i am a new electronic learner
may i know what is the output from current transducer ?
is that voltage or digital signal ? and which model of transducer is suitable for deal with PIC , i am asking for doing a project for power monitoring system such like kill-a-watt meter.

especially for this model
http://pdf1.alldatasheet.com/datasheet-pdf/view/172078/LEM/LTS25-NP.html

 

The LTS25-NP produces an output voltage related to the primary current.

The primary can be configured to have Np=1, 2 or 3Turns. The nominal rated ampere-turns is 25AT. The connection table is shown on page 3 of the data sheet.

So if you have Np primary turns then the output voltage as a function of primary current ±Ip will be

Vout=2.5±0.625*IpNp/25

Say you have Np=2 turns and Ip=+12A

Vout=2.5+0.625*2*12/25=2.5+0.6=3.1V

If you have Np=3 turns and Ip=-5A

Vout=2.5-0.625*5*3/25=2.5-.375=2.125V

With Ip=0A, Vout is always 2.5V, irrespective of the primary turns.

and so on ...

Yes this could be used with a micro-controller for the current input variable of a power monitoring system.

 

The LTS25-NP produces an output voltage related to the primary current.

The primary can be configured to have Np=1, 2 or 3Turns. The nominal rated ampere-turns is 25AT. The connection table is shown on page 3 of the data sheet.

So if you have Np primary turns then the output voltage as a function of primary current ±Ip will be

Vout=2.5±0.625*IpNp/25

Say you have Np=2 turns and Ip=+12A

Vout=2.5+0.625*2*12/25=2.5+0.6=3.1V

If you have Np=3 turns and Ip=-5A

Vout=2.5-0.625*5*3/25=2.5-.375=2.125V

With Ip=0A, Vout is always 2.5V, irrespective of the primary turns.

and so on ...

Yes this could be used with a micro-controller for the current input variable of a power monitoring system.

thx for your help
hmm... if that is Np=3 the Ipn shown 8A, is that means the maximum current apply only 8A?
and it show that Vout=2.5±0.6
so in the calculation i should count as if say (5A)
Vout = 2.5 +0.6 *5 *3/8 = 2.5 +1.125=2.5+1.125=3.625
is that correct?

 

hmm... if that is Np=3 the Ipn shown 8A, is that means the maximum current apply only 8A?

Since the device is rated for 25AT I guess you would have a rated input current of 25/3 =8.33 A with Np=3. That would be the maximum value for guaranteed accuracy. I didn't check the data sheet for absolute maximum primary current per winding.

and it show that Vout=2.5±0.6
so in the calculation i should count as if say (5A)
Vout = 2.5 +0.6 *5 *3/8 = 2.5 +1.125=2.5+1.125=3.625
is that correct?

No - the formula is [as I stated earlier] ....

Vout=2.5±0.625*IpNp/25

So if Ip=+5A and Np=3

Vout=2.5+0.625*5*3/25= 2.875

At Ip=+8A and Np=3

Vout= 2.5+0.625*8*3/25=2.5 + 0.6 = 3.1V

At Ip=-8A

Vout= 2.5-0.6= 1.9V

which is what the data sheet indicates.

 

i am currently doing the power monitoring system for home appliance, for increase the accuracy purpose is that the larger the the Np value .. the accurate output i can get ? is this model suitable for me? because i have see some other else such as 50AT ,100AT

 

i am currently doing the power monitoring system for home appliance, for increase the accuracy purpose is that the larger the the Np value .. the accurate output i can get ? is this model suitable for me? because i have see some other else such as 50AT ,100AT

The primary ampere turns Np*Ip is fixed for a specific transducer type. The transducer output range is directly related to the primary ampere turns.

So within the constraint of the value of Np which can be configured by the primary connections, you have a range of sensitivity options. A larger value of Np gives greater sensitivity to Ip, with the limitation that the maximum Ip range you can monitor through the primary decreases in proportion.

You mention your intention to use the transducer as part of an appliance power monitoring system. Is this for a single appliance or several? Do you actually want to measure the true power consumption or just the current draw?

The choice of transducer will depend upon the dynamic range of current you want to measure. What are your mains supply voltage and appliance current ranges?

