1.5 VDC Pseudocell - Page 1
Forum section: electronics / projects

author: DonKu
author_rank: Regular Contributor
timestamp: November 08, 2020, 09:13:08 pm
content: This project shows how to build a 1.5 VDC pseudocell to take the place of an AA sized battery. The mechanical aspect of this project offers the greatest challenge. The electronic portion utilizes an easy to use LM317 adjustable voltage regulator to convert a 5 VDC source to 1.5 VDC.(excerpt)https://crcomp.net/pseudocell/index.php(Thank you In Advance to readers who alert me to typos and whatnot.)

author: Doctorandus_P
author_rank: Super Contributor
timestamp: November 09, 2020, 07:31:31 am
content: R3 does not work as intended.1V5 / 15k = 100uASo you'll need 100 of them to get to your minimum of 10mAA simpler option would be to remove R3 and use lower values for R1 and R2.Also:The minimum load current for the LM317L is 2.5mA (Checked TI datasheet) Other manufacturers of LM317 variants may deviate.Another easy project with an LM317 is to make a current source / sink with a single resistor.If you calibrate it to 1A or 100mA, you can use it to make 4 point measurements on low value resistors with a simple power supply and a regular DMM.

author: srb1954
author_rank: Super Contributor
timestamp: November 09, 2020, 09:52:56 am
content: Quote from: DonKu on November 08, 2020, 09:13:08 pmWith the resistor values in this circuit the minimum load current is too low. Assuming there is minimal standing current drain from the clock the voltage setting resistors R1 and R2 and the load resistor R3 only draw 1.2mA, well below the recommended figure of 10mA for an LM317. Potentially the LM317 will lose regulation and the output voltage will rise above its setting of 1.5V.The easiest way to correct the situation is to reduce R1 to 120and R2 to 24. R3 is not necessary at all as the current drain through R1 and R2 is sufficient.If this circuit is connected to an electro-mechanical clock there is a potential for any inductive spike voltages from the clock motor to damage the LM317. It is strongly recommended that reverse protection diodes be placed from the LM317 output to ground and between LM317 input and output pins. A suitable output capacitor, say 100uF, would also help soak up any inductive transients.

author: Gyro
author_rank: Super Contributor
timestamp: November 09, 2020, 11:06:20 am
content: This is one of those weird applications where the supply current of the clock is so much lower than the minimum load requirement of the LM317.It might actually be more efficient to just use a series resistor and a couple of forward biased silicon diodes in shunt configuration, together with an electrolytic to handle the pulsed current waveform of the clock. The series resistor value could be set to provide a bit more than the average current consumption of the clock (maybe up to double) in order to keep a bit of bias on the diodes.I bet its consumption would work out at less than the minimum load current of the regulator, with less parts and cost.

author: DonKu
author_rank: Regular Contributor
timestamp: November 09, 2020, 11:46:30 pm
content: In one of those strange coincidences of nature, a poster nymed Piglet made an identical suggestion on sci.electronics.design:https://sci.electronics.design.narkive.com/yUTTcxTv/1-5-vdc-pseudocell#post17My reply to him there works here too:Looks like my worst fear was realized as sloppy discipline alloweda decimal point to slip down from mils to micros. Your hint about howto eliminate R3 is much appreciated too.My approach to technology previously untapped by me is to approachit with caution. And a R2 of hundreds of ohms somehow seemed safe, untilyou helped me see things clearly in my mind's eye.Anyhow, the perfboard will be quickly rebuilt with correctcomponents. It's mounted on an easily accessed backboard and pluggedinto a strip connected to the UPS used for the phone system. The lastbit doesn't matter much, given how a long pole will be used to correctthe clock's drift.