Electric Vehicle C

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Matthew
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Re: Electric Vehicle C 2009

Post by Matthew » December 21st, 2008, 12:18 pm

So I haven't seen electric vehicle run at nationals yet. But I will defer to experience here:

198.5/200 is a good score. But based on what I saw over the years with Mousetrap Car, Scrambler, Wheeled Vehicle, etc. is that at the early rounds, being consistent is most important. If you have a car that can CONSISTENTLY get a very good score, you'll probably win. The microelectronics will help you do that. This is especially true at the states level.

But nationals is a whole different ball game. Many of the teams will show up with a car they've been perfecting over the last few months. By the time everyone takes their run, you need a bit of luck to win. I remember at nationals in 2000, the winning scrambler was within a credit card's width from the wall. The medalling scramblers were all within 1-2 cm of the wall. Based on the range distribution of their two runs, any of those medalling devices also could've been within the winning distance. They also could've all hit the wall. By the time you get to nationals, consistency is pretty much the standard: you need luck to win.

So back to the question at hand: if you need microelectronics to build a consistent car, use them. If you've got something up your sleeve that doesn't use them, those 0.5 points can make the difference between no medal and winning- the scores will be that closely distributed.
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Re: Electric Vehicle C 2009

Post by r00bin » January 3rd, 2009, 10:41 pm

Hi,

I'm currently working on electric vehicle and I'm trying to use the lego RCX micro controller. The problem I'm having is with power. The rules specify that only "4 batteries rated 1.5V" maybe used. The RCX is designed to use 6 AA batteries. I tried running the processor on only 4 batteries shorting the other two sets of battery terminals but was unable to even download the RCX firmware. I read that some of you have used NXT systems which also are designed for 6 AA bateries. How are you guys running the microprocessor on less then 9.6V?

Thanks


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Re: Electric Vehicle C 2009

Post by gh » January 4th, 2009, 7:47 am

Also a couple of images for how you might want to add the TI boost module to your battery case:
4/IMG_4171.jpg4/IMG_4166.jpg
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Re: Electric Vehicle C 2009

Post by r00bin » January 4th, 2009, 10:05 am

gh wrote:Also a couple of images for how you might want to add the TI boost module to your battery case:
4/IMG_4171.jpg4/IMG_4166.jpg

Is that what the sample voltage booster from the TI website looks like? It looks as though the center PCB is the booster and it's connected to a home made circuit wafer with two capacitors. What made you use those two capacitors? Are they required? A quick search on Wikipedia at http://en.wikipedia.org/wiki/Boost_converter says

"Filters made of capacitors (sometimes in combination with inductors) are normally added to the output of the converter to reduce output voltage ripple."

So how did you choose those particular capacitors?

Thanks
Last edited by r00bin on January 4th, 2009, 10:14 am, edited 1 time in total.

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Re: Electric Vehicle C 2009

Post by Dark Sabre » January 4th, 2009, 10:10 am

Check out the product documentation for what the "booster" looks like.

The first page of that documentation also shows a typical circuit, which has those same capacitors (which I believed are being used for voltage smoothing).

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Re: Electric Vehicle C 2009

Post by r00bin » January 4th, 2009, 10:16 am

Dark Sabre wrote:Check out the product documentation for what the "booster" looks like.

The first page of that documentation also shows a typical circuit, which has those same capacitors (which I believed are being used for voltage smoothing).
AH!

Ok, things are making more sense. :D

Thanks

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Re: Electric Vehicle C 2009

Post by gh » January 4th, 2009, 1:11 pm

r00bin wrote:Is that what the sample voltage booster from the TI website looks like? It looks as though the center PCB is the booster and it's connected to a home made circuit wafer with two capacitors. What made you use those two capacitors? Are they required? A quick search on Wikipedia at http://en.wikipedia.org/wiki/Boost_converter says

"Filters made of capacitors (sometimes in combination with inductors) are normally added to the output of the converter to reduce output voltage ripple."

So how did you choose those particular capacitors?

Thanks
Boost converters use a voltage "chopping" action in an inductor to cause magnetic field collapse, a large flux, and subsequently a high(er) voltage in the induced current. The switching frequency is quite high (this one is 650 kHz to 1 MHz), and usually happens to be right in the frequency range that would cause problems with microelectronics. Its output has a ripple because of the above described action and it also draws current from its input at the same frequency, so you get ripples in the input and the output. That is why you need two capacitors to filter out the noise.

The capacitors also serve as buffers that store energy, so if you have sudden change in current (like starting or stopping a motor), it gives the boost converter time to adjust to the load, and prevents the output voltage from suddenly spiking or sinking.

~edit~
Oh, and it'll probably work without the capacitors, but that's really not recommended. It'll wreak havoc on all your circuits and you wouldn't even know where it's going wrong. Bypass capacitors never (well, rarely) hurt anyone.
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Re: Electric Vehicle C 2009

Post by r00bin » January 5th, 2009, 12:31 pm

I have another question about determining what resistor to use.

In the post by nickfastswim, he suggests
Well it was a old converter i found laying around in my stuff, so don't remember where i got it.
Its approximate specs are:
Voltage in (min)~4.8v
Voltage in (max)~12v

Voltage out (min)~15v
Voltage out (max)~18v

About 10W - 20W output power


Basically, you will need a DC/DC converter that has a input voltage of about 3v-6v and output voltage 9v-12v and 10w-15w of power to make the Lego RCX work.

I let P=15 and V=9.6. I then used P=V^2/R to solve for R. My resistance was calculated to be 6 ohms. This is vastly different then the 3.65 Kiloohms specified in the data manual. The data manual also states that I should use the equation R=15000ohms*2/(V-5)-2.94 to determine the resistance based on my desired output voltage. I'm confused as to how this will output will have enough current and power to run the RCX when power is conserved.

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Re: Electric Vehicle C 2009

Post by Dark Sabre » January 5th, 2009, 4:18 pm

Well, I never used A DC/DC converter to take the voltage up for an RCX, since my experience with this was when there was a single battery requirement for the Mission Possible event and I was using a 12V and then down-converting it to 9V for the RCX. What I was using was capped at 500ma, so my speculations on the power you need to be supplying is based on that.

I also used this very interesting page about lego motors for reference back then.
The "Speed and current vs. torque" graphs for the two RCX motors are of particular interest, since you can figure out what the current consumption at stall torque (when RPM reaches 0, torque is at the max). Since the motors I was running off of the 500ma supply were never loaded nearly that heavily and I never ran all 3 at once, I'm comfortable with the graph's assertion that you are not going to be pulling more than 400ma per motor at stall torque.
You can get those same numbers from his "Stalled characteristics" table, but you don't get to see the nice RPM/mA/torque relationships.

He lists 360 mA and 340 mA as the stall torque current draws for the two RCX motors, so assuming you are using two motors to move your EV, the motors would be drawing less than 7W if you placed a large brick on your vehicle such that it couldn't move.

So I don't know why you would as much power as nick suggested. Your margin of safety is partly that you are never going to be operating at stall torque.

As for picking a resistor, I would use the formula that the datasheet provides, though I'm not quite sure where the "3.65 Kiloohms" you cited came from.

Using that power equation like you did is operating under the assumption that the resistive load is what it is dissipating the power. The chip in that converter is using the resistor as a reference for what voltage it should be outputting, which is why it says:
"A 0.05-W rated resistor may be used. The tolerance should be 1%, with a temperature stability of 100 ppm/°C (or
better)."
Since the resistor is not dissipating any real power, it can have a very low wattage rating. However, since it is being used as a reference for the output voltage, the tolerance is very important...hence the 1%.

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