Stalling the BaneBot 775 motors

[Al Skierkiewicz]

Ether,
I am basing the statement on the experience of rise time vs inductance with any motor. Since the Jaguar is switching at 15kHz, current through any motor may not be able to rise to full stall current during a short pulse. I am simply suggesting that an estimate of power dissipated can be made if one ignores rise time for this discussion.

[Ether]

Originally Posted by Al Skierkiewicz

I am simply suggesting that an estimate of power dissipated can be made if one ignores rise time for this discussion.

If you "ignore rise time" and assume that "full power is being applied during the pulse" you will get a very wrong answer if your assumption is wrong.

I do not know what the inductance of the RS775 is, but even if it is as small as 20 micro Henries, then power consumption at 20% duty cycle (with a Jag and locked rotor) will be roughly 40 watts, not 213.6 watts.

If anyone reading this has access to the necessary test equipment and would be willing to measure locked-rotor inductance of the RS775 and post the result here, I will run a simulation and post a graphic of the current waveform (and power calculation).

[fox46]

For the purpose of this discussion I think we are getting too far into the theory behind the component. It would be a very difficult piece of hardware to work with if it only output 4% power when you tell it to give 20%. For the purpose of this discussion and our use, I am quite sure that the Jaguar compensates for the exponential effects of Inductance/switching frequency/rise time and changes its pulse frequency so that the output power varies linearly with the input instructions. In this case, an instruction of 20% output would result in an intuitive 20% power (W) output and thus ~200W.

If the jaguar input controlled the switching frequency- linearly, then I think Ether is right. After one of our team members hooked up a jaguar backwards today, I opened it up to take a peek inside. The output has a capacitor on it- I believe this would effectively "smooth" the output power and generate a physically lower voltage/amperage. With the capacitance/inductance/resistance and switching frequency you effectively have an RLC circuit. Relative to the switching frequency, output would indeed change exponentially just like Ether's calculation.

I believe we are just thinking of two different parameters of which we are commanding 20% thereof- My thinking is that the input commands control the power output wheras Ether's thinking is that the input commands control the switching frequency. I'm going to stick with the former so tell Regis my final answer is ~200W!

If you know what the frequency is (someone with an oscilliscope?), we know what the percieved output voltage, current and resistance is at full power, I can take a look at the guts of the Jag and see what the capacitance is. Can we not deduce the inductance?

[Ether]

For the purpose of this discussion I think we are getting too far into the theory behind the component.

This is pretty basic AC circuit theory, and it has a profound effect (40 watts versus 200 watts) on the answer you get, so it's worth understanding.

I am quite sure that the Jaguar compensates for the exponential effects of Inductance/switching frequency/rise time and changes its pulse frequency so that the output power varies linearly with the input instructions.

Sorry, it doesn't work this way. First, the output frequency does not change at all. It is a constant 15000 Hz. Only the duty cycle changes. The duty cycle varies linearly with input command (unless, of course, you are running closed loop, but that's another story entirely).

In this case, an instruction of 20% output would result in an intuitive 20% power (W) output and thus ~200W.

It doesn't work that way. A 20% command gives a 20% duty cycle.

If the jaguar input controlled the switching frequency- linearly, then I think Ether is right.

The Jag does not control switching frequency. The frequency is held constant at 15000 Hz, which gives a pulse width of 66.7 microseconds. What the Jag changes is the "on" portion of that 66.7 usec.

After one of our team members hooked up a jaguar backwards today, I opened it up to take a peek inside. The output has a capacitor on it- I believe this would effectively "smooth" the output power and generate a physically lower voltage/amperage. With the capacitance/inductance/resistance and switching frequency you effectively have an RLC circuit. Relative to the switching frequency, output would indeed change exponentially just like Ether's calculation.

Not sure what you mean by "exponentially" here. The effective voltage changes linearly with input command, and the effective current also changes linearly. The effective power thus changes as a square function.

I believe we are just thinking of two different parameters of which we are commanding 20% thereof- My thinking is that the input commands control the power output

They do, but not linearly.

wheras Ether's thinking is that the input commands control the switching frequency.

No, it controls the duty cycle, and thus the effective voltage and current.

I'm going to stick with the former so tell Regis my final answer is ~200W!

Are you a betting man?

If you know what the frequency is

It's 15000 Hz.

Can we not deduce the inductance?

If you knew how many windings, and the core material magnetic properties, you could make a rough calculation.

[AustinSchuh]

Originally Posted by fox46

... The output has a capacitor on it ...

http://www.luminarymicro.com/index.p...8&Itemid=59 1

This user manual for the black Jaguar has the schematic. I'm not sure which capacitor you were seeing, but there is no capacitor on the output of the Jaguar. There is a bootstrap capacitor in the circuit, but will not generate the effect you are describing.

[fox46]

Whatever the capacitor is in the middle of the ring of transistors is.

For the purpose of this discussion I think we are getting too far into the theory behind the component.

This is pretty basic AC circuit theory, and it has a profound effect (40 watts versus 200 watts) on the answer you get, so it's worth understanding

I agree but seems like this is much more than a basic AC circuit when we are apparently not dealing with changes in frequency but rather changes in duty cycle. The original purpose of the OP's question was to determe if it was acceptable to stall the RS775 at 20% power. My OPINION is that it is not acceptable, but by all means, go for it- you are only burning up 40W right? Last I checked, a 20% position of a joystick did not result in a 4% robot speed.

The duty cycle varies linearly with input command

Okay but we are not concerned with the duty cycle but rather the output power which according to you is not linear to the input command. I disagree with this.

Not sure what you mean by "exponentially" here. The effective voltage changes linearly with input command, and the effective current also changes linearly. The effective power thus changes as a square function.

Sorry, this is what I meant- Power is not linear to input using your method, as I stated earlier. "Exponentially" was the wrong term to describe this.

In any case, I still think the output is closer to 200W than 40W. I'll bet you the OP's RS775 . Let me know what your simulation shows. Until we come up with some actual Jaguar output wattages relative to control inputs lets just agree to disagree. When we get our software up and running and if I get around to it. I'll measure the percieved voltage and current going to our RS775s and plot it relative to input value in the program. Then we can see if the output power is linear or nonlinear to input command.

[Ether]

Originally Posted by fox46

Whatever the capacitor is in the middle of the ring of transistors is.

Here's the FET bridge portion of the Black Jag. As you can see, there is no filter capacitor on the Jag's output.

I agree but seems like this is much more than a basic AC circuit

With the rotor stalled (which is what we're discussing) it's just a switched LR circuit.

The original purpose of the OP's question was to determe if it was acceptable to stall the RS775 at 20% power.

Correct.

My OPINION is that it is not acceptable, but by all means, go for it- you are only burning up 40W right?

I do not know how long the RS775 can tolerate 40W without overheating. I suggested to Leav that he run some low-power tests to determine at what point the motor starts to become uncomfortably warm to the touch.

Last I checked, a 20% position of a joystick did not result in a 4% robot speed.

I don't know where you got that from. I hope you do not think I am claiming that.

Let me know what your simulation shows.

I already ran the simulation with 20 uH and it shows 40 watts. I do not know what the actual inductance is but I suspect it is higher than that.