I might be wrong but I've got the datasheet open.
The maximum voltage isnt 18, its +-18(36VDC)
I haven't seen a description of your power supply.
Are you using one or two batteries?
If one, use two to get a split supply.
When using two connect them in series i.e. one's plus to the second's -.
Where the + and the - join is your center tap. now. everything that sais earth or ground connects to this point. Reducing noise we can do later. what i'd do is take a couple of wires an solder them all together at one end. Thats your ground star.
If you dont connect the power supply to other gnds then you'll get a floating ground. In my first experiment that put a hell of a dc offset on the speakers, but signal came through.
The maximum voltage isnt 18, its +-18(36VDC)
I haven't seen a description of your power supply.
Are you using one or two batteries?
If one, use two to get a split supply.
When using two connect them in series i.e. one's plus to the second's -.
Where the + and the - join is your center tap. now. everything that sais earth or ground connects to this point. Reducing noise we can do later. what i'd do is take a couple of wires an solder them all together at one end. Thats your ground star.
If you dont connect the power supply to other gnds then you'll get a floating ground. In my first experiment that put a hell of a dc offset on the speakers, but signal came through.
Lol Wynand, we will get there... for now, we need the gentleman to connect the PSU arrangement he built to the actual power pins and not to V- and NI input of second channel as he has it now..
Provided he didn't use insane gain, 9V should be enough to cover basic functionality testing.... even if at the price of some headroom... but I think you may be right, not using a split supply could pass DC to the output... just put a big electrolitic in series with the output for testing...
Provided he didn't use insane gain, 9V should be enough to cover basic functionality testing.... even if at the price of some headroom... but I think you may be right, not using a split supply could pass DC to the output... just put a big electrolitic in series with the output for testing...
First things first. Still not working 
This is proof of my old saying about trying to make mistakes in small quantities. At a couple bucks each I'm not going to cry over burnt chips.
Anyway, I am using two 9V in series so I rewired the board like this using red/black leads from the battery holders:
(R)--(B)--(G)--(R)--(B)
Voltage measures 16V across the correct pins (duh!) on the IC and the LED now magically lights up. Sound, however, remains elusive, coming through very, very faintly.
As an experiment, I disconnected the batteries and the sound did not change at all, which suggests that the only sound I am getting is coming through the feedback resistor. I had a spare unopened IC so I swapped it in just in case I really cooked the previous one and you guessed it, no difference.
WTF?
Thanks again all for the help!
This is proof of my old saying about trying to make mistakes in small quantities. At a couple bucks each I'm not going to cry over burnt chips.
Anyway, I am using two 9V in series so I rewired the board like this using red/black leads from the battery holders:
(R)--(B)--(G)--(R)--(B)
Voltage measures 16V across the correct pins (duh!) on the IC and the LED now magically lights up. Sound, however, remains elusive, coming through very, very faintly.
As an experiment, I disconnected the batteries and the sound did not change at all, which suggests that the only sound I am getting is coming through the feedback resistor. I had a spare unopened IC so I swapped it in just in case I really cooked the previous one and you guessed it, no difference.
WTF?
Thanks again all for the help!
Power first.
Ok. Maybe you should strip everything off the pcb.
Maybe you can wire the circuit without a pcb. It's an easy circuit.
First take the batteries and connect them as we discussed. One positive to one negative, leaving the unconnected positive as your +9 and the unconnected negative as the -9 and....the connected ones as 0V or GND.
I usually make the + red, gnd blue and - black. I'm not sure if it follows standard, but I think green and yellow are usually left for signals.
Following my colours, connect the red wire to pin 8 and the black to pin 4.
pin 4 is on the bottom left if the notch or mark is on top and the pins facing down.
pi8 is then top right.
right we've got power. Measure. bewteen pins 8 and 4 for 18 volt.
leave the led for now, it draws more power than we want to now.
Ok. Maybe you should strip everything off the pcb.
Maybe you can wire the circuit without a pcb. It's an easy circuit.
First take the batteries and connect them as we discussed. One positive to one negative, leaving the unconnected positive as your +9 and the unconnected negative as the -9 and....the connected ones as 0V or GND.
I usually make the + red, gnd blue and - black. I'm not sure if it follows standard, but I think green and yellow are usually left for signals.
