Potentiometer

[Rockgirl]

Hi, I need help. Can I find any potentiometers where the rotation angle is smaller than usual? Usually it's I guess 290 degrees but can somewhere find about 90 degrees? I need a 100k potentiometer that adjusts the 90 degree rotation, I don't know if they are anywhere.

 

[Harald Kapp]

Can you use a 330 kΩ potentiometer and limit the rotation mechanically to 90 °?
I'm not aware of 90 ° potentiometers off the shelf.

 

[Rockgirl]

Can you use a 330 kΩ potentiometer and limit the rotation mechanically to 90 °?
I'm not aware of 90 ° potentiometers off the shelf.

Oh I'm not sure. Maybe I need to open up more about what this is all about. I bought a bidirectional DC-motor controller. I would like to use a potentiometer on the joystick. But the joystick can't turn it enough. If there was any trick to make this work?
Maybe this brings even more difficulty, but I would supply this controller with either 12vdc or 24vdc. Does it make this more difficult?

I tried to put a picture of the connection but it doesn't work

I hope you could help me with this problem.

 

[Harald Kapp]

I tried to put a picture of the connection but it doesn't work

Resize to < 300 kB.

I bought a bidirectional DC-motor controller.

Which model? Can you provide a link to the technical data / datasheet of the controller?

 

[Rockgirl]

Resize to < 300 kB.


Which model? Can you provide a link to the technical data / datasheet of the controller?

Okay. Now pics work. This is also link https://quasarelectronics.co.uk/Item/3166v2-bidirectional-dc-motor-speed-controller-5-32vdc-5a






[mod note - resized oversize images and place inline]

 

[bertus]

Hello,

Why do you need a 90 degrees potmeter?
The circuit uses a standard 100 k potmeter:

Bertus

 

[Rockgirl]

Hello,

Why do you need a 90 degrees potmeter?
The circuit uses a standard 100 k potmeter:
View attachment 52768
Bertus

I already mentioned it above in the post. Because I want use joystick and the trajectory of the joystick is not enough.

 

[bertus]

Hello,

You can adjust the range for the circuit with the 100k and 10k resistors on adjusting stack.
You could make the circuit and measure the voltage on pin 3 of IC1a.
Then you can calculate the needed resistors in stead of the 100k and 10k.

Bertus

 

[Rockgirl]

Hello,

You can adjust the range for the circuit with the 100k and 10k resistors on adjusting stack.
You could make the circuit and measure the voltage on pin 3 of IC1a.
Then you can calculate the needed resistors in stead of the 100k and 10k.

Bertus

Oh I guess I don't know how to do that

 

[Externet]

Something happened to AliExpress site and I get it only in korean, you will have to dig deeper or learn it... :-(

----> https://ko.aliexpress.com/item/32760020962.html?src=bing&albslr=222322000&isdl=y

----> https://ko.aliexpress.com/item/3301...#3564#16062#853&pdp_ext_f={"sceneId":"23416"}

----> https://www.ecosia.org/images?q=joystick potentiometer#id=074DD0ADDD0C72CC286EE647FD37EC329C9E7C1F

 

[bertus]

Hello,

@Externet , Is this the item:
https://www.aliexpress.com/item/32760020962.html?src=bing&albslr=222322000&isdl=y
https://www.aliexpress.com/item/330...#3564#16062#853&pdp_ext_f={"sceneId":"23416"}



Bertus

 

[Harald Kapp]

You're not alone in searching a pot with 90 ° rotation angle. All references I can find mention standard 270 ° pots with mechanically limited rotation angle. The ones that have "90 °" in the title of the sales offer refer to the angle between the pins and the potentiometer axle as 90 °, not the rotation angle.

