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ChrisR
Joined: Feb 22, 2009
Posts: 24
Location: Dark side of the room
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 Posted: Sun Feb 22, 2009 8:58 am Post subject:
 Best practises for scaling voltages and buffering outputs?
Subject description: First post from a beginner, so beware - stupid questions ahead! |
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Hi there,
i have been reading this great forum for some time now and thought i might just as well register and participate (read: ask lots of stupid questions). 
I got some "experience" by playing around with an electronics kit as a young boy (the sine oscillator was fun!), and had a brief flirtation with circuit-bending a toy keyboard about 10 years ago. Apart from that, i'm clueless...
Lately i've decided to finally put my dream of having a modular analog synth into reality. Since i already own a micromoog (which, with all it's ins and outs, would make a nice control center for a modular) i figured i'd start by building a simple and small step sequencer for it. After some initial problems related to my cluelessness i got a basic baby10-like sequencer running on breadboard, clocked from a 555 and powered from a cheapo 13,8V walwart.
Here's where the problems start:
The CV-bus outputs about 5-13V. Ideally i want it to output -10V to +10V. I guess i'd need a bipolar psu (+-15V) to also get a negative voltage out of it, right? Would it be possible to power a circuit like fonik's baby 10 with a bipolar supply?
Ok, back to the main problem, what does one need to do in order to scale a cv to the desired range? E.g. how would i scale the 5-13V to 0-10V, or any range for that matter?
Second problem (last one for today ):
The baby 10 cv outputs seem to be unbuffered, so when i connect more than one device to a cv out (using a multiple), the voltage drops. this can be circumvented with a buffer on the output, right? or am i on the wrong track here? Is there a standard solution which could be applied in situations like this (one output to multiple inputs without any "loss" of signal)?
Any help is much appreciated as i'm making my baby steps in synth diy (with the baby 10 - how fitting ).
If all goes well then the next step is probably to have a closer look at some oscillators (based on some Thomas Henry designs for the ne566, xr2206 or xr8038). Or maybe i'll play around with a 40106 lunetta-style. Or all of it - i already feel the addictiveness of synth diy.
Chris |
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cthulu
Joined: Feb 07, 2009
Posts: 56
Location: Göteborg, Sweden
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Posted: Sun Feb 22, 2009 11:21 am Post subject:
Re: Best practises for scaling voltages and buffering outputs?
Subject description: First post from a beginner, so beware - stupid questions ahead! |
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| Chris Rogue wrote: | Hi there,
Here's where the problems start:
The CV-bus outputs about 5-13V.
Chris |
Actually you don't have a path to ground from the schematic so it's a bit awkward to measure correctly. So the pot's are 100k. Put a resistor(pull down) from the CV out to ground and measure the output again. The resistor could be something like 1M.
| Quote: | | Ideally i want it to output -10V to +10V. I guess i'd need a bipolar psu (+-15V) to also get a negative voltage out of it, right? Would it be possible to power a circuit like fonik's baby 10 with a bipolar supply? |
You would power the Baby10 from one part of the bipolar like -15 - 0V or 0 to +15V. But first get a proper measurement from the CV out.
| Quote: |
Ok, back to the main problem, what does one need to do in order to scale a cv to the desired range? E.g. how would i scale the 5-13V to 0-10V, or any range for that matter?
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http://www.allaboutcircuits.com/vol_3/chpt_8/index.html
| Quote: | Second problem (last one for today ):
The baby 10 cv outputs seem to be unbuffered, so when i connect more than one device to a cv out (using a multiple), the voltage drops. this can be circumvented with a buffer on the output, right? or am i on the wrong track here? Is there a standard solution which could be applied in situations like this (one output to multiple inputs without any "loss" of signal)? |
Correct, but as you'll need a summing amplifier for the job to convert the level out, you won't need a buffer as the output from the summing amp will be LowZ...
Enjoy the wrrrr...zzzzz....boing!
Oskar |
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fonik
Joined: Jun 07, 2006
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Location: Germany
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Posted: Sun Feb 22, 2009 3:35 pm Post subject:
Re: Best practises for scaling voltages and buffering outputs?
