Hi,

I'm an amateur enthusiast on electronics.
I'm trying to build the following schematic
It is a "log amp" converting voltages from Hz/V to V/oct for analog synthesizers.




I have some basic queries regarding building this on a breadboard.

1) All the points circled in red are + and - points.
Are that supposed to be connected on my power supply (+12V / -12V)
Am I right?

2) Circle in purple, is an arrow that is "connected" to the positive of the input jack.
How should I connect this point exactly ?

3) On the green arrow, seems that the middle leg of the potentiometer goes back to the one of the other legs.
Should I just bridge the middle and the third leg?

To be honest I ve tried to built this on a breadboard with an external adjustable laboratory power supply.
Sadly, the results are not as should be.

When I connect a 5V input, the output was around 3.6V.
When I connect a 8V input the output was around 5.6V.
Also unfortunately, any of the potentiometers didn't change anything at all on the output...

You can take a look below what voltages should be right :


I hope that my queries are clear.

Any suggestions, advice or comments are welcome.

Jim

 

That appears to be a +ve, 0, -ve supply.
The green arrow it seems is just indicating a pre-set pot symbol..
VR5 is a set as a var resistance not a potentiometer.
i.e. slider is connected to one side.

 

Hi,

I'm an amateur enthusiast on electronics.
I'm trying to build the following schematic

I have some basic queries regarding building this on a breadboard.

1) All the points circled in red are + and - points.
Are that supposed to be connected on my power supply (+12V / -12V)
Am I right?

Yes,

2) Circle in purple, is an arrow that is "connected" to the positive of the input jack.
How should I connect this point exactly ?

That is a normal, the jack has a contact that is closed when the plug is not inserted and is lifted when it is.

3) On the green arrow, seems that the middle leg of the potentiometer goes back to the one of the other legs.
Should I just bridge the middle and the third leg?

Yes, the outside part of the potentiometer is wired like a fixed resistor in series and the wiper is shorting a varying portion of it.*

*I am not certain about this anymore, the schematic's symbology is a little different than I am used to. I think it's right but I'd wait for someone more certain.

 

That is not +12 and ground, it is +12V and -12V...

 

With also a chassis connection, that presumably is the 0v reference?

 

With also a chassis connection, that presumably is the 0v reference?

It would seem reasonable. There is just enough information in the schematic to be right or wildly wrong.

 

Hi,

Thank you very much for your help!
I really appreciate it.

To be honest, I can't understand the term "variable resistor". I thought it was just a potentiometer.

One more thing.
When I plug a 5V input to the jack, I got 3.6V at the output either if I power my circuit or not.
Power has nothing to do with the output... (?)
So, I assume that I do something wrong on the power supply.

I use this power supply with the 3 pins (+ - and ground)


Is it that strange?

Any suggestions, advice or comments are welcome.
Jim

 

Your pic shows a shorting link between -ve and ground. Try removing the link.
On the maker's web-site, that supply is described as 'single channel' and appears to be a single polarity one with the option to ground the -ve. Your schematic calls for a dual-polrity supply.

 

This is my interpretation of the power source:

 

Hi,

Thank you very much for your help!
I really appreciate it.

To be honest, I can't understand the term "variable resistor". I thought it was just a potentiometer.

One more thing.
When I plug a 5V input to the jack, I got 3.6V at the output either if I power my circuit or not.
Power has nothing to do with the output... (?)
So, I assume that I do something wrong on the power supply.

I use this power supply with the 3 pins (+ - and ground)
View attachment 232618

Is it that strange?

Any suggestions, advice or comments are welcome.
Jim

This is not the supply you need. The requirement is for a dual supply providing both negative and positive relative to the common. That jumper bar in the picture will not help at all. Disconnecting it will bring no benefit in this application.

 

Hi,

Thank you for your replies and your help.
I will try to remove the link to the above laboratory power supply to check if it will work.

But generally, I'm planning to use this circuit to my eurorack modular synth.
Modular synth provides this power supply for the other modules (see picture)
You can see that each slot can provide +5 / +12 / -12 / ground.

Do you believe that kind of supply would work?

Thank you in advance
Jim

 

If you have a supply of both +12 V and -12 V and the common between them that should work very well with the circuit you showed. AND it became clear that those resistors with the oval around them and the "MF" designation were specified as Metal Film types, which are lower noise devices. That is reasonable because noise on a pitch control line would certainly be very objectionable.

 

Hi,

I really appreciate your replies.
Really push me to search & learn more in electronics.


So, I'm back to my thread because I finally realise that the power supply from my eurorack modular synth (see my previous post) is working good with my schematic.

However a quick google for "dual power supply" gets me more confused on how I can make my own power supply for +12V, -12V, ground.

One more thing.
I 'm using this power supply for my "LOG AMP" circuit and in the same time there are some other (expensive) modules connected in the same power strip. Is there any danger to burn my other modules? or you think it's safe?

So, if anyone knows some kind of "protection" circuit for my schematic I would be happy to see it
But for sure I don't want to burn any of my other modules...


Any advices, comments or suggestions are welcome,
Jtel

 

This is a valid concern. Being careful to not short circuit or overload the power supply is the way to protect it. And your power supply is probably well regulated and well filtered, considerably more complex than the circuit in post #9.
It is not likely that damage to the other modules would be caused, but for experimenting it is usually more convenient to have a separate power supply.

 

A module would probably be burned if its power supply voltage is too high or the polarity of the power supply is connected backwards.