tag:blogger.com,1999:blog-7276377053120174333.post5676620691823642578..comments2017-05-19T09:38:15.813-04:00Comments on EEG Hacker: Estimating OpenBCI Battery LifeChiphttp://www.blogger.com/profile/10352943033779293161noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-7276377053120174333.post-6457997744286661242015-07-11T01:02:51.890-04:002015-07-11T01:02:51.890-04:00Hi Chip,
We are in agreement, thanks.
Regards,
Er...Hi Chip,<br />We are in agreement, thanks.<br /><br />Regards,<br />Eric<br />Eric Tianhttp://www.blogger.com/profile/09238086019846686696noreply@blogger.comtag:blogger.com,1999:blog-7276377053120174333.post-23999197243491893682015-07-10T21:06:04.340-04:002015-07-10T21:06:04.340-04:00I think that we're in agreement. OpenBCI has ...I think that we're in agreement. OpenBCI has effective resistance of 96 ohms. We were applying 6V to OpenBCI and we measured 62 mA going into OpenBCI. Therefore, the OpenBCI board has an effective resistance of 6V / 0.062mA = 96 Ohm. We agree.<br /><br />In using the datasheet, the datasheet only speaks to one battery. We cannot use the discharge curve for a 96 ohm device because OpenBCI isn't being powered by one battery. It's got four batteries in series boosting up the voltage (the current through each battery still 62mA, however). There are several ways of thinking about this configuration. I chose to think of it as "distributed resistance per battery", like you said. If that doesn't click for you, that's fine. "Equivalent resistance" isn't actually the greatest way to model OpenBCI's power consumption. It should actually be modeled as a constant current draw. So, I've added additional material to my post using the constant current model. I get basically the same answer.Chiphttp://www.blogger.com/profile/10352943033779293161noreply@blogger.comtag:blogger.com,1999:blog-7276377053120174333.post-37056401149500799072015-07-10T11:30:04.225-04:002015-07-10T11:30:04.225-04:00Hi Chip,
Sorry but I am still not quite following ...Hi Chip,<br />Sorry but I am still not quite following your opinion.<br />I think we can simplify the system like this:<br />----------------------+1.5V+1.5V(I1)+1.5V+1.5V---------------------<br />| |<br />---------------------------------------R(I2)---------------------------------------<br /><br />(Sorry I don't know how to insert an image in comment...)<br />I1(from where you measured)=I2=0.062A, and the voltage between R should be 6V, so R=6 / 0.062 = 96 ohms<br />If we try to use the discharge curve from the datasheet, the calculated R= 24 ohms should be regarded as "the distributed resistance per battery", and the whole resistance is 96 ohms.<br /><br /><br />Regards,<br />Eric<br />Eric Tianhttp://www.blogger.com/profile/09238086019846686696noreply@blogger.comtag:blogger.com,1999:blog-7276377053120174333.post-3156718746234117452015-07-10T07:10:10.839-04:002015-07-10T07:10:10.839-04:00Hi Eric,
If we're talking about characterizin...Hi Eric,<br /><br />If we're talking about characterizing the electronics from the perspective of the whole battery pack, you are correct...the load seen by whole battery pack is 96 ohms. <br /><br />But, I was trying to use the discharge curve from the datasheet, and that discharge curve was for a single battery. Since what is most important is the amount of current coming out of the single battery, I calculated the apparent load that would result in the current that I actually measured. I measured 62 mA. One cell makes 1.5V. Therefore, from the perspective of one cell in my battery pack, it thought that it was seeing 1.5V / 0.062A = 24 ohms. IMO, that's the right number to use to lookup the discharge life in the datasheet for a single battery.<br /><br />ChipChiphttp://www.blogger.com/profile/10352943033779293161noreply@blogger.comtag:blogger.com,1999:blog-7276377053120174333.post-86811066264933494562015-07-10T05:21:44.937-04:002015-07-10T05:21:44.937-04:00Hi Chip,
Thanks for the reply.
I think the resist...Hi Chip,<br /><br />Thanks for the reply.<br />I think the resistance of the whole electronics system should be R = 1.5*4 / 0.062 = 96 ohms because you use four cascaded(not paralleled) batteries as the power supply.Please correct me if I am wrong.<br /><br />Regards,<br />Eric<br />Eric Tianhttp://www.blogger.com/profile/09238086019846686696noreply@blogger.comtag:blogger.com,1999:blog-7276377053120174333.post-37237723224868324252015-07-09T07:39:59.959-04:002015-07-09T07:39:59.959-04:00Hi Eric,
The effective load is simply another way...Hi Eric,<br /><br />The effective load is simply another way of expressing how much power a system draws. The "effective load" is the apparent resistance of the whole electronics system...as if the system were replaced by a single resistor.<br /><br />During my testing, the OpenBCI board was drawing 62 mA (aka. 0.062 A). Each battery was at a voltage of 1.5 V. So, the effective load can be calculated as R = V / I...R = 1.5 / 0.062 = 24 ohms.<br /><br />On the OpenBCI board, there's not a whole lot that one can do to reduce the power consumption (ie, reduce the current draw, which is another way of saying "increase the effective load"). There might be low power (quick sleep) modes that you could invoke in the microprocessor, or there might be lower power modes that you could invoke in the wireless transmitter, but both of those approach would require you to dive into the lowest-level code...not an easy thing to do.<br /><br />Sorry I can't be more help.<br /><br />ChipChiphttp://www.blogger.com/profile/10352943033779293161noreply@blogger.comtag:blogger.com,1999:blog-7276377053120174333.post-43154621481609989922015-07-09T07:13:11.696-04:002015-07-09T07:13:11.696-04:00Hi Chip,
I am trying to build my own EEG circuit a...Hi Chip,<br />I am trying to build my own EEG circuit and what your work gives me many inspirations. But I have some questions:<br />1. You wrote "the effective load is (1.5/0.062) = 24 ohms", what does effective load mean? The load of each battery of the load of the whole system? I think it should be the former.<br />2. What can be done to extend the lifetime of the battery? According to your graph, the bigger the load gets, the longer lifetime we will achieve, so can adding some resistors to the system extend the lifetime?<br /><br />Regards,<br />Eric Eric Tianhttp://www.blogger.com/profile/09238086019846686696noreply@blogger.com