Small Stone LFO's P2P Current?

Started by ExpAnonColin, July 30, 2004, 01:45:26 PM

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ExpAnonColin

Am I missing something or can someone rattle a number off the top of their head?   I'd measure it with my multimeter, but my multimeter has trouble measuring A's, and I'd rather not cut traces or desolder anything yet.

-Colin


ExpAnonColin

Quote from: Jason Stouthttp://diystompboxes.com/sboxforum/viewtopic.php?t=23614&highlight=stone+lfo
Right, I followed that thread, but the P2P current was never mentioned, just the P2P voltage... unless I=V/R and the volts are 1.2 to 3.6 or 4.2, and the resistor mentioned was 270k...  4.44 *10^-6 to 1.3*10^-5... or 4.44 * 10^-3 mA to 1.3*10^-2 mA.  That doesn't make any sense.

-Colin

ExpAnonColin

Oh wait!  It's a 100ohm resistor... so the P2P Current =
12 -> 36
or
12 -> 42
mA's

Right?

-Colin

gez

Quote from: anonymousexperimentalistOh wait!  It's a 100ohm resistor... so the P2P Current =
12 -> 36
or
12 -> 42
mA's

Right?

-Colin

I still don't know how the single amp version works, but you can bet the output current isn't that high!  The max these chips can take for bias current is 2mA (from memory, I might be slightly off here) so the amps used for the phase stages would fry.

The figures given by PT must be for the dual amp version.  I don't have a schematic, but the Ross phaser is dual amp and is pretty intuitive to understand (and easy to work on).  I scoped its output and it goes from 2 diode drops (roughly 1.2V) to about 4.6V.  So that's 3.4V divided by 10k and split between the four phase stages.  I'll leave you to do the number crunching... :)
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter

ExpAnonColin

Well, he does also say that there are series resistors of 12k there...  But then we would be dividing the voltage by 12000, which is 120 times more than 100... meaning that it would be modulating from .1ma to .3ma or .1ma to .38 or so mA.  Is it really that low?

-Colin

ExpAnonColin

To answer my own question, the datasheet uses 500uA as the test rating, so I guess not.  Can someone confirm these values?

-Colin

puretube

#7

ExpAnonColin

Interesting info, even though it wasn't answering my Q.  Thanks PT.

-Colin

michael_krell

I took the voltage across my 100 ohm resistor using my scope and i found that i had a p2p voltage of 50mv. so the peek to peek current is .5mA which sounds pretty close to what someone said before.

puretube

#10

gez

Thanks for that Michael.

PT, I've messed around with the Hyper-triangle in the past but used a diode shaping network to do this and it was for a wah circuit.  The wave form was flipped so that it resembled a positive going FWrectified sine - works opposite to the phaser in that the concentration is on the higher part of the sweep, which sounds more like a pedalled wah accenting the beat (well, I think so).  The OTA way is a lot simpler and I'm starting to become a convert!  :)

I've also messed around with the 2 amp OTA triangle/square oscillator you mentioned (from the LM13700 data sheet) before and the other day I played around with the bias of a LFO version I once did (makes a great VC LFO).  Just shifting the bias down (and feeding the amp bias input from its output) compresses the bottom of the sweep and you get a pretty good hyper-triangle.  The Ross version is a slight variation of this circuit and you get exactly the same result from both circuits.
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter

michael_krell

Your welcome. I actually need some help with mine. I have a good LFO signal but I get no sweep. When I drive a 440 hertz wave to the input I get output on all four op amps. The output looks like the wave is dancing arond and it gets faster as I increase the rate. THis looks like its working When I measure the ouput of the pedal I get the same signal as the input with a 75% drop in amplitude.

I would appreciate if someone could help me diagnose this problem.

gez

Would probably be an idea to connect the input buffer's output to each phase stage in turn (taking the output of each stage in turn to your amp) and trying a guitar signal thru it.  You should hear slight vibrato.  This will allow you to check if all the stages are working ok.  If they are, then check the output connections.

Is this something you've breadboarded or are you trying to revive a dead one/get a kit working?  If it's on a breadboard there are all sorts of possibilities.
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter

michael_krell

This is an old broken one that i am trying to get working.

DO you mean the output (pin 6) of each of the 4 opamps?

also another question. Is the fifth IC (the one at the top of the board) the LFO? How does the LFO work. I am pretty sure mine is working the way it is supposed to.

gez

Ah, as your circuit is not on a breadboard this makes life slightly harder.  Hold on and I'll post in a couple of hours time (busy at the moment).
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter

puretube

#16

gez

Damn, pipped to the post!  :)

Follow PT's advice.  Because this isn’t on a breadboard, what I suggested earlier is too fiddly.  

I'm assuming this is your version as you mentioned the 100R resistor:

http://www.montagar.com/~patj/smlstone.gif

If you want to test the phase stages aurally then take a large cap (1u or bigger) from pin 6 (yes, this is the output) of the first stage to your amp. If you hear vibrato all is well.  Disconnect the cap and try the same from the output of stage 2, and so on.  

I’d have preferred to test each stage in isolation because if one stage isn’t working right but still passing signal, you might mistakenly think it’s ok as you’re hearing vibrato from the stage before.  For this reason it’s important that you hear an increase in vibrato as you go down the chain of phase stages.  There should be no change in volume as you do this.  

If your phase stages are ok, suspect the circuitry used for mixing the signal at the output (as PT pointed out).

QuoteAlso another question. Is the fifth IC (the one at the top of the board) the LFO? How does the LFO work. I am pretty sure mine is working the way it is supposed to

It's the one at the bottom of the page in the above schematic with its pin 6 connected to the 100R (I don't have a SS so I can't tell you the position on the board).
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter

michael_krell

i do have a scope and I get a shift from left to right that changes speeds when the rate knob is changed. what is the signal supposed to look like at the output, at what point do the wet and dry signals mix? how do I adjust the leve because it looks like im getting all dry and no wet.

gez

Quote from: michael_krelli do have a scope and I get a shift from left to right that changes speeds when the rate knob is changed

I take it that this reading is from pin 6 of the last phase stage?

Check everything after this.  There's a 27k leading to a .1u then there's a 470k pulldown resistor.  The dry signal is coming via that 30k from the input buffer.  The junction of this 30k and the 27k coming off pin 6 of the last stage is where the mixing is done (it's passive).

Follow both chains, dry & wet, and see if you can find where the breakdown is.
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter