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MRA:
I'm definitely not the first to do this. I've been wanting to try this on something reasonably "safe", before biting the bullet and re-powering a lathe or mill. So far its' had its' ups and downs, but getting a more powerful motor in a smaller form factor with variable speed to boot... I think that so far my biggest gripe is that wile you CAN vary the speed, it's in 100rpm steps. When you consider that this thing was designed to power an industrial sewing machine, I guess that 100rpm steps was close enough. It's obviously close enough for lathe, mill, and sanding belt applications, as I've seen it used to re-power all those devices.
I took my foot pedal control apart last night and checked the signal voltages. My VEVOR servo voltages agree with the voltages listed in the Hobby Machinist thread link that I provided above and my "manual" matched his.
At 2.4V on the signal wire, my signal wire was yellow, the motor started running at the minimum speed setting.
At 4.2 volts the motor was running at the maximum speed setting.
If you want to know how I arrived at my resistor values, read the long-winded explanation below - if you're not interested I'll see you later.
We building a voltage divider circuit. Knowing the required voltage values at the various points in my voltage divider, I can now calculate the required resistor values. They probably WON'T match standard resistor values, so I'll need to get close then run the voltage calculations again. I need to be sure that the voltage at the wiper of the pot will be close to but still below 2.4V at the minimum setting, and close to but still above the 4.2V at the maximum setting. Then I need to find a standard sized resistor that will give me the 0.8V drop that I need between the 5.0V source voltage and the 4.2V needed at the pot. Eazee-peazee, a 1st year electrical engineering student can do it. The problem is that I haven't been one of those in about 40 years, so I need to remember how to do that - but I'll get it figured out.
What we are building is a voltage divider that has 3 resistors and we know that voltage values at each end of the resistor and we know the value of one of the resistors - the pot, mines 10K. If we play fast and loose with this, and assume and/or ignore a lot of stuff that the 1st year electrical engineering student wouldn't be allowed to do, we can get close enough without too much brain sweat.
We know that the voltage drop across the 10K pot is 1.8V, (4.2V-2.4V), that gives us a rough number of a 0.18V per 10K of resistor. We want a 2.4V drop after the pot, 2.4V/1.8V tells us that we need a 13.3K resistor. We can get a 13K or a 15K, I don't know what I've got in my box right now, let's go with a 13K as the will give us a little lower voltage and ensure that when the pot is turned to 0 that the speed is 0. On the other end we need a 0.8V drop above the pot. 0.8V/1.8V tells us we need a 4.4K resistor, the closest standard resistor valve is probably 4.7K. OK, let's look at this voltage divider using these resistor values and the 5.0V input power and see what our new high/low voltages are for the pot wiper. Using those resistor values we get an approximate voltage drop of 1.80V per 10K of resistance. Using this value we can do a sanity check to see if this thing will turn off at the minimum pot position. Our 13K resistor will give us a 2.34V drop, this is definitely below the measured 2.4 volts so it will turn off. What about our pot wiper voltages, what will it see? We know on the bottom end it will be at 2.34V, on the upper end it will be at 4.14V which is CLOSE to 4.2V, don't know if it's close enough though. I'll have to look through my resistor stash and see if I can bread-board this and see if it's CLOSE ENOUGH. Sometimes calculated values just DON'T work in the real world, but not often.
I am going to include the Run/Stop switch option, and the effects of this is one of those things that I'm going to ignore that the 1st year electrical engineering student couldn't. The resistor used for it is a parallel circuit and WILL affect the speed control side. I'm going to do the same thing the OP in the Hobby Machinist thread did, make it a large resistor. (10-20 times the resistance of the speed pot side should work. Like us electricity is lazy and seeks the path of least resistance.) That does 2 things for us, first we can effectively ignore the resistor in OUR calculations, and second it ensures the when the Run/Stop switch it turned off the voltage that the signal wire sees will definitely be below the 2.4V minimum speed voltage.
Just for funzies we can determine the wattage our resistors need to handle. Watts=(V^2)/R, if V=5, and R=4.7K+10K+13K then Watts=25/27700=0.9mW, 1/4W or 1/8W resistors will work JUST fine.
