0
\$\begingroup\$

I have a circuit that has a variable draw of current based on how many revolutions of a wheel is made per hour.

Every other revolution of the wheel my circuit logs data. Which draws around 50 milliamps of current.

I've made this little table to break it down:

Constants

RPM per one-mph on 27.5 inch wheel = 12.2RPM
Current draw per activation = 50 Milliamps
activation uptime = 10 Milliseconds

Variables

Average speed over 1 hour = 20 mph
RPM @ 20mph = 244 RPM
Revolutions per hour = 14640 RPH
Activations per hour (divide by 2 as im only logging every other)   = 7320 Activations

Values

Total activation uptime per hour    73.2 Seconds (7320*10/1000)

But now im struggling to see how I convert this value into a meaningful value. Given my circuit is running at 5 volts I know that drawing 50 milliamps is going to mean im using 0.25 watts. Do I then multiply this by 73.2? to figure out how many wattseconds im using?

Also from that can I then work out how many milliamp hours this is? The reason im trying to work this out is so that I can spec a battery. I think im close but im getting a bit confused.

\$\endgroup\$
  • \$\begingroup\$ Apologies, but his answer was correct was it not? I can still upvote it :) \$\endgroup\$ – Festivejelly Apr 20 '17 at 0:01
  • \$\begingroup\$ You have a point though, i've unmarked it as an answer until I receive more answers then I can decide which one best answers the question. I just got a bit over excited that I possibly had an answer to a question thats been plaguing me for hours. \$\endgroup\$ – Festivejelly Apr 20 '17 at 0:03
2
\$\begingroup\$

But now im struggling to see how I convert this value into a meaningful value. Given my circuit is running at 5 volts I know that drawing 50 milliamps is going to mean im using 0.25 watts.

Correct.

Do I then multiply this by 73.2? to figure out how many wattseconds im using?

Yes. (Note that Watt-seconds are also called Joules)

Also from that can I then work out how many milliamp hours this is?

To get amp-seconds divide by the voltage again; to get milliamp-seconds multiply that by 1000; to get milliamp-hours divide that by 3600.

The reason im trying to work this out is so that I can spec a battery. I think im close but im getting a bit confused.

If your battery isn't 5V I'm guessing you have some sort of voltage converter in-between the data logger and the battery. This calculation will tell you the amount of milliamp-hours drawn from the converter, not from the battery. (If it's a linear regulator they will be the same; if it's a switching regulator they will not be)

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Hi thanks for the answer. It steps up from 3.7 volts to 5v. So yeah theres an efficiency loss there. They reckon its about 90% efficient, so ill factor that in. I guess im more interested in what the circuit itself is doing. So I guess thats 18.3 Joules \$\endgroup\$ – Festivejelly Apr 19 '17 at 23:53
  • \$\begingroup\$ And approx 5 milliamp hours? If my calculations are correct :) \$\endgroup\$ – Festivejelly Apr 19 '17 at 23:56
  • \$\begingroup\$ Yes, I get about the same answer. (Assuming that's milliamp hours from the battery) \$\endgroup\$ – user253751 Apr 20 '17 at 0:07
  • \$\begingroup\$ Also note that "milliamp-hours per hour" is equivalent to average milliamps. \$\endgroup\$ – user253751 Apr 20 '17 at 0:07
  • \$\begingroup\$ Then you gain the answer thats fandabbydozy cheers. \$\endgroup\$ – Festivejelly Apr 20 '17 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.