 

The primary ampere turns Np*Ip is fixed for a specific transducer type. The transducer output range is directly related to the primary ampere turns.

So within the constraint of the value of Np which can be configured by the primary connections, you have a range of sensitivity options. A larger value of Np gives greater sensitivity to Ip, with the limitation that the maximum Ip range you can monitor through the primary decreases in proportion.

You mention your intention to use the transducer as part of an appliance power monitoring system. Is this for a single appliance or several? Do you actually want to measure the true power consumption or just the current draw?

The choice of transducer will depend upon the dynamic range of current you want to measure. What are your mains supply voltage and appliance current ranges?

yes... i planing to use the transducer as a part of the appliance power monitoring, i would like to only monitoring for single appliance so far because i am just a beginner to try out hardware designing and i need to measure the true power consumption by the help of transducer output voltage to PIC. and the accuracy will be concern.

my supply voltage in my country is 240V and i think the maximum range of current will not over 15A. but i am wonder is that the voltage need to be concern? or i can assume as all the appliance are using 240V?

I am really glad for your help

 

yes... i planing to use the transducer as a part of the appliance power monitoring, i would like to only monitoring for single appliance so far because i am just a beginner to try out hardware designing and i need to measure the true power consumption by the help of transducer output voltage to PIC. and the accuracy will be concern.

my supply voltage in my country is 240V and i think the maximum range of current will not over 15A. but i am wonder is that the voltage need to be concern? or i can assume as all the appliance are using 240V?

I am really glad for your help

With respect to the voltage the issue of whether you are measuring "true power" will depend on this and a couple of other factors.

If the load has a high power factor (i.e. very close to unity) and the supply voltage is a reasonably stable 240V, then you can assume the load power is just Pload=Vrms*Irms=240*Irms. Also depends on how you compute the RMS current from the transducer output signal.

If the load is non-unity or low power factor, then you have a problem because you have to take account of the phase difference between the voltage and current. This is usually done by direct multiplication of the current and voltage transducer time based signals with some back-end low pass filtering. Then you also have to consider the micro-controller A-D input sampling rate and resolution required to achieve the necessary accuracy - normally not a problem with the latest devices available.

There are probably many examples on the web of just such a PIC based device.

 

thx for your help, i will go find out the current transducer and test it when i am free

 

may i have your msn?

 

Sorry miloboy I can't do that - no offence meant.

I offer assistance only on the AAC forum to other forum members and that's it.

Please keep posting to AAC if you have other questions - there are plenty of other outstanding members with the necessary skills at hand.

 

ok.. its ok
i have some question to ask for the transducer.
as i know the current pass trough the transducer might be large, and i will like to connect the transducer on the on bread board or PCB ,but what is the allow maximum current can be pass through for PCB and bread board we use normally? and what is ur suggestion?

 

Sorry miloboy I can't do that - no offence meant.

I offer assistance only on the AAC forum to other forum members and that's it.

Please keep posting to AAC if you have other questions - there are plenty of other outstanding members with the necessary skills at hand.

ok.. its ok
i have some question to ask for the transducer.
as i know the current pass trough the transducer might be large, and i will like to connect or mount the transducer on the on bread board or PCB ,but what is the allow maximum current can be pass through for PCB and bread board we use normally? and what is ur suggestion?

 

i am confuse that how transducer work ? is that current must go trough (connect to) the transducer coz what i refer to data sheet it look like need to connect the current into the component ? or i check other transducer that is only pass through the wire to the hole provided and it will detect and convert it to output voltage.

 

The data sheet clearly shows the 6 solder leads which enable connection of the primary current. You wire the leads according to the no. of primary turns you require. This would normally be done on a PCB or solder prototype board. You would size the PCB lands and track to the current carrying requirements. One could have link solder lands on the top side of the PCB to allow the user to select any of the permitted primary connection options.

 

The data sheet clearly shows the 6 solder leads which enable connection of the primary current. You wire the leads according to the no. of primary turns you require. This would normally be done on a PCB or solder prototype board. You would size the PCB lands and track to the current carrying requirements. One could have link solder lands on the top side of the PCB to allow the user to select any of the permitted primary connection options.

so what shud i do ? i need to direct connect the current into the pin 1 provided on datasheet? or i just pass through the hole and take the output voltage ? i am confuse, because i saw some of the ppl said do only pass through the wire to the hole and come out with the output?

