Armature Current Limiting

[fechter]

You could use the bidirectional one, but to feed an analog meter, you would need to put the equivalent of a 2.5v zener diode in series with the output to offset the zero voltage. There are other ways of doing this with an op amp (more complicated).

In a brushed system, the bidirectional feature is not needed. Even in a brushless system, you could get by with a unidirectional sensor since you can easily filter the output. The actual motor current won't change that fast unless it shorts out or something.

I guess you should have bought the cheaper one, but the one you have will work.

 

[safe]

the part that fechter specified is the ACS755 not the ACS754 that you purchased.

I wanted to buy the ACS755 but Digikey was out of them and they couldn't get them in stock for a few months. It was a case of:

"Oh heck, I can proceed with the alternative or remain stalled in inaction for several months."

...so I just went ahead and bought them.

If anyone can find the older ACS755 current sensors ANYWHERE let me know... they seem to no longer be in circulation and are probably being phased out.

Found one:

http://www.cavalierstock.com/advanced_search_result.php?keywords=ACS755SCB-050-PFF&p=ALLEGRO+MICROSYSTEMS+Quantity%3A+130

...only it's a ripoff. They are asking $17.76 per item.

 

[safe]

You could use the bidirectional one, but to feed an analog meter, you would need to put the equivalent of a 2.5v zener diode in series with the output to offset the zero voltage. There are other ways of doing this with an op amp (more complicated).

What I really want to do is remap the current voltage like this. This way the pot control knob would have more precision within the range of values that I care about. The precision of the current sensor is about a percent or two at most, so the only area that really matters is the ability to set the control knob well. However... sometimes "Keep It Simple Stupid" is the easiest thing to do and it might be "good enough" to just trim the knob based on the original values if there was a digital readout that was also accurate. This is a calibration issue more than anything else. Another advantage of this is that the current sensor uses a little of it's own current and the starting voltage might not be exactly 5 volts. So if the digital readout told you that before you started to ride the voltage was say:

2.36 volts

...then you could do the quick calculation of:

2.36v + 0.80v = 3.16 volts

...to get to 40 amps.

Temperature on a hot day might mean that the starting voltage for the current sensor is:

2.38 volts

...then you could do the quick calculation of:

2.38v + 0.80v = 3.18 volts

...to get to 40 amps.

So you get the idea.

 

[rkosiorek]

it looks like these are just so popular that they are becoming a non-stock item for all of the suppliers. digikey will even stop stockingt the ones you bought once they run out of stock. Allegro will still be producing them but we would have to buy them in lots of 510 pieces each. and it is for all of the ACS75X series of chips. which for people like us who only need onesies and twosies as good as out of production.

worlwide the only ones i could find in stock are the ACS755 in the 150A version. digikey still has 644 of those.

as convenient as designs based on these suckers might be looks like other alternatives should be examined for future use.

still need to think a little on how to use up the ones that you bought though. (you got almost the last of those)

rick

 

[safe]

ACS750SCA-050 — ALLEGRO MICROSYSTEMS — Hall Effect IC

$5.97 1639 in stock

http://www.newark.com/81H6588/semiconductors-prototyping/product.us0?sku=ALLEGRO-MICROSYSTEMS-ACS750SCA-050

While this is +/- 50A it still is better precision at 40mV/A verses the 20mV/A that the +/- 100A does. The best is without a doubt the ACS755 at 60mV/A but as you said they aren't easy to get in single quantities.

Besides... the way I have it now if for some reason I want to go up to 60 amps or more the circuit can handle it without modification.

The easy way is just to use my multimeter and calibrate it once and then forget it. The more sophisticated way would be to give it it's own digital readout so that I could watch the voltage. The top of the line idea would be an actual ammeter.

The whole point is heat reduction... I really am not going to want to watch any meters while riding because in an Armature Current Limiting situation the current is a constant anyway. (there's not much to watch at full throttle)

Here's how this different sensor might look...