Following my colours, connect the red wire to pin 8 and the black to pin 4.
pin 4 is on the bottom left if the notch or mark is on top and the pins facing down.
pi8 is then top right.
right we've got power. Measure. bewteen pins 8 and 4 for 18 volt.
leave the led for now, it draws more power than we want to now.
Testing opamp + power
Connect the opamp as shown. This is a buffer and wont give any gain, but the volume should be the same.
Short points 1 and 2. Pin 1 is the output and then pin 3 is the input.
remember to connect the gnd from the incomming signal to the battery ground and the ground for the output.
If everything is correct you should have sound on the output and have proved your batteries is connected the right way.
If I've made a mistake please correct me before he builds this everyone!!!
Connect the opamp as shown. This is a buffer and wont give any gain, but the volume should be the same.
Short points 1 and 2. Pin 1 is the output and then pin 3 is the input.
remember to connect the gnd from the incomming signal to the battery ground and the ground for the output.
If everything is correct you should have sound on the output and have proved your batteries is connected the right way.
If I've made a mistake please correct me before he builds this everyone!!!
Attachments
![buf3[1].gif](/community/data/attachments/64/64931-7c9582f91547dd28f40cf96ebc8db36e.jpg)
changing to inverting circuit
Remove the battaries from the circuit.
Now remove link between pins 1 and 2, but leave the output wire on.
Remove the input signal wire. solder the input resistor to the input signal wire and then solder it to pin 2.
next solder the feedback resistior between pins 1 and 2.
Solder a wire from pin 3 ( + input or non-inverting input) to the group of ground wires.
You should have a working circuit if you were carefull.
Couple of notes. You don't have a cap in for dc protection, but after this, it's a minor mod.
The ground wires you should have are from one batteries -, one batteries +, pin 3 of the op-amp, gnd from the input you're using and gnd from the output your using.
This should all be joined at the same place for now. NOT in a circle or loop.
There are more improvements to be made but as said before. First steps first.
Once again if i've erred please correct me!!!
Remove the battaries from the circuit.
Now remove link between pins 1 and 2, but leave the output wire on.
Remove the input signal wire. solder the input resistor to the input signal wire and then solder it to pin 2.
next solder the feedback resistior between pins 1 and 2.
Solder a wire from pin 3 ( + input or non-inverting input) to the group of ground wires.
You should have a working circuit if you were carefull.
Couple of notes. You don't have a cap in for dc protection, but after this, it's a minor mod.
The ground wires you should have are from one batteries -, one batteries +, pin 3 of the op-amp, gnd from the input you're using and gnd from the output your using.
This should all be joined at the same place for now. NOT in a circle or loop.
There are more improvements to be made but as said before. First steps first.
Once again if i've erred please correct me!!!
Whooooooooaaaaaaahhhhh...
The chip is for a single rail supply!
The 'Ground' in this case is the -ve rail. As it stands, you have an 18V supply (fine), with a 9V feed into the non-inverting input. That probably will kill the channel, as it has no current limiting resistor (something we will not need to worry about, but something you can look up later
).
By the way, the LF353 is a dual op-amp, so you can 'kill' each chip twice if you like (the power input will be ok, as will the other channel)
So, remove the centre tap from your battery supply. Connect pin 3 to pin 4 using a jumper, and pin 5 to pin 4 with a jumper.
The PSU +ve must go to pin 8.
Then follow the instructions regarding the pin out that I set out in my previous post.
I really want you to 'get there', as it's a great place to be sonically (in my opinion)..
Ok?
Owen
Ps - how about you post a photo before you apply power next time, so we can check its all ok...?
The chip is for a single rail supply!
The 'Ground' in this case is the -ve rail. As it stands, you have an 18V supply (fine), with a 9V feed into the non-inverting input. That probably will kill the channel, as it has no current limiting resistor (something we will not need to worry about, but something you can look up later
).By the way, the LF353 is a dual op-amp, so you can 'kill' each chip twice if you like (the power input will be ok, as will the other channel)
So, remove the centre tap from your battery supply. Connect pin 3 to pin 4 using a jumper, and pin 5 to pin 4 with a jumper.
The PSU +ve must go to pin 8.
Then follow the instructions regarding the pin out that I set out in my previous post.
I really want you to 'get there', as it's a great place to be sonically (in my opinion)..
Ok?
Owen
Ps - how about you post a photo before you apply power next time, so we can check its all ok...?
Thanks, gracias, vielen danke, xie xie, etc...
So I am doing my best to follow Edison's principle of discovering all the ways to not make a light bulb
Owen et. al. thanks again. When I'm done I am going to put together a "newbie guide" for this to help others which assumes roughly zero experience. Parts are cheap enough so I'm not sweating it too much. Fortunately I have a good store in the area so I can run there after work and replace all the stuff I smoked the night before.
Re: the power, single vs. double rail... little unclear on who is what. If I have two 9V in series like so:
(R')--(B)--(C)--(R)--(B') (R = +, B = -)
I can put my probes across as follows:
R'->B' : 18V
R'->C : 9V
C ->B' : -9V
Question: Where in this is the 0V rail exactly? If it's (C) and I only wire up the positive (R') then I will only get 9V, and then why am I using two batteries again?
FYI I had put an LED in so the voltage across 4 and 8 was 16V.
Also, just to make sure I've got the stupid stuff right, on the input signals and the pot... Here is what my 1/8" stereo jacks look like in glorious ASCII art:
(1)
---------\/--
|<--|--|-------
---^ ^-----
(2) (3)
Now, I understand this is carrying two channels, so these represent neutral, left, and right.
Nordic's diagram shows IN- and IN+ on the volume pot, which in this example is doing just one channel. How do these actually map to each other? I.e., I need to know:
1. Which is neutral and and which are signal
2. For the signal channel I will be using, this goes to the IN+ on the pot, yes?
Again, thanks a lot, you guys have the patience of the saints.
So I am doing my best to follow Edison's principle of discovering all the ways to not make a light bulb
Owen et. al. thanks again. When I'm done I am going to put together a "newbie guide" for this to help others which assumes roughly zero experience. Parts are cheap enough so I'm not sweating it too much. Fortunately I have a good store in the area so I can run there after work and replace all the stuff I smoked the night before.
Re: the power, single vs. double rail... little unclear on who is what. If I have two 9V in series like so:
(R')--(B)--(C)--(R)--(B') (R = +, B = -)
I can put my probes across as follows:
R'->B' : 18V
R'->C : 9V
C ->B' : -9V
Question: Where in this is the 0V rail exactly? If it's (C) and I only wire up the positive (R') then I will only get 9V, and then why am I using two batteries again?
FYI I had put an LED in so the voltage across 4 and 8 was 16V.
Also, just to make sure I've got the stupid stuff right, on the input signals and the pot... Here is what my 1/8" stereo jacks look like in glorious ASCII art:
(1)
---------\/--
|<--|--|-------
---^ ^-----
(2) (3)
Now, I understand this is carrying two channels, so these represent neutral, left, and right.
Nordic's diagram shows IN- and IN+ on the volume pot, which in this example is doing just one channel. How do these actually map to each other? I.e., I need to know:
1. Which is neutral and and which are signal
2. For the signal channel I will be using, this goes to the IN+ on the pot, yes?
Again, thanks a lot, you guys have the patience of the saints.
The resistors and caps are to help keep the two rails equaly far from gnd (0v)
The caps can be any value in 10v or more rateing as long as they are the same....
In- is the "ground" normaly the outside copper shield on audio cables...
remember your source is a voltage source (kinda like a small AC produceing battery), you need to connect both terminals on a battery for any current to flow.
on the headphone plugs the 2 front sections are left and right and the collar part gnd.
Your scematic shows 2 stages, sorry typeing from memory here...
the first opamp part acts as a buffer and is at unity gain... the second is at probaly too high gain considering the voltage ceiling of the supplies... maybe something towards 180k would be good in that position. (your schematic shows 2 resistors in parallel for power handling - not needed).
But those 2 opamps are in 1 chip.... not the best solution as you are limited to unity gain stable opamps to handle the first stage...
I did ask you before how you wanted to split your chips... and the solution given was based on your answer, and stripboard you are useing.
So you need to build a second one of these for the second channel.
I see some confusion here.
And a point that I could clarify my schematic 
All we are building is the unity gain buffer on the left hand side of the circuit diagram - the right hand opamp is an LM3875! (duh *slaps head*)
http://mysite.orange.co.uk/fleapit/images/8-picture3.gif?0.0835368280533706
This is what it should look like. Both channels are buffered on one chip. They then feed into a gainclone of your making (or a kit).
All the Resistors are 10k.
I hope this helps to clarify my mistake.
