What could be causing this high pitched whine in my reverb circuit? (beginner)

[deadlyshart]

Hi guys, I'm trying to build a reverb circuit based on the BTDR-3 module: http://www.tubeampdoctor.com/images/File/BTDR-3%20DIGI-LOG%20REVERB%20MODULE.pdf

Since I actually had trouble finding too many schematics online, I'm basing it mostly on the (very vague) mono circuit in the above link, and the circuit here: http://tagboardeffects.blogspot.com/2014/07/rub-dub-reverb-deluxe.html

To be honest, there seems to be strangely little info about it on the internet. Anyway, it seems fairly straightforward, so I put it together. For the op amps I'm using a dual TL082 op amp. To power the rails of the op amps, I'm using a ICL7660CPA charge pump, which inverts the +9V I supply it with. I also use a 7805L LDO to produce the +5V to power the BTDR-3.

So, I assembled it on the breadboard and it works. You can hear the reverb effect, it's cool (honestly, I wish it had more decay time, but whatever...). However, there's this high pitched whine when I'm not playing anything.

I'm pretty new to this. Does anyone know of typical things that could cause this? I might bring it into work tomorrow so I can poke around in the circuit with a scope and see if I can figure out at what stage the whine is generated. Thanks.

[bluebunny]

Welcome!

I'm using a ICL7660CPA charge pump
. . .
However, there's this high pitched whine when I'm not playing anything.

You need the 7660 part with an "S".  This has the facility to hike the internal oscillator out of the audio range: short pins 1 and 8.

[samhay]

There are a few DIY designs that use the BTDR-3 brick.
You can get more depth out of the brick by using a larger depth/decay pot - I would suggest log 100k.

As an example, you can see a schematic of the Rub-a-dub Deluxe on pp3 of the BOM pdf here: http://1776effects.com/product/rub-a-dub-deluxe/

Note that you do not need to use a bipolar supply for the op-amp.

[bluebunny]

Note that you do not need to use a bipolar supply for the op-amp.

Good point, Sam.  So then the need for a 7660 would simply go away.

[deadlyshart]

Welcome!

I'm using a ICL7660CPA charge pump
. . .
However, there's this high pitched whine when I'm not playing anything.

You need the 7660 part with an "S".  This has the facility to hike the internal oscillator out of the audio range: short pins 1 and 8.

thanks!

Wow, great point, I didn't think of that... so basically, the frequency of the oscillation that causes the "charge pumping" mechanism is in the audio range I guess?

Actually yep, I just checked the datasheet:
Oscillator Frequency fOSC, Pin 1 Open or GND 10 kHz
Oscillator Frequency (Note 10) fOSC, Pin 1 = V+  35  kHz

Awesome tip, thanks!

Luckily, I ordered these 7660 chips from a few vendors when I did, and one of them was actually the 7660S! So I can try that later.

[deadlyshart]

There are a few DIY designs that use the BTDR-3 brick.
You can get more depth out of the brick by using a larger depth/decay pot - I would suggest log 100k.

Damn, I can't believe I didn't think of that, it seems so obvious now... if turning the knob up gave me more reverb, and its function was increasing the resistance, then using a bigger pot would do it more...

Why do you say a log pot though?

As an example, you can see a schematic of the Rub-a-dub Deluxe on pp3 of the BOM pdf here: http://1776effects.com/product/rub-a-dub-deluxe/

Note that you do not need to use a bipolar supply for the op-amp.

Awesome, thanks, I hadn't seen that schematic before.

I'm still trying to get my head around powering op amps. When you say that you don't need a bipolar supply, do you just mean that you can use 0 and 9V for the rails? As I understand it (which very may well be wrong), the rails are relative; they don't know positive or negative. But they'll be relative to the input signal, which will definitely be with respect to 0V. So I've heard you can bias the input signal so it's now sitting at 4.5V and with respect to that, but I don't totally understand how, if the input signal and power supply have the same ground. Is that what you mean?

thanks!!

[midwayfair]

An op amp cares about the highest and lowest voltage it sees.

The Rub-a-Dub has a voltage divider to make 4.5V, and then pin 4 sees ground (0V) while pin 8 sees 9V. The rest of the pins sit in the middle after being referenced to 4.5V with a large resistor for non-inverting stages or, in the case of an inverting stage, by connecting the + pin directly to 4.5V. This is basically identical to running it at +-4.5V. Nearly every guitar pedal uses this method because they run on +9V supplies.

I'm not sure exactly how you were hooking up the -9V to the rub a dub layout you provided, but not only is it not needed it is likely to cause some sort of issue if you tried to just stick it in there without reworking the design for +-9V operation.

[Groovenut]

There are a few DIY designs that use the BTDR-3 brick.
You can get more depth out of the brick by using a larger depth/decay pot - I would suggest log 100k.

Why do you say a log pot though?

A log pot will give a more even reverb depth response as it's turned up. You could stick with the lin pot, it would just make the depth increase seem to come on faster as the pot is rotated CW. YMMV.

[samhay]

^what he said.

I have a stereo (well pseudo-stereo) BTDR-3 circuit in the skunkworks with a log 100k decay/depth pot.
This works quite nicely if you wan an 'effect' (lots of reverb), yet with some means of reigning it in for more sane stuff.

[deadlyshart]

An op amp cares about the highest and lowest voltage it sees.

The Rub-a-Dub has a voltage divider to make 4.5V, and then pin 4 sees ground (0V) while pin 8 sees 9V. The rest of the pins sit in the middle after being referenced to 4.5V with a large resistor for non-inverting stages or, in the case of an inverting stage, by connecting the + pin directly to 4.5V. This is basically identical to running it at +-4.5V. Nearly every guitar pedal uses this method because they run on +9V supplies.

