voltage drop indicator - Page 1
Forum section: electronics / beginners

author: dimoune
author_rank: Newbie
timestamp: March 12, 2016, 06:38:06 am
content: Hello,I am a layman in electronics. I want to build a led indicator when my voltage drops below 10V. My power supply is 28V 500ma DC. It passes through a current regulator and the output is about 12V and 3ma. I want to devise a led indicator which lights up when the current drops from 12V to below 10V.I try to do some search on internet and find an ic called MC34161. I guess I can achieve what I am looking with this IC but I cannot figure how to do it. Any help please? Thank you.

author: Joule Thief
author_rank: Regular Contributor
timestamp: March 12, 2016, 07:48:37 am
content: This may provide 2 options as a starting point for you - one is based on a single supply comparator circuit and the second easily built from readily available discrete components.http://www.bristolwatch.com/ele2/img/comparator_lm358.pnghttp://www.bristolwatch.com/ele2/comparator.htmor another option based on transistor circuitryhttp://www.circuitdiagramworld.com/monitor_circuit_diagram/Alternator_And_Battery_Monitoring_Circuit_3982.html

author: dimoune
author_rank: Newbie
timestamp: March 12, 2016, 09:20:20 am
content: I want the circuit to draw current from my power supply which is 28V but measures the voltage drop after passing through my current regulator. The voltage after passing through the current regulator is 12V. I need to monitor it for drop below 10v.Also, the current reaches max of 5ma with the current regulator. I do not think it is enough to power the led. This is why I prefer to draw the current from the power supply.

author: Seekonk
author_rank: Super Contributor
timestamp: March 12, 2016, 09:43:58 am
content: Easiest option is the 3 terminal LM431. Only load will be voltage divider to obtain 3.5V. Led can run off 28V supply.

author: dimoune
author_rank: Newbie
timestamp: March 12, 2016, 10:25:07 am
content: Thanks for your input. The LM431 seems to power the led when voltage increases to a certain level. However, I am looking for the opposite.

author: danadak
author_rank: Super Contributor
timestamp: March 12, 2016, 12:34:58 pm
content: Fig 17 is the appropriate circuit. So you power the part off your28 Vdc. The comparator, because of its internal reference, basicallytrips at 1.27 Volts. So1.27 V = [ 10V x R1 ] / (R1 + R2), so pick R1 as 4.7K, a standard valuethen 1.27 x R1 + 1.27 x R2 = 10 x R1 or R2 = ( 10 - 1.27 ) x R1 / 1.27R2 = ( 8.73 / 1.27 ) x 4700 ~= 32307 or 32KDouble check the trip1.27 trip = Vx x ( R1 / [R1 + R2] ) or Vx = 1.27 x ( R1 + R2 ) / R1 = (1.27 x 36700) / 4700 ~=9.91 VSo trip checks.If you wanted more precise trip replace R1 & R2 with a 5 or 10 turn pot, say a 10K pot,and adjust for exact trip.Regards, Dana.

author: mariush
author_rank: Super Contributor
timestamp: March 12, 2016, 01:29:19 pm
content: See, what you want is kind of backwards, a device powered by a battery shouldn't turn on a led when battery voltage is too low, because the additional power consumption of the led will drain the battery even faster. You may find some integrated chips that work with two leds, for example to keep a led green above a certain voltage and turn it off and turn on a red led below a voltage but i can't think of any from memory.It's a bit overkill, but a small microcontroller would do just fine. For example, aPIC12LF1571is only 8 pin soic, works from 1.8v to 3.6v so you could power it from 12v using a 4:1 voltage divider (using two resistors) since it uses less than 0.1mA of current at less than 1Mhz (don't need more to light up a led), so the voltage will be 12/4 = 3v when powered from 12v, or 10/4 = 2.5v and will keep running even if voltage drops below 8v.Basically, all you do is feed the voltage through another voltage divider to an input pin and then the code in your microcontroller would be:initialize internal oscillator, initialize ADCfor as long as the mcu runsstart to measure the input voltage on the ADC , wait a second or so until the process is doneif the voltage is below a threshold, turn on the led, otherwise turn off the ledwait some timerepeatAnother advantage of using a programmable mcu would be that you could even make the led blink if the voltage is.. let's say below 12v but above 11v.The downside to microcontrollers is that you need a programmer for them, and a basic knowledge of C (or follow tutorials or ask on forums, the code is so short it can be written in around 6-10 lines on the forum).Genuine Microchip PIC programmers are kind of expensive (around $40-60) but you can find clones on eBay for much less, for examplehere's one for <$15

author: dimoune
author_rank: Newbie
timestamp: March 12, 2016, 01:46:16 pm
content: Quote from: danadak on March 12, 2016, 12:34:58 pmI don't see fig 17. Please provide a link. Thanks.

author: nowlan
author_rank: Frequent Contributor
timestamp: March 12, 2016, 01:50:29 pm
content: >a device powered by a battery shouldn't turn on a led when battery voltage is too lowheaps of low battery indicator circuits on google.

author: Seekonk
author_rank: Super Contributor
timestamp: March 12, 2016, 01:53:55 pm
content: Quote from: dimoune on March 12, 2016, 10:25:07 amDUH, doesn't take much to stop you. Just invert it. Emitter follower NPN with LED in emotter with limiting resistor. LED is on edge of ghosting so I would add an additional diode in series.

author: danadak
author_rank: Super Contributor
timestamp: March 12, 2016, 02:02:33 pm
content: The MC34161 datasheet. Just google it.Regards, Dana.

author: Zero999
author_rank: Super Contributor
timestamp: March 12, 2016, 02:35:05 pm
content: Quote from: Seekonk on March 12, 2016, 01:53:55 pmLike this:https://www.eevblog.com/forum/beginners/voltage-level-indicator/msg875244/#msg875244The values for R1 & R2 can be calculated using the site linked below:http://solderer.tv/tl431-calculator/And if the LED doesn't fully turn off, a resistor in parallel with it should do the job.

author: dimoune
author_rank: Newbie
timestamp: March 12, 2016, 02:49:28 pm
content: Quote from: Hero999 on March 12, 2016, 02:35:05 pm

author: Zero999
author_rank: Super Contributor
timestamp: March 12, 2016, 02:52:54 pm
content: Quote from: dimoune on March 12, 2016, 02:49:28 pmThen why not modify the circuit so it doesn't draw much power from the circuit being monitored and another power supply is used for the LED? Hint: it involves changing the connections to R1 & R2.