Owen
And a point that I could clarify my schematic
All we are building is the unity gain buffer on the left hand side of the circuit diagram - the right hand opamp is an LM3875! (duh *slaps head*)
http://mysite.orange.co.uk/fleapit/images/8-picture3.gif?0.0835368280533706
An externally hosted image should be here but it was not working when we last tested it.
This is what it should look like. Both channels are buffered on one chip. They then feed into a gainclone of your making (or a kit).
All the Resistors are 10k.
I hope this helps to clarify my mistake.
Owen
Who's on first?
OK, so I think I know where my signals are coming from.
I think I know how to count pins from 1-8.
I understand the purpose of the feedback resistor and the operating principle of an op-amp, well, at least a little bit.
Where I'm still groping in the dark a little is the power supply vs. ground. With two batteries in series I see +9V, 0V, and -9V.
Looking at Owen's schematics, is the following correct:
+9V goes to +Ve
-9V goes to -Ve
0V goes to GND
Also, I take it the 47K resistors specified n the previous schematic are supposed to be 10k. You also called for a 40k pot and a 2.2uf cap. Are those values correct?
I'd like to see one of these work, then I can worry about optimizing it a little in terms of power, etc.
OK, so I think I know where my signals are coming from.
I think I know how to count pins from 1-8.
I understand the purpose of the feedback resistor and the operating principle of an op-amp, well, at least a little bit.
Where I'm still groping in the dark a little is the power supply vs. ground. With two batteries in series I see +9V, 0V, and -9V.
Looking at Owen's schematics, is the following correct:
+9V goes to +Ve
-9V goes to -Ve
0V goes to GND
Also, I take it the 47K resistors specified n the previous schematic are supposed to be 10k. You also called for a 40k pot and a 2.2uf cap. Are those values correct?
I'd like to see one of these work, then I can worry about optimizing it a little in terms of power, etc.
Re: But does it blend?
The circuit is the one you linked.... can't say I have ever seen a 40k pot, its probably 47k.... R1 and R2 are both 47k.
There is a 2u2 cap between the unity gain and output stage... that is correct... anything up to 10uf wouold proabably be ok... but don't leave it out... Build everything on the left side of the chip and connect the power... there should be output between pin 1 and pin 4...(0V rail)... it will be less loud than the signal comming in, because it is a unity gain stage and there are some practical losses... amplification is in the second stage...
sansbury said:
The circuit is the one you linked.... can't say I have ever seen a 40k pot, its probably 47k.... R1 and R2 are both 47k.
There is a 2u2 cap between the unity gain and output stage... that is correct... anything up to 10uf wouold proabably be ok... but don't leave it out... Build everything on the left side of the chip and connect the power... there should be output between pin 1 and pin 4...(0V rail)... it will be less loud than the signal comming in, because it is a unity gain stage and there are some practical losses... amplification is in the second stage...
The +ve from your battery stack (after the LED), goes to the +ve on the chip.
The -Ve from your battery stack goes to the -ve on the chip.
The ground is a technical ground that connects the input ground to the output ground, and is 'zero' volts.
What may be a little confuzzling is the battery you have built could be described as +18V, implying that the base is zero, it could also be equally described as -18V, implying that the top is zero.
In actual fact the battery has a potential of 18V.
In this case the preamp is a single rail supply, so we need all 18V (16 or so after the LED) applied to the +ve, and the -ve connected to the technical ground (0V)
link to big picture of the 'guts' of the buffer (warning it is full size, and will take a while to open on dial up...)
and this may guide you a little further.
Ok..?
Owen
The -Ve from your battery stack goes to the -ve on the chip.
The ground is a technical ground that connects the input ground to the output ground, and is 'zero' volts.
What may be a little confuzzling is the battery you have built could be described as +18V, implying that the base is zero, it could also be equally described as -18V, implying that the top is zero.
In actual fact the battery has a potential of 18V.
In this case the preamp is a single rail supply, so we need all 18V (16 or so after the LED) applied to the +ve, and the -ve connected to the technical ground (0V)
link to big picture of the 'guts' of the buffer (warning it is full size, and will take a while to open on dial up...)
An externally hosted image should be here but it was not working when we last tested it.
and this may guide you a little further.
Ok..?
Owen
OK, so I dug out my good camera and decided to start over. before I start soldering stuff into place, here is where I am.
