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I need an ammeter for a six volt battery and i need it to show 0-15 amps via ten leds. I was wondering if i could use like a lm3114 driver? As little power dissipation as possible would be best. Thanks!
here is what i think ypu should try.
use a shunt reistor of a small vlaue (0R1), so at maximum curent you would loose 0.1*15*15=22.5W quite much isnt it, try using 5 of these resistors in paralel, so the power you would loose would be 5 times smaller.
i couldnt find a datasheet for lm3114, but dont you mean to use lm3914 or 3915? these are indicators as well.
i think that you should use a differential amp to amplifi the voltage on the shunt resistor and then use it ti drive the indicator.
but what do you power from a 6 v batt that needs 15 amps?
Yeah, you are correct about the 3914, sorry about that. Is there any other way to monitor the current draw without a shunt resistor. The reason is that, when current of ten amps is drawn from the battery, the voltage falls from 6.24 volts to 5.7 volts and this varies greatly with current drawn, also, high wattage resistors can be expensive and bulky especially when a heat sink is needed for power dissipation. The project I am making is a portable psu which delivers 0-40 volts from a 6v battery. I was also considering using a current divider across the step-up transformer. If a current of 8-20ma could be somehow be directly proportional to the current draw i could light an led and have that shine on a light sensitive resistor, and make a voltage divider...ect. If you could tell me how to do that i would greatly appreciate it.
Why don't you go for a Hall-effect sensor? It is an easy and accurate method of mesuring current through any conductor without any phisical/ohmic contact with the conductor.
Have a look at this page:
**broken link removed**
Your idea sounds good, but can you explain, or perhaps show me a schematic of how to integrate a common(on that i can purchase at Digikey) hall sensor into a current monitor. I understand why you said to use a hall sensor, but i just don't understand how to use one unless i buy the ic that your website referred me to. Thanks
The above given Hall effect sensor provides ratiometric supply voltage output proportional to the current flowing through pin 4 and 5. You can connect its output to a 2V digital panel meter and limit the supply voltage to the Hall IC in such a way that it will give output voltage that is directly in trems of current. For example for 2A current, output voltage of the IC should be 200mV or whatever you desire. In this case the panel meter will read 2.00 (adjust your decimal point) and thats 2.00 A for you.
There's one problem though, there isn't any way i can purchase that item. Even if i could, i would probably have to buy a large quantity at one time. I was just wondering if you could integrate an ordinary hall sensor from digikey and make a device like the one listed at that web page.
Hey how about a regular ammeter from a car parts house
I know the circuit you are looking for has good potential, but a very practicle method would be a 6-volt ammeter for tractor/motorcycles that cost around $20-30. The analog reading is more accurate than digital method if you need to respond to immediate changes in amperage. Then again I use a cheap Digital Ammeter found in MCM electronics that goes up to 20 amps and is quite accurate (plus/minus 3%). Just a suggestion, firstname.lastname@example.org
you could use three 0.15Ohm resistors in paralel, fivint you a resistance of 0.05 ohms, and a voltage drop of 0.75 volts(still high huh?) and an amplifier(with a gain of 2) and an anlog voltmer(for 1,5 volts, or something like that and you would read 1 amp for each 0.1 V), could work, but still the drop is great in the resistor.