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 Forum index » DIY Hardware and Software » Lunettas - circuits inspired by Stanley Lunetta
Line level
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CHRISKELLY



Joined: Apr 08, 2018
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PostPosted: Mon Aug 13, 2018 2:44 am    Post subject: Line level
Subject description: Total noob question
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Hi guys

For a line level output do I just need a voltage divider after my main volume pot to achieve the 1V approx?

Bit embarrased to ask Embarassed

Cheers
Chris
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PHOBoS



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PostPosted: Mon Aug 13, 2018 5:17 am    Post subject: Reply with quote  Mark this post and the followings unread

You can make a voltage divider by putting a resistor in series with the potentiometer. Adding an extra buffer is not a bad practise either,
you could use an opamp for this or a transistor. And last but not least a capacitor to temove any DC offset.

Have a look at the line level output of the NANDulator:

Posted Image, might have been reduced in size. Click Image to view fullscreen.

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CHRISKELLY



Joined: Apr 08, 2018
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PostPosted: Mon Aug 13, 2018 11:23 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks again Phobos

Just out of curiosity - why did you choose 2*470R against 10k? And for the maths is it 940/10,940 or 470/10,470?
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PHOBoS



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PostPosted: Mon Aug 13, 2018 1:07 pm    Post subject: Reply with quote  Mark this post and the followings unread

CHRISKELLY wrote:
Just out of curiosity - why did you choose 2*470R against 10k?

ok let's see, the 10K resistor (R8 ) is used to reference the capacitor to GND and it discharges it when nothing is connected.
It's not really needed as whatever you connected it to should have some resistance to GND but it should prevent loud pops
when you plug the jack in.

The 2 470 ohm resistors are added to be able to use both stereo and mono jacks. The signal is of course mono but the socket
is stereo, so if you connected a stereo jack to it you wil get the signal on both channels. However if you would plug in a mono
jack the right channel will get shorted to GND. Shorting would be bad of course so I added the resistors in series. The value
wasn't really calculated but if it's too low it will draw too much current, too high and the signal could get attenuated too much.
470 seemed like a good value.

Quote:
And for the maths is it 940/10,940 or 470/10,470?

It depends if you plug in a mono or stereo jack and on the impedance of what you connect it to. With a mono jack one of the
470 resistors is in parallel with the 10K so that's about 450 ohms. I only just realized that this means it will change the cutoff
frequency of the high-pass filter that's made by the capacitor and resistor(s). 35Hz* doesn't seem too bad though. Laughing

* with nothing connected to the output

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CHRISKELLY



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PostPosted: Tue Aug 14, 2018 12:20 am    Post subject: Reply with quote  Mark this post and the followings unread

The more I learn, the less I realise I don't understand!

So the transistor and the 1k act as the voltage divider to reduce the 9-12V down to line level?
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Steveg



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PostPosted: Tue Aug 14, 2018 3:42 am    Post subject: Reply with quote  Mark this post and the followings unread

No, that is just a voltage follower to act as a buffer between the CMOS and the audio output. It keeps the CMOS from potentially being overloaded.
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CHRISKELLY



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PostPosted: Tue Aug 14, 2018 4:15 am    Post subject: Reply with quote  Mark this post and the followings unread

So is P6 acting as the voltage divider?
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Steveg



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PostPosted: Tue Aug 14, 2018 5:15 am    Post subject: Reply with quote  Mark this post and the followings unread

Yes, P6 is your master volume control. Q1 then acts as a buffer and keeps a constant impedance on the output.
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CHRISKELLY



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PostPosted: Tue Aug 14, 2018 5:49 am    Post subject: Reply with quote  Mark this post and the followings unread

Cheers Steve. I think what I'm not grasping is how line level is achieved Confused

I thought line level was about 1V so i assumed (wrongly) that you just need a divider down from whatever the peak system voltage is.

It sounds like i need to do more research Laughing
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Steveg



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PostPosted: Tue Aug 14, 2018 6:45 am    Post subject: Reply with quote  Mark this post and the followings unread

Well you haven't been wrong so far. P6 IS a divider. You can use use a fixed divider if your input is square waves of a known voltage. But you need a different setting if your input is a sine wave because the average voltage is different even if the peak voltage is the same. Then if you are putting your signal through a filter or VCA what comes out could be anything so you use an adjustable voltage divider which is what P6 is. But if you have P6 on its own the ability to deliver current to the DC blocking cap C6 varies with the setting and the P6/C6 combination becomes a high pass filter with a cutoff that varies with the volume setting (technically it is C6/R8 that is the filter - R6 and P6 change the effective value of C6) . So then you put Q1/R7 in the circuit to decouple C6 from P6 and provide a constant drive to the headphones. Does that make it any clearer?
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CHRISKELLY



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PostPosted: Tue Aug 14, 2018 9:54 am    Post subject: Reply with quote  Mark this post and the followings unread

I think I get it now:

- Phobos used 470R against a 10k pot to get up to approx 0.4V at output (470/10470)*9V = 0.4V ish
- the transistor is a buffer to prevent the master volume acting with the DC removing capacitor to create a high pass filter
- my circuit is square waves so peak voltage can be measured to help set line level voltage of 0.447V (not 1V as i said earlier Embarassed )

Thankyou guys for helping.
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PHOBoS



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PostPosted: Tue Aug 14, 2018 10:24 am    Post subject: Reply with quote  Mark this post and the followings unread

CHRISKELLY wrote:
So is P6 acting as the voltage divider?

yes, but there is also a voltage divider made by the potentiometer in combination with R6. Since both are 10K this already
reduces the level by 50%. This is what I meant when I said:
PHOBoS wrote:
You can make a voltage divider by putting a resistor in series with the potentiometer


CHRISKELLY wrote:
I thought line level was about 1V so i assumed (wrongly) that you just need a divider down from whatever the peak system voltage is.

Not all line levels are created equal pig There is some good information about it on the wikipedia page.
The output section of the NANDulator is made for pro line level which is about 3.5Vpp. The recommended supply voltage is
between 9..12V. There is a voltage drop of about 0.7V caused by protection diode D5, so with 9V supply that leaves about
8.3V. The combination of R6 + P6 reduces this to a maximum level of about 4.15V. Considering that there might be some
small losses caused by the CMOS chips, variations in supply voltage and the rest of the output circuitry it should still be
high enough for pro line level. Of course this level can be attenuated with P6 if needed.

Having said all that is said, you can just use a potentiometer, although I would place a resistor in series otherwise you will
probably only be able to use use a very small section of the pot before the level becomes too high. But the rest of the
circuitry might prevent some problems in the future.

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PHOBoS



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PostPosted: Tue Aug 14, 2018 10:45 am    Post subject: Reply with quote  Mark this post and the followings unread

CHRISKELLY wrote:
- Phobos used 470R against a 10k pot to get up to approx 0.4V at output (470/10470)*9V = 0.4V ish
- the transistor is a buffer to prevent the master volume acting with the DC removing capacitor to create a high pass filter
- my circuit is square waves so peak voltage can be measured to help set line level voltage of 0.447V (not 1V as i said earlier Embarassed )


- the 470R value is arbitrary and if you just have a single mono output you don't need those resistors at all. (although it won't hurt as extra protection)
- The transistor is indeed a buffer but that's all it does. It has a high impedance (meaning it doesn't draw much current) and it has a low output
impedance (meaning it can supply a lot of current). The high pass filter is not intended but just something that is created whenever you have a
capacitor in series followed by a resistance to GND. This is why you shouldn't make the value of the capacitor too small otherwise you could loose
some of the low frequencies.
- it depends if you have a squarewave with a 50/50 dutcycle or if the pulsewidth can be adusted as that will change the RMS voltage.

In the end it depends on what you connect it to. Adjusting by ear and comparing it with another signal can often work better and faster than
calculating things. This is also because there is something as perceived loudness but let me not get into that. Laughing

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Last edited by PHOBoS on Tue Aug 14, 2018 3:16 pm; edited 1 time in total
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CHRISKELLY



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PostPosted: Tue Aug 14, 2018 2:42 pm    Post subject: Reply with quote  Mark this post and the followings unread

Thanks for taking the time to explain it all, I appreciate it!

Getting near to finishing my project so hopefully soon I'll have something to show you guys.
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Richarius



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PostPosted: Thu Aug 16, 2018 5:19 am    Post subject: Reply with quote  Mark this post and the followings unread

Steveg wrote:
Well you haven't been wrong so far. P6 IS a divider. You can use use a fixed divider if your input is square waves of a known voltage. But you need a different setting if your input is a sine wave because the average voltage is different even if the peak voltage is the same. Then if you are putting your signal through a filter or VCA what comes out could be anything so you use an adjustable voltage divider which is what P6 is. But if you have P6 on its own the ability to deliver current to the DC blocking cap C6 varies with the setting and the P6/C6 combination becomes a high pass filter with a cutoff that varies with the volume setting (technically it is C6/R8 that is the filter - R6 and P6 change the effective value of C6) . So then you put Q1/R7 in the circuit to decouple C6 from P6 and provide a constant drive to the headphones. Does that make it any clearer?


Clear as a button for me, without a schematic in front of me to follow. I've been self-taught electronics wise since the early 90's. Still so many things I don't completely understand. Lol.

You're one of the rarer people Steve, that I am able to understand instructions from, without any troubles. Appreciated. I've always communicated a 'little differently' from the seeming norm. (Prolly why I could rarely tolerate school.) So when I am able to read someone's instructions and understand them clearly - it's greatly appreciated. Smile

I'm glad that I've followed this entire thread as well.

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Steveg



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PostPosted: Thu Aug 16, 2018 6:04 am    Post subject: Reply with quote  Mark this post and the followings unread

Embarassed Smile
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