Green -> pin 1 (signal out)
Yellow -> pin 2 (signal in)
Blue -> pin 3 (ground)
Red -> pin 4 (+16V)
Violet -> pin 8 (0V)
Just to make sure I get this straight...
(+16V) is the hot positive end from my battery series thingy. (0V) is the other end of that thingy. By "thingy" I mean my 2 batteries in series with the LED. I understand Nordic's more sophistimacated thing; assuming it is not critical to make the thing make meaningful noise, but advisable perhaps for good noise.
I will take one of the L/R signal pairs and run that to one end of the pot. The center pin from the pot goes to the Pin 2 track. The remaining pin of the pot goes to the (0V) end as described above. Pin 3 also goes to the (0V) end. Pin 4 gets +16V. Pin 8 is left cold (since this is a single-rail chip, right?)
Pin 1 is the signal coming back out, hopefully making some noise.
How am I doing?
Green -> pin 1 (signal out)
Yellow -> pin 2 (signal in)
Blue -> pin 3 (ground)
Red -> pin 4 (+16V)
Violet -> pin 8 (0V)
Just to make sure I get this straight...
(+16V) is the hot positive end from my battery series thingy. (0V) is the other end of that thingy. By "thingy" I mean my 2 batteries in series with the LED. I understand Nordic's more sophistimacated thing; assuming it is not critical to make the thing make meaningful noise, but advisable perhaps for good noise.
I will take one of the L/R signal pairs and run that to one end of the pot. The center pin from the pot goes to the Pin 2 track. The remaining pin of the pot goes to the (0V) end as described above. Pin 3 also goes to the (0V) end. Pin 4 gets +16V. Pin 8 is left cold (since this is a single-rail chip, right?)
Pin 1 is the signal coming back out, hopefully making some noise.
How am I doing?
Attachments
Incorrect pin numbers
Notch on the IC should match with the notch on the IC base.
if the IC is inserted properly into this ic base, then
violet- pin 4
green- 5
yellow- 6
blue- 7
red- 8
The blue lines on the drawing are copper tracks on the back side. That means the copper tracks have to be on the other side and not as in your picture. Hope it is not double sided.
Notch on the IC should match with the notch on the IC base.
if the IC is inserted properly into this ic base, then
violet- pin 4
green- 5
yellow- 6
blue- 7
red- 8
The blue lines on the drawing are copper tracks on the back side. That means the copper tracks have to be on the other side and not as in your picture. Hope it is not double sided.
Attachments
HELL, talk about starting at the beginning...
As said above, the layout given is top down, and traces are beneath the board... you'll get it right today I think... not much more that you can do wrong...
As I understand (didn't look it up) you are useing a single rail chip, so you will not be connecting power to 0V rail...( I would suggest a nice dual rail chip (opa2132 or opa2134 -little cheaper). Then we can connect the dual rail setup... with no changes needed to the board except for soldering in some wires...
battery positive goes to pin 8 (red)
battery negative goes to pin 4 (purple)
other pins you marked
NI left (non-inverting) input pin 5 (green)
I left (inverting) input pin 6 (yellow)
Out left pin 7 (blue)
I swapped left and right on my drawing, in the description of inputs, (not a problem, just remember which is which), if it is important to you.
The unmarked pins next to purple are in the order from left to right
NI-R (3)
I-R (2)
Out- R (1)
ON initial testing only use ONE battery - if you do something wrong there is less chance of it destroying your chip... as in theory it should be able to handle that even on the input...
As said above, the layout given is top down, and traces are beneath the board... you'll get it right today I think... not much more that you can do wrong...
As I understand (didn't look it up) you are useing a single rail chip, so you will not be connecting power to 0V rail...( I would suggest a nice dual rail chip (opa2132 or opa2134 -little cheaper). Then we can connect the dual rail setup... with no changes needed to the board except for soldering in some wires...
battery positive goes to pin 8 (red)
battery negative goes to pin 4 (purple)
other pins you marked
NI left (non-inverting) input pin 5 (green)
I left (inverting) input pin 6 (yellow)
Out left pin 7 (blue)
I swapped left and right on my drawing, in the description of inputs, (not a problem, just remember which is which), if it is important to you.
The unmarked pins next to purple are in the order from left to right
NI-R (3)
I-R (2)
Out- R (1)
ON initial testing only use ONE battery - if you do something wrong there is less chance of it destroying your chip... as in theory it should be able to handle that even on the input...
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