Don
I'm definitely not the first to do this. I've been wanting to try this on something reasonably "safe", before biting the bullet and re-powering a lathe or mill. So far its' had its' ups and downs, but getting a more powerful motor in a smaller form factor with variable speed to boot... I think that so far my biggest gripe is that wile you CAN vary the speed, it's in 100rpm steps. When you consider that this thing was designed to power an industrial sewing machine, I guess that 100rpm steps was close enough. It's obviously close enough for lathe, mill, and sanding belt applications, as I've seen it used to re-power all those devices.
I took my foot pedal control apart last night and checked the signal voltages. My VEVOR servo voltages agree with the voltages listed in the Hobby Machinist thread link that I provided above and my "manual" matched his.
At 2.4V on the signal wire, my signal wire was yellow, the motor started running at the minimum speed setting.
At 4.2 volts the motor was running at the maximum speed setting.
If you want to know how I arrived at my resistor values, read the long-winded explanation below - if you're not interested I'll see you later.
We building a voltage divider circuit. Knowing the required voltage values at the various points in my voltage divider, I can now calculate the required resistor values. They probably WON'T match standard resistor values, so I'll need to get close then run the voltage calculations again. I need to be sure that the voltage at the wiper of the pot will be close to but still below 2.4V at the minimum setting, and close to but still above the 4.2V at the maximum setting. Then I need to find a standard sized resistor that will give me the 0.8V drop that I need between the 5.0V source voltage and the 4.2V needed at the pot. Eazee-peazee, a 1st year electrical engineering student can do it. The problem is that I haven't been one of those in about 40 years, so I need to remember how to do that - but I'll get it figured out.
What we are building is a voltage divider that has 3 resistors and we know that voltage values at each end of the resistor and we know the value of one of the resistors - the pot, mines 10K. If we play fast and loose with this, and assume and/or ignore a lot of stuff that the 1st year electrical engineering student wouldn't be allowed to do, we can get close enough without too much brain sweat.
We know that the voltage drop across the 10K pot is 1.8V, (4.2V-2.4V), that gives us a rough number of a 0.18V per 10K of resistor. We want a 2.4V drop after the pot, 2.4V/1.8V tells us that we need a 13.3K resistor. We can get a 13K or a 15K, I don't know what I've got in my box right now, let's go with a 13K as the will give us a little lower voltage and ensure that when the pot is turned to 0 that the speed is 0. On the other end we need a 0.8V drop above the pot. 0.8V/1.8V tells us we need a 4.4K resistor, the closest standard resistor valve is probably 4.7K. OK, let's look at this voltage divider using these resistor values and the 5.0V input power and see what our new high/low voltages are for the pot wiper. Using those resistor values we get an approximate voltage drop of 1.80V per 10K of resistance. Using this value we can do a sanity check to see if this thing will turn off at the minimum pot position. Our 13K resistor will give us a 2.34V drop, this is definitely below the measured 2.4 volts so it will turn off. What about our pot wiper voltages, what will it see? We know on the bottom end it will be at 2.34V, on the upper end it will be at 4.14V which is CLOSE to 4.2V, don't know if it's close enough though. I'll have to look through my resistor stash and see if I can bread-board this and see if it's CLOSE ENOUGH. Sometimes calculated values just DON'T work in the real world, but not often.
I am going to include the Run/Stop switch option, and the effects of this is one of those things that I'm going to ignore that the 1st year electrical engineering student couldn't. The resistor used for it is a parallel circuit and WILL affect the speed control side. I'm going to do the same thing the OP in the Hobby Machinist thread did, make it a large resistor. (10-20 times the resistance of the speed pot side should work. Like us electricity is lazy and seeks the path of least resistance.) That does 2 things for us, first we can effectively ignore the resistor in OUR calculations, and second it ensures the when the Run/Stop switch it turned off the voltage that the signal wire sees will definitely be below the 2.4V minimum speed voltage.
Just for funzies we can determine the wattage our resistors need to handle. Watts=(V^2)/R, if V=5, and R=4.7K+10K+13K then Watts=25/27700=0.9mW, 1/4W or 1/8W resistors will work JUST fine.
Don