 

[rkosiorek]

you can do pretty much the same with the chip that you already have, an op amp and a few resistors.

with this circuit R1 and R2 are set up as a voltage divider that splits the supply voltage in half. and this value is subtracted from the output of the ACS754 chip in the OP Amp. the OP Amp is also used to amplify the resolution of the ACS754 to scale the output to the range that you require. except for the fact that this circuit output starts at 0V instead of the 0.6V that happens with the ACS755 and it is not limited to 50A range, the output would be identical to the 50V unit.

since you would now be using one amp for this conversion and the second for the throttle you would need a third for the ammeter if you still wanted to add one.

rick

 

[safe]

Hmmmm... that does a good job of remapping the one to the other.

I just bought two of the ACS750 circuits from Newerk so I'll have those to play with too.

More options are always better... so with the ACS750 circuits I'll have 40 mV/A verses the 20 mV/A and that means more precision. Every time you do an expansion like that you will amplify the error. (at least that's what I would expect) The ideal would be the ACS755 because at 60 mV/A you have a lot to work within.

This electrical components stuff gets frustrating because the important parts seem hard to obtain. It would be nice if you could go to one place and get all you need. (ideally like at a RadioShack store so that you could just go and get it)

Again, nice job on that remapping circuitry.

Any chance you could post the math equations that define the resistors?

(so that I can understand what is going on?)

So let's say I wanted the scaling to go from 0v to 2v... how would I calculate that?

Do it like it were in a classroom environment... assume I'm an idiot. (which is not that far off when it comes to this stuff)

 

[safe]

Voltage Dividing, Slicing, Shifting?

At the end of the day all that matters is that the pot control knob is sensitive within a range that matches the current sensor output. There are two ways to achieve that, you can remap the current sensor to get it to fill out a 0-5v range. Or you could remap the pot control knob so that it matches the current sensor.

In some ways I'd be happy to just remap the control knob.

So if there was an easy way to make the pot control knob have a range of from 2.5v when the knob is closed to either 3.5v (ACS754) or 4.5v (ACS750) for when it's opened then that's really about all I need. When the comparator sees that the motor current is high all it does is "dump" the throttle current until things come down. The throttle current only "sees" that it's current is being consumed elsewhere and doesn't really care about the decision making process that made it happen.

So either way would work... I suspect that remapping the pot control knob will be the easier way. Plus, if I mess with the current sensor voltage then I'll have a harder time figuring out what it's doing. There's a strong argument for simply doing nothing and just measure the raw output... it's guaranteed by the factory to be within 1% of it's intended target. (you have the highest accuracy before you start to modify things)

Thought

Why couldn't I just do Voltage Dividing like was done above (one resistor on the source and another on the ground) and use that as one half of my pot voltage and then use the full source voltage on the other?

Wouldn't that give a pot that went from 2.5V to 5.0V?

That's all I really want...

 

[fechter]

Right, it would be much easier to make the control pot range match the current sensor. You can do this with a voltage divider, as you suspected.

In the drawing below, I think is how you would do it:


The voltage across each resistor is Rx/(R1+R2+R3) x 5v, where Rx is the resistor in question. So, to get 2.5v at one end of the pot, for example, R1+R2 should = R3.
The voltage across all three resistors = the source voltage (5v in this case).

 

[rkosiorek]

the parts store business that i was thinking of would be precisely for those hard to get oddball bits that our e-bikes and vehicles require. may be not quite as specialized as these current sensors but the other annoying things.

DC to DC converters, FETs, Brake Sensors, odd ball size axle nuts and serrated lock washers, weird connectors, plastic enclosures that can be handle bar mounted some switches, diodes to parallel battery packs, relays for dual speed motors, etc, etc. i might even include some things like this throttle circuit.

since it will deal with an odd assortment of parts i am really leaning towards the Acme Diabolical Devices name.

but i digress.

though there seems to be a lot of stock around for these ACS750 series chips. they don't seem to be a currently manufactured product. they were not listed as part of the 2007 lineup, nor are they listed in the selection guide.

i am only pointing this out as it may be very difficult to duplicate this a year or two from now if we cannot get the parts. i have sent an email to Allegro to see what their future plans are for these parts and find out what their future upgrade path will be.

as to the math here goes.

