# NEED HELP FOR CURRENT MEASUREMENT...

Discussion in 'General Electronics Discussion' started by BHARGAV SHANKHALPARA, Nov 6, 2013.

1. ### BHARGAV SHANKHALPARA

35
0
Nov 6, 2013
Hello everyone...

i am new in current measuring concept, and i make one of control application type project...

actually i am controlling one ac induction motor using microcontroller (ATTINY88). and i want to add one extra function into it is over-current protection.(when current more than 6 amp, automatically gets turn off by microcontroller) for that i decide to use current transformer for current measurement. so i need to interface it with microcontroller.

i have current transformer with 1 (one) turn of primary winding and 350 turn of secondary winding. i know that to convert secondary current into voltage i need to connect burden resistor across it. i need to obtain voltage between 0-5v DC from that to apply microcontroller. also for that i need to signal conditioning for that before apply it to microcontroller.

problem is that...

i don't now that which value of burden resistor is best for my application and after obtaining well voltage signal condition circuit require which include (rectifier and amplifier).

after searching about signal conditioning circuit i found this document and circuit.

http://tinyurl.com/oqr52d3

in this circuit i understand about all resistor and op-amp but initial two zener diodes and at last RC time constant make question in my mind that why its here and if here than what should be its value.

thank you...

2. ### Harald KappModeratorModerator

10,771
2,427
Nov 17, 2011
A current transformer with 1:350 windingg ratio will produce an output current that is 1/350*input current. Assuming that your 6A are RMS, the peak-to peak-value of the input current is 2*sqrt(2)*6A=17A (rounded).Assuming further that you want to be able to measure at least twice the nominal motor current you have 2*17A=34A. I'd suggest designing for 50A peak-to-peak, that gives you some headroom in case of overcurrent.

50A translates into 50A/350=0.143App (pp = peak-to-peak) on the secondary side.

The ADC of an Attiny88 has an input range of 0V...Vcc. Assuming Vcc=5V, the current on the secondary side needs to be tzranslated into a voltage in the range 0V...5V. From Rburden=5V/0.143A you get Rburden= 35Ohm.
The voltage across the burden will be -2.5V...+2.5V. You need to add a level shifter to bring the signal in the range 0V...5V. Such a circuit is described here in figure 6.5.
YOu now can digitize the input signal and evaluate in software the relevant parameters (Irms, Ipeak, di/dt etc.) to protect the motor.

With this setup you will not need the rectifier circuit you linked. To answer your questions concerning that circuit anyway:
1) the zener diodes protect the rest of the circuit from overvoltage in case of high input currents. It is a good idea to add these to your circuit, too.
2) The RC network is explained in the last paragraph of the document you linked to.

3. ### BHARGAV SHANKHALPARA

35
0
Nov 6, 2013

i understand all the concept that you have written...!!

one question i want to ask is why you convert RMS value into Peak to Peak value and after that calculate sendnary current and R burden resistor based on it. we can not obtain this value directly from RMS...?

i have one more question that at last section of my circuit RC circuit, in my application actual AC frequency is 50 Hz than what would be the value of R and C or how can i calculate. for pure DC.

Last edited: Nov 6, 2013
4. ### Harald KappModeratorModerator

10,771
2,427
Nov 17, 2011
RMS is calculated from teh integral of the current over time. Therefore RMS is <= peak value. But the ADC has to digitize the actual signal including peaks and troughs. Therefore I prefer to use peak values for the calculation.
Of course you could do the calculation using RMS values, but then you will have to reduce the input range of the ADC accordingly by converting the fulls-scale input range to an equivalent RMS input range.

In any case, your software will see the real signal, not RMS. RMS has to be computed from the sampled values.

5. ### DLD

24
0
Nov 5, 2013
E. current is measured in Amps, But in most electronics work
purpose have to measure current milliamps, To measure current, Must two leads connects the ammeters port in the circuit,so that the current flows throw the ammeter, in spite of , the ammeter become
into part of the ckt. Ckts for a changeable linear current sources and a programmable 0–5A current source are included. Initially determine how much current flowing in the circuit, which may be used for make sense about turning off peripheral loads to preserve power or to return operation to normal limit. It depends on current limit is available in variety of measurement methods:-
A. Resistive
1.Current Sense Resistors
2. Inductor DC resistance
B. Magnetic
1. Current Transformer
3. Hall Effect Device
C. Transistor
1. Ratio-metri  