0
\$\begingroup\$

Tell me please how best to power the circuit which has an analog part and the part which, with the aid of arduino and digital potentiometers, controls the analog part. The supply voltage V(in) of the analog part is 9 volts (this is a guitar effect). Is it enough to put the voltage regulator after the same power source to get 5 volts (for example 78L05) or is it necessary to put a stabilizer and 9 volts too? I do not know how to do this most correctly. Please advise how to properly supply this scheme. enter image description here

Update: It turned out that the analog circuit works fine from 5v, I tried to draw three possible options for powering the analog and digital parts, tell me which one is better, or suggest improvements or your own version.

enter image description here enter image description here enter image description here

\$\endgroup\$
  • \$\begingroup\$ For professional advice: Draw input and output arrows on some of your components so that nothing is overlapping. It just makes it easier to read the schematic so that there are not a bunch of wires going every where. You did it one your 5v regulator, which is nice :) I'm not mad, just giving a friendly tip. Your manufacturers will be eternally grateful as well once you get out into the workforce. \$\endgroup\$ – KingDuken Sep 18 '18 at 2:30
  • \$\begingroup\$ I am curious why you replacing analog pots with digital, controlled by same analog pots? I'd understand this if you had some kind of digital controls, like encoders or up/down buttons. Or if you had additional controls for storing current positions as presets and recalling them later (that'd be cool, I think). But you have none of that, so the only result is reduced resolution \$\endgroup\$ – Maple Sep 19 '18 at 0:39
  • \$\begingroup\$ Please check my video instagram.com/p/1JT3Ffnyuk I have saved presets and plan to do sequencer on some of the parameters and also use to control for example an infrared rangefinder. \$\endgroup\$ – Maxim Kulikov Sep 19 '18 at 9:53
  • \$\begingroup\$ Sorry, watching shaky video with annoying beat is not my kind of fun. If you do use MCU for presets then great, it explains the digital pots. I'd use encoders instead of analog pots, if I were you though. Unlike pots, they allow you to do adjustements after loading presets (pots can't rotate to stored position by themselves), while at the same time allow big fast changes impossible with Up/Down buttons. \$\endgroup\$ – Maple Sep 19 '18 at 11:08
  • \$\begingroup\$ thanks, I also thought about connecting the encoders \$\endgroup\$ – Maxim Kulikov Sep 19 '18 at 11:52
1
\$\begingroup\$

You cannot use 9v with MCP41XXX powered by 5V. The datasheet specifies voltage range on resistor terminals as 0~VDD, i.e. up to 5V in your case.

The only option I can think of is to make analog part also 5V, and then add 9V powered amplifier stage on the output.

UPDATE

Here, I rearranged your analog part for you. Frankly, I have no idea what it does. I am not an analog guy, but even I know that there should be output somewhere. Also, the polarity of C1 seems to be wrong.

Those 8V from the diode enter circuit through fairly large resistor. Plus, all ground connections also have large resistors in them. IMHO, this circuit should not be sensitive to supply voltage at all.

enter image description here

UPDATE 2

Since you figured out your analog part can work from 5V supply, your suggested supply #2 should work just fine if you remove R19.

Note, that digital potentiometers are digital devices and their resistor terminals are technically not isolated in any way from digital circuitry. So, any tinkering with GND won't do you any good. The best your can do is place all analog parts over separate part of ground plane, making sure it is connected to digital part in one place only. Since MCP41010 will be powered from digital part, you can make separation line passing under the chips around resistor terminals.

BTW, dual chip MCP42010 would be much better choice. It has more convenient location of the resistor pins - all 6 on one side of a chip. And you only need 3 of them instead of 5.

\$\endgroup\$
  • \$\begingroup\$ I’m use 9v only for analog part, mcp41010 power supply 5v(from arduino board now) and I’m plan to change arduino to atmega with boot-loader as on my scheme. \$\endgroup\$ – Maxim Kulikov Sep 18 '18 at 0:47
  • \$\begingroup\$ PA0 of IC6 is connected to VIN, which I assume is your 9V. If you fix this, then using LDO to get 5V and from 9V as already drawn in your schematics is OK. \$\endgroup\$ – Maple Sep 18 '18 at 0:59
  • \$\begingroup\$ thank you, now I understand what you mean, but the fact is that I have PA0 connected to 9v all work fine, even when I use it for a long time. \$\endgroup\$ – Maxim Kulikov Sep 18 '18 at 1:05
  • \$\begingroup\$ Well, if you think that is OK, why did you ask the question at all? "If it hasn't burned out yet it must be alright" attitude basically means you can do whatever you feel like doing anyway \$\endgroup\$ – Maple Sep 18 '18 at 4:31
  • \$\begingroup\$ I apologize, I do not think that everything is in order, I just noted the fact that, strangely enough, everything works. And I definitely do not want to burn anything, I want to do everything right. But now I do not know what to do, the effect scheme is designed specifically for 9 volts, I do not think that it will work well from 5 volts. \$\endgroup\$ – Maxim Kulikov Sep 18 '18 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.