Analysis of 4-20mA and HART communication circuits
HART communication is based on the Bell 202 standard, utilizing frequency shift keying (FSK) and communicating at a rate of 1200bps.
The signal frequencies representing logic 0 and logic 1 are sine wave current waveforms with amplitudes of ±0.5mA at 2200Hz and 1200Hz, respectively.
Since the average value of the sine wave current signal is 0mA, even though the HART current signal is superimposed on the 4 to 20mA analog measurement signal, it does not cause any interference to the 4-20mA signal. This allows for the transmission of both analog current values and digital data using HART.
HART communication is similar to the way data is transmitted by powerline communication devices. HART is also a commonly used communication method in industrial sites. (The disadvantage is that the communication rate is extremely slow, at 1200bps)
Let's start with the principle block diagram of TI and share with you the HART communication. Since HART is an AC current component superimposed on 4-20mA (HART communication is a ±0.5mA sine wave current), we need to first introduce the circuit part of 4-20mA. First, let's remove the AC component of HART and look at the general current flow direction of the DC part. The main path is as follows:
Due to the establishment of negative feedback, the operational amplifier exhibits virtual short-circuiting, so the voltage at the non-inverting input terminal is equal to the voltage at the inverting input terminal. Since the inverting input terminal is connected to GND, the following equation holds: Vp=Vn=0VI2=VREF/R2I1=VDAC/R1. Due to the virtual disconnection of the operational amplifier, almost no current flows through the non-inverting and inverting input terminals, so the following equation holds:
Since the voltages at the upper ends of R3 and R4 are both equal to 0V (the inverting and non-inverting inputs are virtually shorted, both at 0V), the voltage drops across R3 and R4 are the same. Therefore:
Then the loop current is equal to the sum of I3 and I4, so:
The voltage on the loop is equal to, so it is important to pay attention to the power consumption of the transistor. It is necessary to choose a transistor with an appropriate package and ensure proper heat dissipation
Next, let's analyze the communication part of HART communication, which is also the HART communication section. HART communication is divided into TX and RX parts. You can refer to the following diagram. The C1 and R6 in the transmitting part form a high-pass filter (operational amplifier inverting input terminal is virtual ground), with a cutoff frequency of 1/(2*PI*R6*C1):
Since the set high-pass filter does not attenuate the frequencies of 1200HZ and 2200HZ sent by HART, we have I5=Vhart/R6 (it is necessary to ensure that the peak-to-peak value of Ihart current is around 1mA). Therefore, the HART current in the AC part of the loop is equal to
Since the peak-to-peak current of HART is a fixed value of approximately 1mA, the resistance of R6 can be determined based on the Vhart output by the HART communication chip and the resistance values of resistors R3 and R4 in the loop.
Regarding the HART receiving part, a bandpass filter can be employed for filtering (ensuring that 1200Hz and 2200Hz are within the passband) to reduce environmental interference, followed by demodulation for the HART communication chip:
