How to "sense" 1.5 Amp's of current? - Electronic Circuits Projects Diagrams Free

**Krumlink** · 1st July 2008, 04:35 AM

I have been looking at how to sense current draws, but cannot find any good links, etc.

Anybody have any good links for a device that can sense current, around 1.5 Amps?

dknguyen · 1st July 2008, 04:44 AM

Zetex products - ICs - General power management - Current monitors -
http://www.zetex.com/3.0/appnotes/apps/an39.pdf

Mikebits · 1st July 2008, 04:46 AM

Simple, touch bare wire, if your hand jerks away, there is current. J/K
The way I have seen it done is to use a real low ohmic value resistor. When current flows through your line you can detect the voltage drop across the resistor. @1.5A using a .1 ohm you would have a .15v drop. This would be a 1/4w resistor. If you want more sense volts, you have to increase R, but then your VR drop goes up as well as power through the resistor.

DK beat me to the post. The IC method sounds cool, but old fashion methods work too.

**Krumlink** · 1st July 2008, 04:52 AM

Would there be any way to measure it? as in I would be able to distinguish 1.5Amps from say .75 Amps?

dknguyen · 1st July 2008, 04:54 AM

For the current sense resistor method:

You measure the voltage drop across the resistor. It is proportional to the current flowing through it. Lower R = less losses, less sensitivity. Higher R = higher losses, more sensitivity. It's a trade off and doesn't work in high power applications since R would need to be much too low.

**Krumlink** · 1st July 2008, 04:57 AM

Well it may help If I tell you what Im up to:

I am trying to sense when a outlet is using 1.5 Amps of current, it will turn on a relay .When the 1.5 Amps of current drops down to 0.75 Amps of current, it will turn off the relay. Perhaps that will help.

Mikebits · 1st July 2008, 04:58 AM

E=IxR so 1.5Ax1ohm =1.5v .75Ax1ohm = .75v scale resistor as needed. If you plan on using a pic, the A/D could do the math.

**Krumlink** · 1st July 2008, 05:02 AM

So a 1 ohm resistor then? How/where would I measure the voltage drop?

**Krumlink** · 1st July 2008, 05:11 AM

OH forgot to mention, 120VAC. Sorry.

Mikebits · 1st July 2008, 05:17 AM

Oh boy, that it different, and scary. Mains can be a hazard. I think this may be better suited for someone else to answer, as I am too poor for a law suit

Mikebits · 1st July 2008, 05:29 AM

Maybe you should have a look at current sense transformers for AC.

**Krumlink** · 1st July 2008, 05:30 AM

What if I stepped down the voltage, say to 12VDC?

Mikebits · 1st July 2008, 05:36 AM

That's a cool idea, but then you have to make sure your transformer primary is rated for the current of your AC breaker. A current sense transformer is non-invasive to your main circuit.
One example, but many more companies make these things so just search.
AC Current Sensing Transformers - Toroid Corporation of MD

**Krumlink** · 1st July 2008, 05:39 AM

That would definately work, how would I use it though

Mikebits · 1st July 2008, 05:54 AM

Well, I am not an expert in this sort of thing, but I believe they work like any ordinary transformer. That is to say the mains line is the primary, the sense xformer is the secondary. There is a ratio of current exchanged between the primary and secondary as I am sure you already know. So you can figure 1A through primary X amount of amps will flow through secondary. This info is given in the current sense xformer data sheets.

So with x current flowing through secondary rectify AC, run voltage across a resistor, and using E=IxR you know voltage. Of course you must take into account E drop through rectifier.

This may not be the best way, but it gives you a general idea. I have a feeling someone will smack me, as I probably made it over complicated.

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1st July 2008, 04:35 AM	(permalink)
Krumlink	How to "sense" 1.5 Amp's of current? I have been looking at how to sense current draws, but cannot find any good links, etc. Anybody have any good links for a device that can sense current, around 1.5 Amps? __________________ MechTronics World of Warcraft is a disease

1st July 2008, 04:44 AM	(permalink)
dknguyen	Zetex products - ICs - General power management - Current monitors - http://www.zetex.com/3.0/appnotes/apps/an39.pdf __________________ NO, that picture isn't me so don't bother asking if we can be friends. Last edited by dknguyen; 1st July 2008 at 04:46 AM.

