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#### jerano

• Guest
« on: August 16, 2008, 17:13:13 PM »
Hello forum

Please i need your help. I want to make a check engine for the lamps of my motorcycle. This will be a PIC micro, that will open instantly all the lamps of the motorcycle (one after the other of course) and will check in each lamp what if there is a current flowing. if not, this means that the lamp is broken and must be replaced.

Now my problem. I use a PIC that has an ADC. This will convert 5 volts to a 12bit digital. But i do not want to measure voltage. I want to measure current.  The lamps are of range from 5 to 35Watts at 12 volts.

I wil apreciate any help. Thank you

#### jerano

• Guest
« Reply #1 on: September 02, 2008, 18:09:53 PM »
I have seen some transformers used in electric installations that makes a reduction for example 1/100.
They have a coil and the wire passes through this coil. You do not need to cut the wire, just to cross it from the center of the transformer.
The outputs are the amperes of the wire divided.

Can i use one of this? Are there any with less division? Like 1/10 or even less?

#### spic0m

• Global Moderator
• Hero Member
• Posts: 841
« Reply #2 on: September 02, 2008, 18:52:32 PM »
In electric industry you won't find one that suits you, to big and with big reductions cause there ment to measure hundreds of amperes, in electronics i don't know if something like that exists in the sizes you want.
Maybe check in automotive parts, older cars had ampmeters but you have to put them in series.

#### kam

• Hero Member
• Posts: 1849
« Reply #3 on: September 02, 2008, 20:54:36 PM »
I have seen some transformers used in electric installations that makes a reduction for example 1/100.
They have a coil and the wire passes through this coil. You do not need to cut the wire, just to cross it from the center of the transformer.
The outputs are the amperes of the wire divided.

Can i use one of this? Are there any with less division? Like 1/10 or even less?

No you cannot. Minimum may you find 1/50, and refers to AC. You need DC, and i suppose just a couple of amperes, maybe 3Amps. You will put a resistor, 1ohm or maybe less, 0.5ohms is good, 5 watts. Across the resistor you will measure the voltage drop. Use the ohms law calculator to calculate the current from the voltage drop...

#### Mercury

• Guest
« Reply #4 on: September 02, 2008, 22:34:58 PM »
I have seen some transformers used in electric installations that makes a reduction for example 1/100.
They have a coil and the wire passes through this coil. You do not need to cut the wire, just to cross it from the center of the transformer.
The outputs are the amperes of the wire divided.

Can i use one of this? Are there any with less division? Like 1/10 or even less?

No you cannot. Minimum may you find 1/50, and refers to AC. You need DC, and i suppose just a couple of amperes, maybe 3Amps. You will put a resistor, 1ohm or maybe less, 0.5ohms is good, 5 watts. Across the resistor you will measure the voltage drop. Use the ohms law calculator to calculate the current from the voltage drop...
Won't this resistor reduce also the current through the lamp?

#### jerano

• Guest
« Reply #5 on: September 03, 2008, 22:40:47 PM »
I have seen some transformers used in electric installations that makes a reduction for example 1/100.
They have a coil and the wire passes through this coil. You do not need to cut the wire, just to cross it from the center of the transformer.
The outputs are the amperes of the wire divided.

Can i use one of this? Are there any with less division? Like 1/10 or even less?

No you cannot. Minimum may you find 1/50, and refers to AC. You need DC, and i suppose just a couple of amperes, maybe 3Amps. You will put a resistor, 1ohm or maybe less, 0.5ohms is good, 5 watts. Across the resistor you will measure the voltage drop. Use the ohms law calculator to calculate the current from the voltage drop...

#### kammenos

• Guest
« Reply #6 on: September 07, 2008, 10:31:38 AM »
Here you are. The schematic you want is attached.

#### kam

• Hero Member
• Posts: 1849
« Reply #7 on: September 07, 2008, 17:11:46 PM »
kammenos the schematic is ok.

Mercury, a 0.1 ohm resistor may reduce a 12volt battery down to......120 amperes... That will be no problem i suppose

#### Mercury

• Guest
« Reply #8 on: September 10, 2008, 00:21:26 AM »
kammenos the schematic is ok.

Mercury, a 0.1 ohm resistor may reduce a 12volt battery down to......120 amperes... That will be no problem i suppose

of course

#### jerano

• Guest
« Reply #9 on: September 10, 2008, 07:58:13 AM »
Kammenos, the output to PIC will measure voltage? And what is the analogy? Because this is a complicated schematic.

#### kammenos

• Guest
« Reply #10 on: September 15, 2008, 21:26:13 PM »
Kammenos, the output to PIC will measure voltage? And what is the analogy? Because this is a complicated schematic.

you are measurig the voltage drop on the resistor. Using the ohms law:

R = U / I => I = U / R

You know the resistanse (R) that is 0.1 ohm, you also know the voltage drop (that you measure with the PIC A/D U). So you can calculate the current if you divide those two. And it happens that the current in a series circuit (like the attached i gave you) is the same for all the circuit.

#### jerano

• Guest
« Reply #11 on: September 15, 2008, 22:31:20 PM »
And how about the R2 and D1? What the use of them? And do i need to add them into calculation?

#### jerano

• Guest
« Reply #12 on: October 07, 2008, 22:39:35 PM »
And how about the R2 and D1? What the use of them? And do i need to add them into calculation?

#### kam

• Hero Member
• Posts: 1849
« Reply #13 on: October 11, 2008, 14:49:33 PM »
R2 is to limit the current in case of short circuit the load to keep PIC and zener safe.
D1 is to limit the voltage to 5volts as PIC will be in great danger if more voltage is applied to it.

You do not really need to calculate R2 and D1 as what you realy measure with pic is the voltage drop across R1.

The max current to flow within the circuit will be P=U x I => I=P/U => I = 50/12 => I = 4.1Amps through the 50W load. I do not add the 0.1 ohms resistor to calculation because ... you know why...

This current will also flow within R2 causing a voltage drop of UR2 = I * RR2 => UR2 = 4.1 x 0.1 => UR2 = 0.4 volts. This is what you will measure to your PIC. So choose an appropriate comparator voltage.

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