In your circuit the pot is used as a voltage divider together with the 10 kΩ resistor at the lower end and the 100 kΩ resistor at the upper end. It thus delivers these voltages to pin 3 of IC1a:
0.05 × V+ when in the lower position,
0.52 × V+ when in the upper position
You cannot achieve this exact ratio by simple means. You can approximate it, however:

• Remove the upper 100 kΩ resistor by a short circuit.
• Change the lower 10 kΩ resistor to 5 kΩ (use 2 × 10 kΩ in parallel or use 4.7 kΩ, good enough)
• Change the gain of IC1a from 1×to 2× by adding two resistors in the feedback as shown e.g. here. The value of the 2 additional resistors is not important. both shall have the same vale in the range 1 kΩ to 10 kΩ, whatever you have in your parts bin.
• Mechanically limit the potentiometer to move from the left side (lower pin in the diagram) to left side + 90 °.

Using 4.7 k will give you these divider ratios:
0.1 × V+ in the lower position
0.72 × V+ in the upper position.
You can tweak (fine tune) these values by changing the gain f the opamp IC1a from 2 to another (lower) value. Do this by changing the resistor values in the feedback circuit according to the equations given in the link I providd above.

 

[VenomBallistics]

https://hobbymatehobby.com/products...MI26Hekt3R8gIVRBh9Ch240QrdEAQYAiABEgIx-fD_BwE
Cannibalize this for its gimbals and replace the pots as needed

 

[Rockgirl]

You're not alone in searching a pot with 90 ° rotation angle. All references I can find mention standard 270 ° pots with mechanically limited rotation angle. The ones that have "90 °" in the title of the sales offer refer to the angle between the pins and the potentiometer axle as 90 °, not the rotation angle.

In your circuit the pot is used as a voltage divider together with the 10 kΩ resistor at the lower end and the 100 kΩ resistor at the upper end. It thus delivers these voltages to pin 3 of IC1a:
0.05 × V+ when in the lower position,
0.52 × V+ when in the upper position
You cannot achieve this exact ratio by simple means. You can approximate it, however:

• Remove the upper 100 kΩ resistor by a short circuit.
• Change the lower 10 kΩ resistor to 5 kΩ (use 2 × 10 kΩ in parallel or use 4.7 kΩ, good enough)
• Change the gain of IC1a from 1×to 2× by adding two resistors in the feedback as shown e.g. here. The value of the 2 additional resistors is not important. both shall have the same vale in the range 1 kΩ to 10 kΩ, whatever you have in your parts bin.
• Mechanically limit the potentiometer to move from the left side (lower pin in the diagram) to left side + 90 °.

Using 4.7 k will give you these divider ratios:
0.1 × V+ in the lower position
0.72 × V+ in the upper position.
You can tweak (fine tune) these values by changing the gain f the opamp IC1a from 2 to another (lower) value. Do this by changing the resistor values in the feedback circuit according to the equations given in the link I providd above.

Hi, I would like to try this. Would it be possible for you to show (draw) how this is done for example in this picture you added?
I’m a girl and this isn’t my strongest skill, not yet at least I'm afraid if I destroy this controller by doing wrong I want to be sure what I do. I know how to solder and I know some basic stuff about electronics but I’m not sure about this.

 

[Harald Kapp]

Here you go:

 

[Rockgirl]

Here you go:
View attachment 52782

Okay, I'll try this tomorrow. I still have to ask if I understood correctly, so can I add those new two resistors to something between 1k -10k? Thanks

 

[Rockgirl]

The solution was found closer than I could imagine. I found an old joystick potentiometer. It looks bigger compared to other potentiometer. I check those multimeter and I get 110k ohms. This potentiometer work those angle which I want. But why do these pots work differently? Are joystick pots some way different?

Another issue related to this bi-directional control. Is it a feature of this control that it does not start running the engine completely at zero? When I rotate the potentimeter the motor starts to buzz at the beginning and then it suddenly starts to rotate. I don’t mean it starts rotating at full, but about 3-4 volts. Why doesn't the motor run in the 1-3v range?
No matter which potentiometer is used for this, the potentiometer will not affect this. I would like this to work in the same way when I give power to the motor power supply unit, when I start from zero and add volts the motor starts a little quietly spinning up to 1 volts.