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Potentiometer sizing question

CountryBoy19

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Aug 14, 2012
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Bedford, IN
I have a Hitachi NES1 drive that I'm trying to connect to an external control box so the inverter/drive can be located in a more protected location.

The problem I'm having is that the manual calls for a 1k-2k potentiometer for the control.

From my experience this seems quite low. A Toshiba drive I've worked with before calls for 1k-10k.

From my knowledge on circuits/electronics, I don't see how putting in a higher resistance value could cause a large negative effect. There is likely some high level resistor (Mohm) sinking this analog signal to ground within the circuitry. Using a 10k potentiometer in place of the 2k cannot effect the interaction between the 2 drastically enough to change the drive operation.

What am I missing?

The reason I ask is that a 1k pot with 10 vdc applied is .1 watt. Finding a 1k pot with a power rating of .1 watt or higher across the entire signal range drastically increases cost. I already tried a $1 .2 watt pot, it burned out. We're talking about a $20 pot vs a $1 pot. And I already have high quality, USA pots available to me going down as low as 10k (but not 1k).

FWIW, my inclination is to try it. More resistance can't really do damage to it, if anything it will just act funny (already happening with the cheap pot).
 
If it is just setup as a variable divider (both ends of the pot and the wiper tied to the drive) I agree with you. If you have something laying around in 5k or 10k try it. A bigger resistor would not hurt anything.
 
If it is just setup as a variable divider (both ends of the pot and the wiper tied to the drive) I agree with you. If you have something laying around in 5k or 10k try it. A bigger resistor would not hurt anything.

It is setup as a variable divider according to the schematics... I think I'll give the 10k a shot and see how it does. Just wish I hadn't wasted $3 on the cheapo pots and waited for shipping...
 
The digital section designer knows what the optimum swing of the logic gates are. He also know what the logic is before any inputs to the micro-controller.

Have fun experimenting.
 
If you are not getting the right output with a 5k pot, try putting a couple of 2.5k resistors in series with each other and in parallel with the pot. The junction of the resistors goes to the wiper. That will give you a roughly 2.5k pot with a somewhat non-linear, but probably tolerable, output.
 
The digital section designer knows what the optimum swing of the logic gates are. He also know what the logic is before any inputs to the micro-controller.

Have fun experimenting.
Can you elaborate?

I'm an engineer, I have built many circuits, I'm simply trying to understand the WHY here.

The potentiometer isn't a digital signal, it's an analog voltage, therefore I don't think logic gates are relevant. If you think so, can you please explain in more detail?

The potentiometer is being used as a voltage divider, where-in the sum from the first terminal to the 2nd, to the 3rd will always equal what the board is putting out (nominal 10 VDC). But the voltage from 1 to 2 or 2 to 3 is variable. If we isolate the potentiometer as it's own system, separate from the drive, it doesn't matter how large or small we make it, the voltage variation will still be the same. The only thing that changes with pot size is current (and power). We must be careful not to exceed the capacity of the components on the board, so we don't really want to go with a smaller resistance value (that would increase current & power), so larger than "spec" is best. OTOH, the most likely way the board is made up is: DC power supply > T1 of pot > T2 of pot* >** High value resistor > common. *T1 to T2 will provide a variable resistance value (which provides a voltage for the circuit to read), **The voltage is read on the line prior to the high-value resistor. This keeps the current flow through T2 very low, but allows a clean voltage signal. This resistor may be inside a chip, or it may be external to a chip, that is mostly irrelevant. Changing the value of the pot without changing the value of this internal resistor will change the proportion of current flowing through T2 (and possibly into the chip). This may initially alarm you, but don't forget that by increasing the value of the pot we're decreasing the total current, so even if T2 is taking more current proportionally, it's still taking less current. The only other possible effect is that it may slightly distort the linearity of the pot more than the design was intended for by changing pot size without changing the internal resistor size, but we're talking about negligible amounts.

That is what my logic and limited circuitry experience tell me. I'm here asking if there is a glaring hole in my logic. If there is, please explain it so I can understand. This isn't just "tell me how to do it", it's "teach me so I know".
 
Just buy one of these: FM-POT6 It works for me.

It is just a voltage divider for a 0-10 volt source. You want a high enough resistance to not overload the source.
You're running that 10k pot in the Hitachi NES1 drive? If so, I don't need to buy one, as I said in the op I already have high-quality USA made 10K and higher pots in my parts bin.
 
Can you elaborate?

I'm an engineer, I have built many circuits, I'm simply trying to understand the WHY here.

The potentiometer isn't a digital signal, it's an analog voltage, therefore I don't think logic gates are relevant. If you think so, can you please explain in more detail?

The potentiometer is being used as a voltage divider, where-in the sum from the first terminal to the 2nd, to the 3rd will always equal what the board is putting out (nominal 10 VDC). But the voltage from 1 to 2 or 2 to 3 is variable. If we isolate the potentiometer as it's own system, separate from the drive, it doesn't matter how large or small we make it, the voltage variation will still be the same. The only thing that changes with pot size is current (and power). We must be careful not to exceed the capacity of the components on the board, so we don't really want to go with a smaller resistance value (that would increase current & power), so larger than "spec" is best. OTOH, the most likely way the board is made up is: DC power supply > T1 of pot > T2 of pot* >** High value resistor > common. *T1 to T2 will provide a variable resistance value (which provides a voltage for the circuit to read), **The voltage is read on the line prior to the high-value resistor. This keeps the current flow through T2 very low, but allows a clean voltage signal. This resistor may be inside a chip, or it may be external to a chip, that is mostly irrelevant. Changing the value of the pot without changing the value of this internal resistor will change the proportion of current flowing through T2 (and possibly into the chip). This may initially alarm you, but don't forget that by increasing the value of the pot we're decreasing the total current, so even if T2 is taking more current proportionally, it's still taking less current. The only other possible effect is that it may slightly distort the linearity of the pot more than the design was intended for by changing pot size without changing the internal resistor size, but we're talking about negligible amounts.

That is what my logic and limited circuitry experience tell me. I'm here asking if there is a glaring hole in my logic. If there is, please explain it so I can understand. This isn't just "tell me how to do it", it's "teach me so I know".

Have you ever designed chip level stuff.
 
I have used them and 1K AB 800T pots with no trouble ever. Most vendors (the good ones) tell you what to use per their design. I try to adhere to that. I used to have a good source for used (ie cheap) AB stuff, but no longer so I usually use the 22MM stuff from various vendors now.
 
It's just a voltage divider circuit, but understand that it is going into an A/D chip on the board. So messing with the resistance changes the resolution of your pot turning. If the A/D chip is looking for 2ohms of resistance max, and you use 10ohms, the voltage change into the A/D chip across the 2ohm pot is achieved by just 1/5the the amount of movement in the 10ohm pot. So assuming a 270 degree pot, min to max on the 2ohm is the full 270 degrees, min to max on the 10ohm pot will be just 54 degrees and further movement become irrelevant. So you will lose your ability to fine tune your speed command. Some drive will allow you to customize the gain on that signal, but not all of them.
 
Nobody has yet mentioned what is probably the real reason the manufacturer calls for a relative low ohm pot.

Chip level designs or, not.

VFDs often run in electrical environments that are pretty noisy. There's an odds-on chance the manufacturer
deliberately put a shunt resistor across the AD input (which by itself is most likely a megohm or more) and also
some bit of filtering or other protection (MOV maybe) simply to keep the analog signal presented at the
input smooth and clean.

If they put a 10K or a 5K resistor at the input as a shunt, then a 10K pot would give a bit of a non-linear behavior.

I'd go with the high end of what they called for. If I were really worried, I't try a 5K pot just to see how it behaved.
 
Just about any of them will have a buffer on the input, but some MAY not. Those may have a voltage divider setup to get the 0-10V reduced to what the A/D can use. So the input of the divider is what your pot will "drive".