Subject description: First post from a beginner, so beware - stupid questions ahead! |
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| cthulu wrote: | | Quote: | Second problem (last one for today ):
The baby 10 cv outputs seem to be unbuffered, so when i connect more than one device to a cv out (using a multiple), the voltage drops. this can be circumvented with a buffer on the output, right? or am i on the wrong track here? Is there a standard solution which could be applied in situations like this (one output to multiple inputs without any "loss" of signal)? |
Correct, but as you'll need a summing amplifier for the job to convert the level out, you won't need a buffer as the output from the summing amp will be LowZ... |
there will always be some voltage drop, as soon as current passes a resistor, right? this is why i removed the 1k resistor after my MIDI2CV converters outputbuffer (thus sacrificing the protection). before i had a scaling problem as soon as i connected more than one VCO to the converters output. - i am talking about 1V/oct CV here. for modulation the voltage drop wouldn't matter at all.
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matthias
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cthulu
Joined: Feb 07, 2009
Posts: 56
Location: Göteborg, Sweden
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Posted: Mon Feb 23, 2009 1:05 am Post subject:
Re: Best practises for scaling voltages and buffering outputs?
Subject description: First post from a beginner, so beware - stupid questions ahead! |
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| fonik wrote: | | there will always be some voltage drop, as soon as current passes a resistor, right? |
Right.
| Quote: | | this is why i removed the 1k resistor after my MIDI2CV converters outputbuffer (thus sacrificing the protection). before i had a scaling problem as soon as i connected more than one VCO to the converters output. - i am talking about 1V/oct CV here. for modulation the voltage drop wouldn't matter at all. |
Yeah. I had the same problem with my doepfer if I connected more than 2 VCO's. It's probably better to add a buffer instead.
I taught music production to teenagers and that experience has left me wanting protection for things. Really good protection...
Oh. The horror! The horror!... |
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fonik
Joined: Jun 07, 2006
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Posted: Mon Feb 23, 2009 4:57 am Post subject:
Re: Best practises for scaling voltages and buffering outputs?
Subject description: First post from a beginner, so beware - stupid questions ahead! |
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| cthulu wrote: | | fonik wrote: | | there will always be some voltage drop, as soon as current passes a resistor, right? |
Right.
| Quote: | | this is why i removed the 1k resistor after my MIDI2CV converters outputbuffer (thus sacrificing the protection). before i had a scaling problem as soon as i connected more than one VCO to the converters output. - i am talking about 1V/oct CV here. for modulation the voltage drop wouldn't matter at all. |
Yeah. I had the same problem with my doepfer if I connected more than 2 VCO's. It's probably better to add a buffer instead.
I taught music production to teenagers and that experience has left me wanting protection for things. Really good protection...
Oh. The horror! The horror!... |
so what do you want to do? adding a 100% unity gain chain of two inverting mixers? if you wanted to use just a voltage follower there would be no protection!? as soon as you add a 1k at the output you would have the voltage drop again.
is there a workaround?
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matthias
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cthulu
Joined: Feb 07, 2009
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Posted: Mon Feb 23, 2009 6:27 am Post subject:
Re: Best practises for scaling voltages and buffering outputs?
Subject description: First post from a beginner, so beware - stupid questions ahead! |
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| fonik wrote: | so what do you want to do? adding a 100% unity gain chain of two inverting mixers? if you wanted to use just a voltage follower there would be no protection!? as soon as you add a 1k at the output you would have the voltage drop again.
is there a workaround? |
I should perhaps say that I'm more or less self taught on electronics. I took a two year course but it was
more focused on electronic production methods.
As for the output I don't think there is any difference between the voltage follower solution or the chained
inverters except that the inverters are good if you want to modify the construction later with added DC
shift, slew, mixers or whatelse...
Next time I build a modular I would probably do like you. Just skip the resistor. I guess the other option is
to read up on OP-amps with short circuit protected outputs... that would probably be a good idea.
Perhaps you could compromise and put in a lower value resistor on the output. Something like a 100R that
also is dimensioned to whithstand a bit of current. 
However I don't like compromises... 
oskar |
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blue hell
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| Posted: Mon Feb 23, 2009 7:52 am Post subject:
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What is wrong with getting the feedback for the output driver from behind the resistor instead of before?
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Jan
also .. could someone please turn down the thermostat a bit.
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fonik
Joined: Jun 07, 2006
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| Posted: Mon Feb 23, 2009 7:59 am Post subject:
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| Blue Hell wrote: | | What is wrong with getting the feedback for the output driver from behind the resistor instead of before? |
nothing in know. this would be a good solution for an inverting opamp solution, wouldn't it?
a simple voltage follower as buffer does not provide any protection, does it? the output of the opamp is connected directly to the positive input, which is connected to the circuitry i might want to be protected. so there is no resistor between input and the sensitive circuitry...