 

It's possible the device will also work if you pass the primary current through the 3.2mm hole. It's not clear from the data sheet if this is OK. You can try it and see if it works or not. Otherwise you would have to use the dedicated solder leads.

If you need more than 1 turn you could loop the primary conductor through the hole more than once - provided the conductor will fit and that you use a conductor size of appropriate current rating.

Is your problem also that you are unsure of how to connect / configure the PCB solder leads from the data sheet information?

 

It's possible the device will also work if you pass the primary current through the 3.2mm hole. It's not clear from the data sheet if this is OK. You can try it and see if it works or not. Otherwise you would have to use the dedicated solder leads.

If you need more than 1 turn you could loop the primary conductor through the hole more than once - provided the conductor will fit and that you use a conductor size of appropriate current rating.

Is your problem also that you are unsure of how to connect / configure the PCB solder leads from the data sheet information?

OK, which means the device work in 2 ways?
i am not sure is that i can direct feed in the input current to the (input)pin1 and output(pin4) , i worry of burn the device.
is that the input current is same as the output current after pass through the device?
i will go check it out this transducer , because i am still in analysis phase for my project. is quite expensive in malaysia

 

Hopefully from the data sheet you are able to see that there are 3 separate direct primary connections as follows:-

=======
| core |
| core |
=======
pin 1 <---------> pin 6
pin 2 <---------> pin 5
pin 3 <---------> pin 4
=======
| core |
| core |
=======

Each connection is capable of at carrying a continuous current of at least 8.3 Amp rms. The data sheet also states that the device is capable of measuring up to 80 AT, which would mean you could put a single current pulse of 80A through the 3 connections wired in parallel without any damage or loss of accuracy. If you applied a continuous train of current pulses through the primary solder links, the effective value of the pulsed current per connection link [1-6, 2-5 or 4-4] should not exceed 8.3A rms. If you wired the links all in series [effectively giving 3 turns] a pulsed current train of 80/3 = 26.7 Amps peak could still be measured accurately. You would just have to ensure the pulse duty in this case kept the effective rms current to 8.3A.

If you want to use the hole through the core. You could possibly run a primary rms current greater than 8.3A rms in a single conductor going through the 3.2mm hole. You would have to take into account the conductor current rating. Depending on your application, an uninsulated (bare) primary conductor might be hazardous - depending on your circuit layout etc. - so you may have to allow for conductor insulation when considering what maximum conductor size you can fit through the hole.

 

Hopefully from the data sheet you are able to see that there are 3 separate direct primary connections as follows:-

=======
| core |
| core |
=======
pin 1 <---------> pin 6
pin 2 <---------> pin 5
pin 3 <---------> pin 4
=======
| core |
| core |
=======

Each connection is capable of at carrying a continuous current of at least 8.3 Amp rms. The data sheet also states that the device is capable of measuring up to 80 AT, which would mean you could put a single current pulse of 80A through the 3 connections wired in parallel without any damage or loss of accuracy. If you applied a continuous train of current pulses through the primary solder links, the effective value of the pulsed current per connection link [1-6, 2-5 or 4-4] should not exceed 8.3A rms. If you wired the links all in series [effectively giving 3 turns] a pulsed current train of 80/3 = 26.7 Amps peak could still be measured accurately. You would just have to ensure the pulse duty in this case kept the effective rms current to 8.3A.

sorry i not really understand this :

if you applied a continuous train of current pulses through the primary solder links, the effective value of the pulsed current per connection link [1-6, 2-5 or 4-4] should not exceed 8.3A rms.

is that means i configure the N=1 connetion?

If you wired the links all in series [effectively giving 3 turns] a pulsed current train of 80/3 = 26.7 Amps peak could still be measured accurately. You would just have to ensure the pulse duty in this case kept the effective rms current to 8.3A.

what means is kept the effective rms current to 8.3A ? the minimum i can detect at least more than 8.3A?