You should figure that the input resistance MAY be in the 25k ohm range. What that does is to drag down the output of the pot.

At half rotation, the output impedance of the pot is 1/4 the total resistance. So, with a 1k pot, it is 250 ohms. That is approximately 1% of the 25k considered as a divider. So your worst error from that source is about 1%. Most likely that is what the designer figured on.

If you use a 10k ohm pot, the source resistance is 2500 ohms. That is 10%, so the voltage at half rotation will be lower. The control will be non-linear, and will, as the pot resistance goes up, start to look more like a log taper pot.

Maybe that is of no consequence to you, and if so, a 10K pot is just as good. But maybe it IS important, and if so, you need to take the input resistance into account. The spec for that should be in the manual.

For Rons:

1) It sill be an A/D input, definitely analog, so digital will not be an issue.

2) Chip level (internal chip design) is not the issue here, all you need are the specs of the device. If you mean design at the level of assembling ICs and parts to make a functional circuit, well, yes, that is what affects the specs of the device. But you still only need the device specs.

And, in my case, yes, I have been doing IC and discrete part level design for almost 50 years now. Started with tubes, actually.

jraef:
That can be true, I suppose, but in most cases I see, the A/D has some sort of passive or active circuitry ahead of it that isolates the iput from the A/D somewhat. A buffer, or just a divider. The impedance of the divider output is the designer's issue, your issue is to drive the divider etc properly, which means a low enough source impedance to avoid nasty non-linearity. (or maybe you were actually meaning the non-linearity? I re-read, and it can be interpreted that way also)

Jim...

Low impedance is good, but of course they "expect" that you will use shielded cable for any sensitive circuits, plus they almost certainly have some sort of HF filter on the input to get rid of garbage.
 
Jraef

Please explain the difference between 40 degree rotation on a 1k or a 10K potentiometer with a constant voltage across it. I see 40/270 X Vref from either. The different resistance will affect the current drawn but not the Vout.
 
"Started with tubes, actually.

Tubes, of what?

:-)

project_2.jpg
 
One consideration I did not see referenced in this discussion is the power rating of variable resistors. If a pot has a power rating of 1 watt, it is 1 watt for the entire resistor element. If the wiper is at 1/4 rotation the power rating may be only 1/4 watt. I realize that this depends on the circuit design and what the pot is controlling. If a 1K pot is specified perhaps the normal operating position is 1/2 to 3/4 (500 - 750 ohms)from the end. If a 10K pot is substituted and only 500 to 750 ohms is needed, you are using 1/10 of the element and the power dissipation capacity could be about .1 watt. If you try to dissipate 1 watt of power in a .1 watt resistor the magic smoke is usually released.

Bob
WB8NQW
 
In the case of VFD speed input I believe the wiper signal would go to a high impedance input and have negligible effect on the power rating. The rating would be for the full resistance dropping the supply voltage, usually 10 volts. So a 1k would need a continuous rating of 0.1 watt etc.

 
In the case of VFD speed input I believe the wiper signal would go to a high impedance input and have negligible effect on the power rating. The rating would be for the full resistance dropping the supply voltage, usually 10 volts. So a 1k would need a continuous rating of 0.1 watt etc.


Yes.

But do not count on the high impedance, depending on what you call "high". Many are in the 25k ohm region, and then will significantly load a 10k control, while retaining very good linearity with a 1k control.

However, another consideration is the source inside the VFD. You can go UP in control resistance, if you can tolerate the loading effects, but the capability of the source determines the MINIMUM resistance. The source is a highly regulated voltage, so that it will not change and affect the speed signal. Such supplies have limited current capability, and you need to avoid using a control of a lower resistance than the spec allows. A low resistance may cause the maximum voltage to drop.
Y
 
1k pot burning out sounds very strange to me. Are you absolutely sure that the connection is correct?
0.1 watts is rather tiny amount of power even for smaller resistors and pot track has typically plenty of real estate to dissipate the power.
If you accidentally connect the 10v supply between the pot wiper and one of the ends it's destined to fry..
 








 
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