Do audio signals 'see' high LDR resistance as similar to an open circuit?

[kimelopidaer]

Hi DIY.

I am experimenting with photocell switching to toggle between two variable output volumes for a simple booster circuit.
I set the output level on each 100k pot, and use an LDR before and after each pot to block the audio path (using two LEDS to switch either pair on or off).

I was experimenting with this method after trying it mechanically using an SPDT switch and getting 'The POP', which I expected - especially when switching while playing.

I noticed that boss style FET switching doesn't ground the effect input, just blocks the unused path with super high resistance.

I rolled some photocells to try out in place of FETs. Works in practice.
When one pair of LDRS are off (megaohms of resistance) I wonder how the audio signal interacts with it while being routed easily through the LDRs that are on (few hundred ohms resistance). ?

If resistance is high enough, will that path be 'ignored'? (no tone loss)
Or is that possible without mechanically breaking the path or grounding it?

Peace and thanks
K

[amptramp]

If you want to cut the signal off with an LDR, the impedance it is working into should be low.  That way, you get several megohms in the signal path from the LDR to work into a low impedance and the voltage divider is several megohms into somewhere near zero ohms.  A suitable zero ohm input would be the summing junction of an inverting op amp stage.  A 100K pot  and a 1.9 megohm LDR would give you an attenuation of 20:1 which would leave the "cutoff" signal quite audible.  Boss FET switching works because while one signal is cut off by several megohms of JFET off resistance, the other signal is fed in by a low-impedance source such as an op amp output or an emitter follower.  Thus, the voltage divider ratio is set by the output impedance of the active source.

One method of getting a drop in sound level is to have one LDR in series with the signal at high resistance and another LDR across the output turned on to a few kilohms, giving attenuation of up to 1000:1.  Pots are not low impedance except at the grounded ends and signals will interact because of the voltage divider effect.

[R.G.]

To amplify (sorry!  ) on AT's answer: it depends on the resistances, perhaps especially the input resistance the LDR(s) feed, and any stray capacitances.

Your question "if resistance is high enough, will that path be 'ignored'?" is pertinent. The answer to that is undoubtedly yes; the problem is now how do you say what "high enough" is as a number?

It depends on the internal resistances of what drives the pots, and what loads the output of the whole mess. If what drives it is a guitar pickup, which is primarily inductive up to about 6-7kHz, where its internal impedance is something like 100K inductive, the 100K pots themselves are a significant load, and there will be some "tone loss"; actually, there will be a tone change, as the loading of the pots on the guitar will affect treble more than bass. This is quite independent of what the LDRs are doing. If the guitar is buffered, so what feeds the LDRs and pot is maybe 100 ohms resistive, there will be no selective treble loss because of the loading. This is, in fact what buffers are for.

If it's for a booster circuit as you mention, the 100K pots by themselves will load the booster. If it's an opamp, there's probably no loss. If it's a single transistor or FET with a collector/drain resistor, the pot itself may load the output, but will cause only a signal level loss, not a selective loss of treble.

If your LDRs change between, say, 2K and 2M, which is a pretty decent range, you can calculate what happens by replacing the LDR with the estimated 2K or 2M values depending on what it on and off, and solve the network for how much leaks through or not.

Say you're trying to turn an input off by turning off the input LDR. The signal runs through 2M of LDR and 100K of pot. The pot will have a signal voltage equal to 100K/(2M+100k) or about 0.047 of what the input voltage is. Sounds like a lot of loss, yes? But that's only 26.4db down. You'll still hear it with the pot turned up. But there's that output LDR too, right?

If it's going into a 1M input amp, and the 100K pot is full up, the additional 2M LDR and 1M amp input only attenuate the signal by another 2/3, so overall, the signal into the amp input is -36db down. That's better, and the 100K pot can turn it down to essentially zero signal fed through, but worst case, you'll still hear it.

For "high enough", you probably need LDRs with 10M or more off resistance, or some kind of signal shunt to pull the signal down, as AT says.

The procedure is to sub in the on and off resistances of the LDRs and calculate what the signal attenuation is.

[kimelopidaer]

I see two replies already while i was fixing up a diagram of what i put together.
I am going to try to absorb those suggestions.

"...attenuation of up to 1000:1. "
So would the LDR tuned to a few kilohms be acting as a shunt?

Much appreciated,
K

[kimelopidaer]

Okay,
I slept on those comments
Thankyou sincerely for the depth of your responses.

My LDRs have a resistance that is too high for my multimeter. They measure in the tens of megaohms before the meter gives the OL sign. I assume that I am measuring properly.

At Very high amp volumes, the 'cutoff' signal is barely audible, but it is there nonetheless. I tested by cutting the volume entirely on the 'cutoff' pot and then switching to it.

Question: by shunting the unwanted signal to ground, can I achieve full attenuation? I was thinking it would be similar to grounding the effect send when performing a true bypass. I did this once to kill oscillations that were coming through in bypass mode.

Thankyou
K

[crane]

you can also take a look at some amp schematics that have LDRs in their switching circuitry (eg dual recto). But remember to keep track on which side of the circuit is low impedance and which is high.