R1 and R2 where chosen more or less arbitrarily. i just wanted the voltage divider to not load the throttle circuit voltage regulator too heavily.
R3 and R4 where chosen to be at least double of R1 and R2 to limit interaction.

the ratio of R4 and R6 set the gain of the amplifier. i wanted the amplifier to multiply the input signal by 3 so R6 = 3X R3. that is just one of the standard design guidelines for standard Op Amps. and R5 is the same as R6. again that is just standard practice for OP Amps.

rick

 

[safe]

The voltage across each resistor is Rx/(R1+R2+R3) x 5v, where Rx is the resistor in question. So, to get 2.5v at one end of the pot, for example, R1+R2 should = R3.
The voltage across all three resistors = the source voltage (5v in this case).

Wow, you beat me to it...

This was the idea that I had. Have I stumbled across the same result? (just so I know my thought process is correct)

This would go from 2.5v to 5.0v right?

 

[safe]

Math Class

There are two voltages that I want to get:

On the lower end I want 2.5v.

On the higher end I want either 3.5v or 4.5v.

We do the lower first:

R2 = R1 / (5/2.5 - 1) = R1 / (2 - 1) = R1

...so for the lower value we need two identical resistors. (any value seems okay)

We do the upper next:

R2 = R1 / (5/3.5 - 1) = R1 / (1.43 - 1) = R1 * 2.3
R2 = R1 / (5/4.5 - 1) = R1 / (1.11 - 1) = R1 * 9

...so for the upper value we would need a resistive relationship of 2.3 or 9 to 1.

Aside from the fact that I'm using four resistors to get this done am I on the right track?

The pot seems to introduce an extra resistor that needs to be calculated differently... hmmmm.... pots don't seem to follow the voltage divider formulas in the same way...

 

[safe]

Kirchhoff's Voltage Law and Voltage Divider Rule

Technically speaking there are two principles taking place here. (I think) Kirchhoff's Voltage Law says that the voltage drop across a circuit adds up to zero. Then the Voltage Divider rule is what gives us the output voltage that we use in the comparator circuit.

So you can do it all at the same time. (as Fechter hinted at, but didn't explain fully for a novice like me)

Now I just need figure out the equations for this.

R1, R2, and R3 need values...

 

[safe]

The voltage across each resistor is Rx/(R1+R2+R3) x 5v, where Rx is the resistor in question.

Any chance you could "cheat" and give me an answer for the resistors so that the voltage range is:

2.5v to 3.5v

or

2.5v to 4.5v

...assume a 10K pot.

Showing your work is even better.

 

[rkosiorek]

one too many 3.3K resistors. you just neet 2 resistors and the 10K pot.

if the top of the 10K pot is to be at 3.5V for example and the bottom is at 2.5V then we are dropping 3.5-2.5 = 1.0V across the 10K pot. so the total series current would be Ipot=Epot / Rpot = 1.0V / 10K = 0.1mA

in a series circuit the current remains the same at any point. and will in this case be 0.1mA

the upper resistor drops 5V - 3.5V = 1.5V. and it drops this at 0.1mA. Rtop = Etop / I = 1.5V / 0.1mA = 15K

the lower resistor drops 2.5 V. Rbottom = 2.5V / 0.1mA = 25K.

so the chain is 15K + 10Kpot + 25K at the bottom.

for the 4.5V case the numbers are calculated in the same way.