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matthias
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blue hell
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| Posted: Mon Feb 23, 2009 8:44 am Post subject:
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| fonik wrote: | | so there is no resistor between input and the sensitive circuitry... |
But a resistor could be added in that path as well, as the input is high impedance there will be nearly no voltage drop over it.
as in image below I mean.
A no no?
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Jan
also .. could someone please turn down the thermostat a bit.
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cthulu
Joined: Feb 07, 2009
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frijitz
Joined: May 04, 2007
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| Posted: Mon Feb 23, 2009 11:19 am Post subject:
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| Blue Hell wrote: | | What is wrong with getting the feedback for the output driver from behind the resistor instead of before? |
Nothing whatsoever. No idea what those other guys are talking about.
Ian |
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fonik
Joined: Jun 07, 2006
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| Posted: Mon Feb 23, 2009 3:57 pm Post subject:
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| frijitz wrote: | | Blue Hell wrote: | | What is wrong with getting the feedback for the output driver from behind the resistor instead of before? |
Nothing whatsoever. No idea what those other guys are talking about. |
WE other guys are talking about things we don't have a clue of, while YOU guys talking about things you understand. so be soft on me.  (not too soft, sometimes i just need a kick in the arse)
nevertheless, i hopefully will learn from what you guys are telling me. so thank you for sharing. BTW i can't remember to have seen such a configuration in any circuit before (the one jan suggests). not that i haven't seen it, i just can't remember.
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blue hell
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| Posted: Mon Feb 23, 2009 4:22 pm Post subject:
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The idea is that you can put a lot of stuff into the output path of the opamp, but when you take the feedback from after that element (and obtain your output from there, in this case) the opamp will try to regulate the voltage at that point, its output will not be ok, but the point you take the feedback from will. The possible output swing will be less too, of course, nothing for free ...
When you take a close look at exp converters you'll see another example of this idea (be it less straight forward).
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Jan
also .. could someone please turn down the thermostat a bit.
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ChrisR
Joined: Feb 22, 2009
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| Posted: Mon Feb 23, 2009 4:27 pm Post subject:
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| cthulu wrote: | | I guess the other option is to read up on OP-amps with short circuit protected outputs... |
For the output buffer i just used a 741 as a voltage follower. It has short-circuit protection on the output. Found out about that on the page you linked above.
| cthulu wrote: | | Put a resistor(pull down) from the CV out to ground and measure the output again. The resistor could be something like 1M. |
Ok, that sounded like a good idea, so i did it.
Now i measure 0-13V before the 741 and 2-13V after it - don't know where the 2 volts are coming from but it's a step in the right direction i guess. Or maybe i'm just too tired to measure correctly, have to repeat that tomorrow.
| Blue Hell wrote: | | What is wrong with getting the feedback for the output driver from behind the resistor instead of before? |
That's interesting! Would something like that still be needed when using an op-amp with short-circuit protection?
| fonik wrote: | | WE other guys are talking about things we don't have a clue of, while YOU guys talking about things you understand. so be soft on me. |
Ok, i guess that puts me in the kindergarten.
Thanks for the suggestions, seems i have to learn a lot more about op-amps... (and everything else - still have to figure out how to scale the cv). |
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frijitz
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| Posted: Mon Feb 23, 2009 4:46 pm Post subject:
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| fonik wrote: | | WE other guys are talking about things we don't have a clue of, while YOU guys talking about things you understand. |
No offense meant -- I was just saying I couldn't understand what you were trying to say.
These circuits can very simply be analyzed using the opamp golden rules. Try it!
One reason you don't see this technique very often is that it really isn't necessary to add protection to modern opamps, since they can withstand a short circuit at their output for an indefinite time.
Mae West: Men are always trying to protect me. Against what, I have no idea.
Ian |
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blue hell
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| Posted: Mon Feb 23, 2009 5:20 pm Post subject:
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| Chris Rogue wrote: | | Blue Hell wrote: | | What is wrong with getting the feedback for the output driver from behind the resistor instead of before? |
That's interesting! Would something like that still be needed when using an op-amp with short-circuit protection?
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Well I'm glad Ian answered that one 
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fonik
Joined: Jun 07, 2006
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| Posted: Mon Feb 23, 2009 6:56 pm Post subject:
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| frijitz wrote: | | fonik wrote: | | WE other guys are talking about things we don't have a clue of, while YOU guys talking about things you understand. |
No offense meant -- I was just saying I couldn't understand what you were trying to say. |
| Quote: | | These circuits can very simply be analyzed using the opamp golden rules. Try it! |
i will try it, thank you for remembering (this is the kind of kick in the arse i need sometimes to move on). hoping it is as easy as you say
| Quote: | | One reason you don't see this technique very often is that it really isn't necessary to add protection to modern opamps, since they can withstand a short circuit at their output for an indefinite time. |
didn't know that either! so what else is the 1k output impedance needed for?
