0
\$\begingroup\$

I have some positioning servo driver and servo with pot feedback. Now I try to build circuit to switch between 3 positions. For testing I just use 3 way switch and 3 pots as voltage reference, but in my destination project my switch will by 10 meters away from servo and driver so current circuit may be noisy. So can I build this without passing ref. voltage thru switch? Or using buttons?

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ "... my switch will by 10 away ...". mm? feet? meters? kilometers? Welcome to EE.SE. You could add a little capacitance on the servo input to act as a low-pass filter to reduce the noise. You might also want to check that it's OK to leave the servo input floating if all the switches are open. \$\endgroup\$
    – Transistor
    Commented Sep 1, 2018 at 16:46
  • \$\begingroup\$ 10 meters of course \$\endgroup\$
    – Dobijasek
    Commented Sep 1, 2018 at 18:45

1 Answer 1

0
\$\begingroup\$

Your servo remains in an undefined state if all three switches are open-circuit. A simpler arrangement using two switches leaves the servo at its mid-position in a fairly low-noise condition, because both remote switches are open-circuit.

R1 defines the default state, while R2 pulls up toward +5V, and R3 pulls down toward ground. The switch chip and variable resistors can lie close to the servo, while the logic control signals can be remote. Be aware that the switch chip requires a logic supply and ground connection (not shown).

schematic

simulate this circuit – Schematic created using CircuitLab
If you really need the servo in an undefined state, a quad-switch like 74HC4066 can be used (one-of-four unused). As before, control signals can be remote, while analog servo voltages can be more closely associated with the servo:

schematic

simulate this circuit

\$\endgroup\$
4
  • \$\begingroup\$ This undefined state is just error in my picture. As I understand analog switch is visible in circuit as some resistance so until there is no load on the other side there there won't be any voltage drift? \$\endgroup\$
    – Dobijasek
    Commented Sep 1, 2018 at 18:40
  • \$\begingroup\$ Yes, analog switch adds some resistance to the variable resistor equivalent resistance. Analog switch resistance ranges from a few ohms for FET switches, up to a few hundred ohms for old CMOS switches. Parametric search is your friend. \$\endgroup\$
    – glen_geek
    Commented Sep 2, 2018 at 11:45
  • \$\begingroup\$ IMO your first circuit, the one with R1 default pot and R2/R3 pull up/down is not reliable nor user friendly. Playing with pots and inputs you may short-circuit power supply and anyhow control functions are not linear and interact each other. I'd stick on your second circuit, eventually introducing a default condition on the logic side of switches. \$\endgroup\$
    – carloc
    Commented Sep 2, 2018 at 14:51
  • \$\begingroup\$ @carloc fair enough - you've pointed out a potential flaw. I had surmised that the default pot was at mid-scale (2.5V), so shorting to Vcc or to gnd by activating the other pots wouldn't likely cause smoke. I still think a hard-wired default to a known state is a good plan should some control wires be disconnected/broken. \$\endgroup\$
    – glen_geek
    Commented Sep 2, 2018 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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