Big differences in current - what value should resistor have? - Page 1
Forum section: electronics / beginners

author: niasie
author_rank: Newbie
timestamp: December 03, 2014, 09:14:52 pm
content: Hey,I'm building around the Bosch BMP085 pressure sensor and planning to use the sensor to measure my quadrocopter's height. The drone is powered by a propietary rechargable battery with 3.7V. The sensor, according to its datasheet, only likes 1.8V to 3.6V. I would like to bring the voltage down to somewhere around 2.5V, considering that the battery will drop in voltage when it's not fully charged anymore.Now here is my problem: The sensor varies in load from 5µA (in standard mode) to a peak of 1000µA, at least according to its datasheet. I want a voltage drop of about 1.2V across the resistor, so I do some simple, relying on the minimum and maximum current:1.2V / 5µA = 240k?1.2V / 1000µA = 1.2k?If I choose 240k?, the voltage drop will be far too big when the load is at peak, but if I choose 1.2k?, the voltage drop will be far too small. What resistor value should I choose, or is there any other solution to my problem?I'm grateful for any response, and sorry for my not-so-perfect English.niasieEdit: I just saw, that the Greek Omega isn't displayed correctly. The "?" are supposed to mean Ohms.

author: Paul Price
author_rank: Super Contributor
timestamp: December 03, 2014, 09:21:29 pm
content: Use a LDO regulator IC or use a resistor-zener diode combo to set the sensor operating voltage.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: December 03, 2014, 09:31:18 pm
content: Use whatever resistor you need to saturate the ADC voltage range at the highest current you expect (plus some margin), and use an ADC with whatever resolution is necessary to resolve (to the desired level of accuracy) the lowest voltage you expect.If the sensor can vary from 5uA to 1000uA, and your ADC range is 0-2.5v, then you would want to use a 2.5k resistor (call it 2k to give yourself some margin). With a 2k resistor, a 5uA current will give you 10mV. If you want to resolve this with a 1% uncertainty, then your ADC needs to be able to accurately resolve 100uV on a 2.5V range, which is about 14.5 noise-free bits. This shouldn't be difficult to achieve with a 16+ bit ADC. If you want to only hit 1.2v at 1000uA, then you'll need a 1.2k resistor and an extra bit out of your ADC.I don't think your dynamic range is extreme enough to require an adjustable gain on the ADC, but that's an option as well.

author: niasie
author_rank: Newbie
timestamp: December 04, 2014, 04:56:10 pm
content: Thank you for your responses! I'm pretty sure your suggestions will solve my problem.

author: mikerj
author_rank: Super Contributor
timestamp: December 04, 2014, 06:49:25 pm
content: Quote from: suicidaleggroll on December 03, 2014, 09:31:18 pmThe BMP085 is an I2C device, no external ADC is required. The 1000uA is the maximum supply current for the device during conversion, though the typical is 650uA.The OP is trying to understand how to choose a suitable resistor to drop the supply voltage, and the answer is that you don't use a resistor in this situation, you'd use a low dropout regulator. A simple zener shunt would work, but would be wasteful of current, requiring a constant 1mA even though the average current of the sensor will be far lower.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: December 04, 2014, 08:03:15 pm
content: Quote from: mikerj on December 04, 2014, 06:49:25 pmMy bad, based on the wording of the post I was thinking the device had an analog current output that needed to be converted to a voltage for sampling. Yes in this case an LDO would be the best option.

author: Seekonk
author_rank: Super Contributor
timestamp: December 04, 2014, 08:20:36 pm
content: If the power doesn't need to be regulated, just use a diode to drop the voltage .65V and an additional resistor load to insure enough current.