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matthias
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Sound
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| Posted: Tue Feb 24, 2009 2:40 pm Post subject:
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| fonik wrote: | | frijitz wrote: | | One reason you don't see this technique very often is that it really isn't necessary to add protection to modern opamps, since they can withstand a short circuit at their output for an indefinite time. |
didn't know that either! so what else is the 1k output impedance needed for? |
Yes I would like to know it also.
Maybe it is there to protect the opamp of an accidental output to output patch? |
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frijitz
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| Posted: Tue Feb 24, 2009 5:49 pm Post subject:
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| fonik wrote: | | didn't know that either! so what else is the 1k output impedance needed for? |
Well, first, the 1k needs to be there for opamps that can't take a short circuit, or if you just don't want to ever stress your chips that much.
The other resason -- less discussed -- is so you can combine two signals by simply connecting their outputs together. I've never found that especially useful, but it's there.
Ian |
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Sound
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| Posted: Wed Feb 25, 2009 3:59 am Post subject:
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| And maybe it is also because must be an impedance ratio within an output and the input which receives its signal? |
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blue hell
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| Posted: Wed Feb 25, 2009 4:17 pm Post subject:
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For voltage driven equipment you'd want the voltage to go through unattenuated, this is best accomplished by having an output impedance of 0 Ohms, or an input impedance that is infinite ... so from that point of view the 1 k resistor should not be present (or be compensated for as in the circuit I posted above - effectively it is considerably lowering the output impedance).
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Sound
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| Posted: Thu Oct 15, 2009 9:27 am Post subject:
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Heheh, again with this...
Well I have made some tests with different output circuits as following:
Circuit A: Normal Inverting Amplifier with 1K resistor output.
Circuit B: Inverting Amplifier with the feedback resistor connected after the 1K resistor output.
Circuit C:Inverting Amplifier with a trim resistor as a feedback resistor.
I applied scaled voltage at the inputs and measured the outputs and I write here as a cents of volts of difference within the input.
Input..................A.......B.......C
+0.08 V.............+1....+1.......0
+1.07 V..............-1.......0.......0
+2.05 V..............-2.......0.......0
+3.03 V..............-2.......1......-1
+4.02 V..............-4.......0......-1
+5.00 V..............-4.......0.......0
+5.99 V..............-6.......0......-1
+6.97 V..............-6.......1.......0
+7.96 V..............-8.......0......-1
+8.94 V..............-9.......0......-1
We can see that Circuits B and C keep a relation with the Voltage input.
On Circuit B the input resistor and the feedback resistor were matched at 0.01%. Same circuit with unmatched 1% resistors give me same values +1. It works also because it keeps the relation with the input and, oscillators have offset pots to tune them.
Yes, I have read here that also is possible remove the 1K output resistor but since I'm learning and one of my sources is the Thomas Henry works, I would like to preserve the 1K resistor output and its impedance, like he does.
Because the B option is the more cheap and easy, I would like to know how calculate its impedance.
So, this is, how could I calculate exactly the Impedance output of an inverting amplifier which its feedback resistor is connected after the 1K resistor output?
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blue hell
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| Posted: Thu Oct 15, 2009 10:38 am Post subject:
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| Sound wrote: | | So, this is, how could I calculate exactly the Impedance output of an inverting amplifier which its feedback resistor is connected after the 1K resistor output? |
It is approximately 1 kOhm divided by the open loop gain of the opamp used, or (very near to) zero ohms to say it more compact. This is not a problem BTW but a feature as it is the reason why it tracks so well in your experiments.
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Sound
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| Posted: Thu Oct 15, 2009 2:04 pm Post subject:
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Aha! Thanks Blue! ...close to zero... yes the truth I don't see a problem, I should study a little about impedance. 
Sorry that I insist but, there is no reason to have 1kΩ impedance like all other modules? There would be any improvement if I trim the feedback resistor in order to preserve that 1KΩ impedance like in circuit C? |
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blue hell
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| Posted: Fri Oct 16, 2009 2:53 pm Post subject:
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I don't know why people would want a 1 k output impedance .. maybe someone using that system or designing for it could explain?
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Jan
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