Some Gecko 201 Questions

[mikie]

Hi guys,

I'm going to use the Gecko 201's in my project but I've managed to confuse myself in the process. I have a couple of questions I was hoping I could get answered to clear things up.

My motors range in voltages/currents, from memory they lie within 5V - 7V and 1.5A to 5A.

Questions:

1. The gecko manuals state that it requires a minimum of 24Vdc to operate correctly and will operate up to 80Vdc. I assume that if I build a supply of around 30Vdc (this figure was obtained from another post as a recommended size) that the drives will regulate the voltage/current to my motors?

2. Is the 30Vdc a good size pwr supply to have? What current should I be looking for when building this supply. I assume if I size it for the combined current ratings of the motors that I want to use, plus 30% for other potential uses and to keep the supply within a good limit.

This is probably a good set of questions to start with.
Any help appreciated

Mike

[ger21]

1. The Geckos regulate the current to the drive. The voltage doesn't matter to the motor, but higher voltages allow the current to not fall off at higher speeds. (see 2 below)

2. If the lowest voltage motors you have is 5V, I'd get at least a 50V supply, minimum. Usually 10x to 20x motor voltage is recommended with Gecko's. 30V would just barely get you over 4x on the 7V motors. The higher the voltage of your supply, the higher top speed you're motors will spin at.

[arvidb]

Hi guys,

*snip*

2. Is the 30Vdc a good size pwr supply to have? What current should I be looking for when building this supply. I assume if I size it for the combined current ratings of the motors that I want to use, plus 30% for other potential uses and to keep the supply within a good limit.

*snip*

The Gecko Drive Power Supply White Paper states that the drive will always draw less than 2/3 of the rated current of a parallell/half-winding connected motor, or 1/3 of the rated current of a series/full-winding connected motor.

That said, since the Geckos are chopper drives, they conserve power and this means you should need less current with a higher voltage power supply than with a lower voltage one. I'm not really sure where this comes into play... would Mariss care to comment? (Or anybody else that know of course.)

BTW, the White Paper doesn't say anything about that the AC current from the transformer of a linear power supply will be 1.4 times the DC current that is drawn (power in = power out, voltage out = 1.4 x voltage in => current in = 1.4 x current out). Also, because of the very spiky current waveform that is the result of the rectification process and the fact that copper losses are proportional to current squared, I've read that it's best to dimension the transformer for 1.8 times the DC current required.

Arvid

[mikie]

Thanks guys,

This has helped, i'll have to design a pwr supply with this in mind.
Looks like some maths this weekend.

Mike