2 of the pot terminals. Now I understand the middle has to be connected to ground with one of the other ones.

The thing at the top is 9v power.

 

You copied my mic and preamp circuit.

The output from the microphone is not a low impedance then its output level at all frequencies are cut in half when the rheostat is turned down to cut most bass and voice sounds. The rheostat must not affect the DC voltage at the input of the opamp so you are missing a coupling capacitor between the rheostat and the opamp input.

The breadboard is not suitable for a microphone circuit because the rows of contacts and long wires all over the place are antennas that pickup lots of hum from electricity wiring in walls.
None of my microphone circuits produce hum because I make the circuit compact with very short wires and use a shielded audio cable for mic in and circuit out.
Maybe your highpass filter will be used to reduce the hum picked up??

Your highpass filter is the "reducing half" of a bass tone control circuit. Like all tone control circuits, it should be at the low impedance output of the mic preamp opamp then the filter should feed the high input impedance of a second opamp.

 

Just put the pot on the plate. I don't really understand the difference between a potentiometer and a resistor. However, a quick internet search shows that the pot is installed in parallel with the power supply and the potentiometer is installed in series with the load. I don't know...maybe I'm using the wrong bowl. I don't know what the amplitude of the mic looks like. Should have more later in this cycle. So I think I need the circuit on post #1 which is the preamplifier. I want to add hpf instead. The circuits after work #6, but the correct HPFs.

 

The center connection (slider) in the pot moves across the resistor from one side to the other as you rotate the pot.

In your circuit, you have not connected the center terminal! So all you have us a fixed resistor, rotating the control will have no effect.

To use a pot as a variable resistor, you connect the center terminal and either of the others. Which one you choose determines whether the resistance increases or decreases as you turn it clockwise.

You don’t need to connect the third terminal, but connecting it to the center terminal ensures that the pot does not go to open circuit if the slider loses contact, it will go to the full resistance.

 

Thanks for the responses! Yes, I realize I was using the pot wrong. I'll try to get back to it sometime this week and test it.

 

Your highpass filter is the "reducing half" of a bass tone control circuit. Like all tone control circuits, it should be at the low impedance output of the mic preamp opamp then the filter should feed the high input impedance of a second opamp.

So to combine the mic preamp shown in post #1 and a variable HPF, I should have the preamp first, then the variable HPF after it? Like combining these 2 circuits as shown: ?

I'd need the HPF opamp TL072 to take the same 9v supply somehow. The HPF shown is a 2nd order. I'd want the range to be ~80-1000 hz. So changing the caps to 68nf and the base resistors to 3k would give 73-1212 hz when accounting for the 2nd order factor.

 

Here is how a Virtual Ground can bias the new Sallen-Key 2nd-order highpass filter:

 

Here is how a Virtual Ground can bias the new Sallen-Key 2nd-order highpass filter:

thanks. Virtual ground is a new concept for me, been reading about it.

In the single pole supply, is IC terminal 4 connected to anything? Connected to the 47k resistor? Or the rest that says virtual ground?

And I guess I don’t understand the virtual ground section of your picture. Sorry, I’m new to this.

 

In the single pole supply, is IC terminal 4 connected to anything? Connected to the 47k resistor? Or the rest that says virtual ground?

And I guess I don’t understand the virtual ground section of your picture.

In the single polarity supply circuit, Pin 4 of the opamp is connected to the circuit "ground" which is 0V of the 9V battery, the output 47k resistor and the shields on the input and output cables.

With a dual polarity supply, the 0V is ground and is at half the supply voltage so that the outputs of the opamps can swing the maximum amount up (positively) and down (negatively).
With a single polarity supply, the opamp DC biasing ground at its + input is also at half the supply voltage so that the outputs of the opamps can swing the maximum amount.

The "virtual ground" opamp has its input biased at half the supply voltage with the 100k resistors and produces a very low output resistance so that its output can supply the half the supply voltage to the + inputs of many signal opamps, and so that the filter circuit has its resistors (that connect to ground on the dual polarity supply circuit) connected to the very low output impedance of the virtual ground opamp. The signal ground and virtual ground have no signal and no voltage changes.