voltage drop across the pot is 4.5V - 2.5V = 2.0V
current in the pot is 2.0V / 10K = 0.2mA

the top resistor handles 5V- 4.5V = 0.5V
the resistor to do so would be 0.5V / 0.2mA = 2.5K

the bottom resistor still handles 2.5V and it will be 2.5V / 0.2 mA = 12.5K

the 4.5V chain would become 2.5K + 10K pot + 12.5K

the circuit that you drew is more complex and is called a series/parallel circuit. the lower 3.3K pot would recieve current from two sources one is a 3.3K resistor and the other a 13.3K equivalent resistor made up from a 3.3k in series with a 10K pot. the total resistance of this would be 1/ (1/13.3 +1/3.3) or 2.64K the bottom of the pot would not be at 2.5V but instead at (5.0V /( 2.64K + 3.3K)) X 3.3K = 2.78V the top of the pot would be
5.0V - (((5.0V-2.78V) / (10K + 3.3K)) X 3.3K) = 4.45V so the top is close to what you need but the bottom is not. the simpler series circuit is what you need.

okay i did the cheat.

rick

 

[safe]

Excellent

I really liked how you went through the thought process of how to solve the equations. Before I took a break I was starting to think that you really couldn't know the value of the resistors until you figured out some common current that would go through the circuit. Your multi-step approach was exactly what I needed to see.

So now I get it...

 

[safe]

So the circuits will look like this...

 

[safe]

Next Step

Figuring out the best way to place the components on the board and how to solder it all together. This would have to be called the "circuit board design" stage...

 

[rkosiorek]

why put the current sensor on board and lead some big-honkin-fat wires up to it. the sensor should be mounted close to if not within the motor. the fat high current wires should be as short as possible.

you should connect some small 24awg or 22 awg signal wires to the sensor and lead those up to the board.

i would connect the wires to the sensor and encapsulate it in epoxy. easiest way is to stuff it and the wires into a short piece of small diameter PVC pipe or even a small PVC pipe cap and epoxy away.

rick

i see that all you are doing is using this as a straight current limiter. not a bad idea.

 

[fechter]

OK, I think he's on to something here. The resistor values look OK now.

As far as mounting the sensor, Rick is right, and I would add that mechanical stress of fat wires bending the leads would be a potential failure. I'd recommend some way to anchor the wires on either side of the sensor. Just a piece of fiberglass or stiff plastic with holes drilled in it so you can zip tie the wires to the board to take the stress off the sensor would work. There may be better ways.

 

[rkosiorek]

fully encapsulating it in epoxy would also strain relief the wires. also provide some weatherproofing.

i like the idea of an add on current limiter.

but for my own personal experimentation i think that i would like to try a current based throttle. but i think that i would leave that for a different thread in the future.

rick

 

[safe]

You might think that this entire circuit would be located up front near the handlebars or somewhere like that. But there's no real reason to locate it there since the only control that this circuit performs is current limiting with a pot control knob. So I was planning to simply place the circuit in between the connectors that attach the controller to the motor... what could be better?

Another reason to give some space between the motor and the sensor is that these sensors are vulnerable to inaccuracy if the temperature changes. So for heat reasons the idea of locating the sensor near the motor is unwise.

Now I need to go out and shovel a couple inches worth of snow off my driveway... I'm looking forward to summer... :|

 

[rkosiorek]

sure you can do it that way. the disadvantage is that you will be adding a long length of cable to the controller / motor loop. the longer the cable here the more it will affect motor performance and the more heat that will be wasted in the wires.

it just makes more sense to mount the sensor remotely and close to the motor.

rick

 

[fechter]

Well, I wouldn't put it too close to the motor since the magnetic field from the motor magnets could affect the reading (a lot). It would be better near the end of the motor rather than the side of the motor. I'd stick it closer to the controller. You want the wiring from the motor to the controller to be as short as possible, but the sensor has to be in between somewhere.

 

[safe]

...the disadvantage is that you will be adding a long length of cable to the controller / motor loop.

No actually I was going to shorten the controller-to-motor wires and insert the ACL circuit in between. The length of the wires is unchanged.

You have to realize that the ACL circuit is NOT going to be on the handlebars or anything like that... it will be in the middle between the controller and motor.

If I decide to add an analog or digital panel meter then I would string those wires all the way up to the handlebars. The most logical location for this circuit is about two inches from the controller and it will simply act like a "controller add-on" that you plug in... (I like the "plug and play" concept of modularity)