1st July 2008, 04:46 AM	(permalink)
Mikebits	Simple, touch bare wire, if your hand jerks away, there is current. J/K The way I have seen it done is to use a real low ohmic value resistor. When current flows through your line you can detect the voltage drop across the resistor. @1.5A using a .1 ohm you would have a .15v drop. This would be a 1/4w resistor. If you want more sense volts, you have to increase R, but then your VR drop goes up as well as power through the resistor. DK beat me to the post. The IC method sounds cool, but old fashion methods work too. Last edited by Mikebits; 1st July 2008 at 04:48 AM.

1st July 2008, 04:52 AM	(permalink)
Krumlink	Would there be any way to measure it? as in I would be able to distinguish 1.5Amps from say .75 Amps? __________________ MechTronics World of Warcraft is a disease

1st July 2008, 04:54 AM	(permalink)
dknguyen	For the current sense resistor method: You measure the voltage drop across the resistor. It is proportional to the current flowing through it. Lower R = less losses, less sensitivity. Higher R = higher losses, more sensitivity. It's a trade off and doesn't work in high power applications since R would need to be much too low. __________________ NO, that picture isn't me so don't bother asking if we can be friends. Last edited by dknguyen; 1st July 2008 at 04:55 AM.

1st July 2008, 04:57 AM	(permalink)
Krumlink	Well it may help If I tell you what Im up to: I am trying to sense when a outlet is using 1.5 Amps of current, it will turn on a relay .When the 1.5 Amps of current drops down to 0.75 Amps of current, it will turn off the relay. Perhaps that will help. __________________ MechTronics World of Warcraft is a disease Last edited by Krumlink; 1st July 2008 at 04:59 AM. Reason: I for no reason at all said open a garage door. Completely random

1st July 2008, 04:58 AM	(permalink)
Mikebits	E=IxR so 1.5Ax1ohm =1.5v .75Ax1ohm = .75v scale resistor as needed. If you plan on using a pic, the A/D could do the math.

1st July 2008, 05:02 AM	(permalink)
Krumlink	So a 1 ohm resistor then? How/where would I measure the voltage drop? __________________ MechTronics World of Warcraft is a disease

1st July 2008, 05:11 AM	(permalink)
Krumlink	OH forgot to mention, 120VAC. Sorry. __________________ MechTronics World of Warcraft is a disease

1st July 2008, 05:17 AM	(permalink)
Mikebits	Oh boy, that it different, and scary. Mains can be a hazard. I think this may be better suited for someone else to answer, as I am too poor for a law suit

1st July 2008, 05:29 AM	(permalink)
Mikebits	Maybe you should have a look at current sense transformers for AC.

1st July 2008, 05:30 AM	(permalink)
Krumlink	What if I stepped down the voltage, say to 12VDC? __________________ MechTronics World of Warcraft is a disease

1st July 2008, 05:36 AM	(permalink)
Mikebits	That's a cool idea, but then you have to make sure your transformer primary is rated for the current of your AC breaker. A current sense transformer is non-invasive to your main circuit. One example, but many more companies make these things so just search. AC Current Sensing Transformers - Toroid Corporation of MD

1st July 2008, 05:39 AM	(permalink)
Krumlink	That would definately work, how would I use it though __________________ MechTronics World of Warcraft is a disease

1st July 2008, 05:54 AM	(permalink)
Mikebits	Well, I am not an expert in this sort of thing, but I believe they work like any ordinary transformer. That is to say the mains line is the primary, the sense xformer is the secondary. There is a ratio of current exchanged between the primary and secondary as I am sure you already know. So you can figure 1A through primary X amount of amps will flow through secondary. This info is given in the current sense xformer data sheets. So with x current flowing through secondary rectify AC, run voltage across a resistor, and using E=IxR you know voltage. Of course you must take into account E drop through rectifier. This may not be the best way, but it gives you a general idea. I have a feeling someone will smack me, as I probably made it over complicated.