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I was trying to make a square wave to triangle or sine wave converter circuit, and it seemed to work, the problem is that I think there's a problem when you want to use it for something that requires a negative and positive input, like a motor or transformer. This is the circuit that I made:

enter image description here

The output seems to work for a piezo buzzer but not for a motor or transformer. It's possible there's some issue with my square wave generator, but it seems to work fine and I think the issue is that I'm not sure where I would connect a second wire as ground. Can this be done, or would you need another converter circuit to act as ground?

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    \$\begingroup\$ Have you considered how much current you can push through two 1.5k resistors? And how much current a motor needs to run? \$\endgroup\$ – JRE May 24 '18 at 18:28
  • \$\begingroup\$ Isn't the efficiency for charging and discharging in an RC circuit 50 percent? \$\endgroup\$ – Tom May 24 '18 at 18:34
  • \$\begingroup\$ The two resistors are in series between your input and output. Current is voltage divided by resistance. Draw the circuit as a schematic instead of a wiring diagram, and it will be more obvious. \$\endgroup\$ – JRE May 24 '18 at 18:39
  • \$\begingroup\$ I understand that. So what you're suggesting is that the problem might not be that I don't have a ground wire, but that too much power is being dissipated? \$\endgroup\$ – Tom May 24 '18 at 18:58
  • \$\begingroup\$ You aren't getting any power through those resistors is what I'm saying. \$\endgroup\$ – JRE May 24 '18 at 19:18
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This is your circuit:

enter image description here

No matter what you do with those capacitors, the current between IN and OUT is limited by those two 1.5k ohm resistors.

If I assume you have 5V going in to IN, then the cannot get more than 1.67mA. That 2/1000 of an ampere.

A motor will need any where from a couple of hundred mA (for a small DC hobby motor) to several amperes (for an AC universal motor like you might find in a kitchen mixer) to hundreds of amperes for a large industrial motor.

If you try to drive the transformer from a wall wart with this, then you will have similar problems. The impedance (sort of like resistance, but it applies to inductors and capacitors) of a typical transformer is so low that it will take much more current than you can get through your filter.

This ignores the fact that your signal generator cannot drive a motor or transformer even if you leave out your filter.

If you connected the motor straight to the signal generator, the motor wouldn't turn.


Also, note how I've drawn the ground connections. That is how you should connect the motor or transformer so that you have a complete circuit.

One last thing:

Those little resistors are rated for 0.25 watt. At 5V, you are only allowed to push 50mA through them. More than that and they will burn up.

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  • \$\begingroup\$ Oh, I think I see what you mean. It's a signal generator, but the power that isn't dissipated by the resistor will just get stored in the capacitor, right? So it doesn't actually transmit power as it is? Could I use another signal converter that's discharging a capacitor in combination with this one to output power? \$\endgroup\$ – Tom May 24 '18 at 20:39
  • \$\begingroup\$ No, you can't. You could work out another filter with much lower resistors but the same cutoff frequency. That will allow more current to pass. But, you are still limited by the signal generator. \$\endgroup\$ – JRE May 24 '18 at 20:56
  • \$\begingroup\$ Those little capacitors won't store enough energy to matter at all. If you use large capacitors but keep the large resistor, you will get something that you can let charge for a (long) time, and then connect your motor and see it turn (a little.) \$\endgroup\$ – JRE May 24 '18 at 20:59

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