LDR based tremolo - depth control question

Started by tss, January 10, 2013, 07:02:34 AM

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tss

I am looking at the Anderton tremolo like shown in this schematic:


I have constructed the circuit and used my own "home made" LDR with an ON resistance of 350 Ohm and OFF resistance of about 10M (hard to measure exactly). Looking at the schematic, R3 & R4 are in in parallel with the LDR.

With R4=0 Ohm & LDR ON (350 Ohm) Rtotal=325 Ohm
With R4=25K & LDR ON Rtotal=345 Ohm

With R4=0 Ohm & LDR OFF (10M) Rtotal=4.7K
With R4=25K & LDR OFF Rtotal=29.6K

So generally speaking the ON resistance does not change much anyway but the OFF resistance changes by a factor of 6.3. From what I can see there is almost no change at all when altering R4. I know people have built this circuit and are probably happy with it, so what is missing here?


~arph

It alters the gain of the following inverting stage.

At minimum the gain is : 4k7 / (325 + 1k) = 3.5
At max the gain is : 4k7 / (29.6k + 1k) = 0.15

So that stage sweeps between those two gains at max depth.

tss

Quote from: ~arph on January 10, 2013, 08:05:36 AM
It alters the gain of the following inverting stage.

At minimum the gain is : 4k7 / (325 + 1k) = 3.5
At max the gain is : 4k7 / (29.6k + 1k) = 0.15

So that stage sweeps between those two gains at max depth.

Oh off course! It sets the gain by Rp/Ri + 1.
But this brings another question, why only 25K, why not a larger VR? At short there will be no difference but at max resistance the depth can be adjusted a bit further.

So when R4=0 Ohms
LDR ON: 4.7k / [(4.7k || 350) + 1k] = 3.54
LDR OFF 4.7k / [(4.7k || 10M) + 1k] = 3.48

when R4=25k
LDR ON: 4.7k / [(29.7k || 350) + 1k] = 3.49
LDR OFF 4.7k / [(29.7k || 10M) + 1k] = 0.15

when R4=100k
LDR ON: 4.7k / [(104.7k || 350) + 1k] = 3.48
LDR OFF 4.7k / [(104.7k || 10M) + 1k] = 0.05

when R4=1M
LDR ON: 4.7k / [(1004.7K || 350) + 1k] = 3.48
LDR OFF 4.7k / [(1004.7k || 10M) + 1k] = 0.005 (divide the signal by a factor of 200 for a chopper tremolo)

~arph

#3
To avoid clipping.

Most opamps are unable to swing to the supply rails. Humbuckers can be a volt pp. times 3.5 would mean a swing between 1.5 and 8 volts

tss

Quote from: ~arph on January 10, 2013, 08:51:55 AM
To avoid clipping.

Most opamps are unable to swing to the supply rails. Humbuckers can be a volt pp. times 3.5 would mean a swing between 1.5 and 8 volts

In what setting it would clip? The max gain is not being affected.

~arph

depends on the opamp.. it can clip at the 0v rail too.

tss

#6
Quote from: ~arph on January 10, 2013, 09:01:13 AM
depends on the opamp.. it can clip at the 0v rail too.

Possible solution: adding a bit of DC offset at the input of the op amp? and run it on 15V or something like that.

tss

I'm just looking at a data sheet of a LM158 op amp and the max delta from the negative supply rail is 20mV and the max delta from the positive rail is 3V. If the signal is fed into an op amp's inverting input and you have Vcc/2 going into the non-inverting input (DC bias), with 15V supply you could have a max amplitude of 4.5V. So you can set the gain to 2 with no worries or a bit more with a higher Vcc.

PRR

> almost no change at all when altering R4

First get a good path with the LFO not-wobbling but sitting at the average DC voltage. Lift one end of R11, put the switch to out of U2A. Verify 4V-5V here.

That's the "middle" of the LDR's wobble (now non-wobble).

Shim _R12_ for unity gain or a trace more. You are biasing the LED/LDR for about 3.7K, so (with 1K and 4.7K) the center gain is nominal.

From what you are saying, 4V in 2K2 into your LED+LDR is giving a much lower resistance, far lower than 25K. So you probably want to try R12 at 3.3K, 4K7, or 10K.

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