[PRR]

> getting 'The POP'

There's at least two POPs:

1) if you switch live audio, even same signal at two gain settings, the transistion is abrupt, "POP". Slower switching _may_ be less offensive. (However LDRs at 100K impedance are already fairly slow.)

2) if you have stray DC on the switching, it will POP. I see your drawing has a cap on one side, but check for DC in the switching anyway.

[kimelopidaer]

Thankyou for pointing out the importance of impedance to the switching.
I see how the buffer in the Boss switching circuit feeds the JFet switch with a low impedance signal...

The second buffer in the Boss scheme, what function does it serve? (besides 'righting' the signal after it was reversed coming out of the first buffer)
What might the effect be if it were omitted?

Regards
K

[kimelopidaer]

Hi DIY,

Buffers ( emitter follower, source follower)
Do not invert phase. I believe that this is correct and I was assuming the opposite in my last post.

Please chime in if this statement is not suitable.
K

[ashcat_lt]

Couldn't you get a whole lot more attenuation if you used the LDR pairs in the legs of a voltage divider?  One would be on while the other is off, and vice versa, and you should get pretty good.  What's the full on resistance of the LDR?

Also, these are the same signal, no?  Do you actually envisage using this thing as a kill switch?  Why else would it matter if there's a little bit of spill from one side to the other?  As long as you're setting two different non-zero volumes with these pots, I'd think the little bit of bleed would just be figured in by ear is a you adjust the individual controls.  Do you really need better than 36db difference between the two?

I could see how this would be a problem if the two signals were significantly different - like if one was heavily distorted or something - but these signals exactly the same!

[kimelopidaer]

Hi,
Thanks for your questions.
Yes, we'll I actually thought I might use the other channel as a mute of sorts. I can't see it being used all the time but certainly I was envisioning the potential. It really only developed as an issue when I realized that at really high amp volumes I couldn't completely kill the signal coming through the "on" LDRs, even with the volume pot turned to minimum.



I am wondering at this time if the culprit might be crosstalk. It's all still on the breadboard at this stage.

LDRs are showing 1k ohms at full brightness.
Let's say I deal with just one channel at a time....at full brightness there should only be 1k ohms before the pot (lug 3) and 1k ohms after the pot (lug2)

Using this scheme, with lug 1 properly grounded, is it reasonable to assume I should be able to achieve a complete cut of the signal as per normal volume pots?

Also, please describe this scheme for using LDRs on the legs of a voltage divider...

Appreciate everything,
K

[ashcat_lt]

You'll only ever get complete silence out of this scheme when both volume controls are all the way down.  That's a period.  To get silence, your LDRs need to either go all the way to 0 or all the way to infinity.  Anything else will fail.  The signal will not ignore any valid path it sees.  No matter how steep the climb, at least a few electrons will struggle their way up.  The signal will take every path it can find to complete the circuit unless it finds one path with absolutely no resistance.

I spiced a sim of this.  I used 2K minimum resistance and 10M max into a 1M load.  Got almost 70db attenuation the way you've got it now.  With the LDRs arranged as a voltage divider I got just about 80db.  With 1K min and max greater than 10M results will be better.  The signal at this point should be down in the noise floor.  Unfortunately our brains are pretty darn good at finding coherent signals amongst noise.  Well, it's a trait which has allowed us to survive and evolve to the point where we can discuss this topic, rather than going extinct somewhere on the African plains, but it's working against you right now.

The voltage divider:  (the following happens twice, once on either side of the split)
Output from the cap goes to LDR1.  Connect LDR2 and the top of the pot to the other end.  The other end of LDR2 goes to ground.  The pot is wired directly to the output.  Switching will have these two LDRs on alternately - when one is on the other is off.  And of course this will be the opposite on the other side.

It sims out to work better than your series resistances, but still doesn't get you silence.

[ashcat_lt]

I honestly can't imagine why the physical switching pop here would be any worse than in other circuits.  Did you try a pull down resistor after the coupling cap?

At this point you've got about as many parts as it would take to implement an FET switching arrangement, which should get you much closer to what you're shooting for.

[kimelopidaer]

hello,
really thanks for the visualization on the voltage divider concept.

although i got into this method of switching to remove POPS from the switching,
I dont find the pop to be an issue with the photocell switching -

what i currently find more challenging is this seeming inability to completely mute either channel using that channel's volume pot.

it switches alright for me, certainly quieter than breaking the circuit mechanically using a dpdt switch. (although this does eliminate my bleed problem)
as you mentioned earlier about using it as a kill switch... well i suppose the way i look at it is that an ability to turn one signal into a mute would be desirable in the range of possible values i could set for that pot. all the way from zero volume (mute switch if pressed) to full volume.  

thanks again,
K

EDIT: as for FET switching, still learning about that one. Was under the assumption that this type of switching was best at certain (lower) signal levels, to avoid distortion. I started a new topic with this is mind.

[ashcat_lt]

But have you tried the mechanical switching with some largish resistor as a pulldown?  You wont be able to get true kill switch functionality using real world LDRs.