40106 line out and speaker

[c1rcu1t]

Hi im building a circuit for a simple instrument based around a 40106 chip with two oscillators mixed through diodes and buffered from other gate on the same chip.
I would like to maintain the circuit simple and easy to build for begginers to follow, so i could teach others. So I used a very simple circuit of the Lm386 and a switch jack to disconnect the speaker when connected to line out.
Im still a newbie, experimenting circuits and still with a lot to learn, so i would like to understand if im doing this correctly for the line output, if it can be a ceramic capacitor for the 104 or maybe it should be a electrolytic? And if i should change or add more things to this part of the circuit, if anyone can help :--)

thank you!

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[dk]

Hi, the 100n cap (104) should probably be larger and a different type. While we're hardly working with precision circuits or audiophile signal paths, ceramic capacitors are a pretty poor choice for removing DC offset from a signal (which is the function of this cap in your circuit) and 100n is small enough to remove a lot of still musical low frequency content. From a functional standpoint, you're aiming to have the cap small enough to remove the DC offset while keeping it large enough to have most of the action in the frequencies below human hearing. I'd reckon a 10-47uF electrolytic would be fine there. You might get away with as small as a 1uF.

[c1rcu1t]

[dk]

The cap should be after the pot.

What voltage are you running this off of? Assuming 5V-9V, there is no issue with connecting this directly to a mixer. You have a volume knob to turn it down to prevent clipping, but unless you connect it to consumer gear running off of a tiny power supply (ie USB), I can't imagine you damaging anything with it.

[Steveg]

Hi piupiupiu,

A coupling capacitor of 1uF to 10uF would be usual. The output through the amp may be disappointing compared to the line out jack because the capacitor and the resistors near pin 3 will form a high pass filter.

That feed to the amp has been designed for a symmetrical power supply +/-15 volts or similar. If you use it on a single ended power supply you will have the input to pin 3 going negative for half the square wave cycle which may damage the amp. For a single ended power supply you need to bias pin 3 to half Vcc rather than ground.

[c1rcu1t]

[c1rcu1t]

[Steveg]

The effect of the two resistors is to divide the input to the amp chip by 11. I assumed they were there because you knew you needed a much higher level going to the line out than the amp. Just removing the resistors won't change the negative voltage situation (it may not damage the chip but you will be relying on the chip's internal protection diodes which are only meant for emergencies).
It may be as simple as removing the 100K resistor and replacing the 10K resistor with a 100K resistor and adding a new 100K resistor from pin 3 to +9 volts. The resistors need to be the same value to balance the input to pin 3 at about 4.5 volts. If the resistors have a too low value they will form a high pass filter with the coupling capacitor. If you are getting too much volume you could replace the volume pot with a 10K pot and put a 100K resistor between the pot and buffer. Even better might be a 1K pot and 10K resistor. Go any lower than that and you would be causing the buffer to supply too much current.
(my memory says 10K is a safe minimum load for a 40106 at 12V so if I have not completely mis-remembered it will be good for 9V. An 11:1 reduction brings the 9V peak to peak (p-p) supplied by the 40106 down to near the 1V p-p expected by standard line out conventions and most amps. With a 1K pot you could probably tweak that 10K down to 8K without problems.)

[PHOBoS]