I'm not sure exactly how you were hooking up the -9V to the rub a dub layout you provided, but not only is it not needed it is likely to cause some sort of issue if you tried to just stick it in there without reworking the design for +-9V operation.

Hi, thanks for the response. I suspected I've been doing this kind of wrong, because it seemed unlikely that these pedals were using more complicated methods to produce the rails voltages. I'm looking at the rub-a-dub schematic right now and I see the voltage divider and what you're talking about, I think (VB is the divided 4.5V voltage, and it's attached to the + pins of the op amps since they're in an inverting configuration).

Here's the op amp part from their circuit:


http://i.imgur.com/mFbgDeJ.png

The part I still don't get is, how does the input signal get biased so it's around 4.5V? Because right now it seems like it's oscillating around 0V (the signal, with respect to the cable jacket). Is that what R1 is somehow doing, and is GND(T) significant? (I've never heard of GND(T) before, what is it?)

The other thing is that I was reading this site: http://tangentsoft.net/elec/vgrounds.html which talks about powering op amps with various methods, and he showed a way that a resistor divider failed, so it kinda scared me off of that. Is what he says not a worry people have? (I guess it isn't if everyone is doing it...)

I should also clarify, I wasn't exactly copying the rub-a-dub circuit I posted; I just followed the schematic in the BTDR manual I posted, so dealt with the power on my own. I would like to do this voltage divider way though, it seems better.

[balkanizeyou]

GND(T) is just an Eagle symbol for ground, which automatically adds a pad for a ground wire to a PCB layout.

Resistor devider method for achieving virtual ground has it's limitations obviously, but it's hard to encounter those limitations in the guitar-pedal small audio amplification, especially with the circuit you posted, which has a gain of -2. Most of the guitar pedals of this world use this configuration with great success. It can sometimes be problematic when using for example ridiculously high gain in one stage, or some careless design, but not in a standard configuration shown here.

The op amp self-biases it's inverting input at 4.5V - remember, an op amp does everything in it's power to keep the voltage at inverting and non-inverting inputs equal by manipulating it's output. In this case, everything the op-amp has to do is set the output DC voltage at 4.5V, which is coupled to the inverting input throught R3 making the op amp happy, because as a result both inputs are at 4.5V.

At which point did you measure the input voltage? If you did it before the C1 capacitor, it's not surprising that it showed 0V, because the capacitor effectively blocks any DC voltage from getting through, and the only reference point there is the ground through the R1 resistor.

[deadlyshart]

Hey thanks for the info.

Yeah it seems like the voltage divider is the way to go, so I'll try that.

I think I get how it works with the op amp getting unipolar bias alone, but I'm having trouble still figuring out how the input signal works into that. I'm probably phrasing my confusion badly because the part I'm missing is what would let me ask it more intelligently...

So let's say my input signal is just a simple sine wave, amplitude 1V, with respect to GND, the outer jacket of the input cable. When I plug it into the circuit, that outer jacket is contacting the chassis, which is also GND for my circuit. So it seems to me like, when I connect this input to the op amp like in the schematic piece above, my sine wave is still going from -1V to +1V, but that's a problem because the rails of the op amp are 0 and 9V, and the op amp can't output stuff outside of the range of its rails voltages... so what am I missing that allows this to work?

thanks!

[Elijah-Baley]

I built the veroboard layout linked in the first post. It has a 250k dual pot for the decay, but I had to replace it with a 100k dual pot because I got oscillation very soon. 100k is perfect for me.

[samhay]

>So let's say my input signal is just a simple sine wave, amplitude 1V, with respect to GND,...

Remember that capacitors block DC, so after the input cap (C1) the DC component of the signal is determined by any DC-coupled (no caps) path(s) to some voltage - i.e. you need to follow the resistors.
Here, R2 acts as a biasing resistor. One end of R2 is connected to the inverting input of the op-amp (as discussed 2 posts above), so this biases the signal at the other end of R2 (where it meets C1) to half supply (4.5-ish V). You now have your 1V signal swinging either side of 4.5V and thus it is kept safely away from the op-amp supply rails.

[deadlyshart]

Ahhhhhhh perfect, okay, this is what I was looking for. This makes sense to me, thank you guys so much.

It's still a little strange to me that it's biased from that op amp node, but it's also giving some input to that node, so it's like those two are "giving" voltage to each other.

Off to add it to my circuit! thanks again!

[balkanizeyou]

that's because in this case you should think about DC operiation and AC (signal) as two slightly different things.

When you assume that there is no input signal in this circuit and you wait a second after turning the power on (actually, much less) for the capacitor to charge, there are no changes is this circuit. The input is at 0V potential, the inputs and output of opamp are at 4.5V, both ends of R2 resistor are at 4.5V.

But - when you couple the 1V p-p AC signal to the circuit, the signal does get through the capacitor (unlike any DC voltage). The 4,5V DC on the "right" end of C1 is added to the 1V AC signal, and you get a signal swinging from 3.5V to 5.5V at the junction of C1 and R2.
The other end of R2 (the inverting input) is firmly held at 4.5V, because that's the job of an op-amp. But in order to achieve that, the op-amp has to swing it's output from 2.5V to 6.5V (so you get a 2V p-p output AC signal), because R2 and R3 form a voltage divider.

So for example when the input signal at some moment is -1V, the junction of C1 and R2 is at 3.5V (4.5V-1V). The inverting input is at 4.5V of course, so the output needs to be at 6.5V (because of the voltage divider).