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Why Need 250 ohms resistor in HART protocol
Need of 250 ohms resistor in HART protocol

Hello everybody,

Why do we need to include 250 ohms resistor in the loop while using HART protocol?

2 out of 2 members thought this post was helpful...

The 250 ohm resistor in HART is a shunt resistor just like the terminator for FOUNDATION fieldbus.

The HART communication signal works this way:

The talking device transmits a current

The listening device picks up on voltage

The purpose of the 250 ohm shunt resistor is to convert the transmitted
current to a voltage the received can pick up

Take a look at chapter 11 in the yellow book "Fieldbuses for Process
Control: Engineering, Operation, and Maintenance" buy online:


3 out of 4 members thought this post was helpful...

Sorry Jonas, I do not believe you are correct. In my previous reply to this question, I pointed out that the 250 ohm resistor converts the 4-20 mA analog signal from the HART transmitter to a 1-5 volt drop to allow the Analog Input card of the DCS or PLC to read the analog signal. The HART digital FSK (Frequency Shift Keying) signal is also present on this line and supplies the HART digital data to the HART modem chip usually present on the same Analog Input card for all new DCSs and PLCs. The Bell 202 protocol used for this FSK
digital circuit needs no terminating or load resistor. The 250 ohm resistor is only for the analog circuit.

Dick Caro
Richard H. Caro, Certified Automation Professional, CEO, CMC Associates,
2 Beth Circle, Acton, MA 01720
Tel: +1.978.635.9449 Mobile: +.978.764.4728
Fax: +1.978.246.1270

2 out of 2 members thought this post was helpful...

-->yes you are right, i like to say another point here that 4-20mA converted into voltage by using 250 ohm Resistor, DCS accept only voltage signal 1-5v

-->if you are sending signal 10-50mA then you should use 100 ohm Resistor to convert into voltage 1-5V

-->if you send the voltage signal 1-5V then no Resistor is needed ,directly given to the DCS according to the Priority wise (+ and -),if the priority is interchanged then in DCS Soft File Error will be coming

Rakesh Kumar.N BE Instrumentation Engineer
Trainee in MFL, Looking for a good job

Thank you

1 out of 3 members thought this post was helpful...

YES. Unless you wish to apply the 24 volts DC of your loop power supply directly across the Analog Input terminals that expect to see a 1-5 volt drop across this 250 ohm resistor. I don't think so.

If you are not using the 4-20 mA analog signal from a HART instrument, why are you using HART's analog transmission. There is also a faster, though less used, PSK multidrop version of HART that does not need a 250 ohm signal because there is no 4-20 mA signal. Read all about HART at

Dick Caro
Richard H. Caro, Certified Automation Professional, CEO, CMC Associates,
2 Beth Circle, Acton, MA 01720
Tel: +1.978.635.9449 Mobile: +.978.764.4728
Fax: +1.978.246.1270

By Dino Lacsamana on 3 September, 2010 - 2:17 am
3 out of 3 members thought this post was helpful...

The 250-ohm resistor is not placed in the loop for the idea of converting the current (4-20mA)from transmission to voltage (1-5V)for reception, and also the value 250 is not a special value for a loop resistance needed for HART communication.

Explaining first the true purpose of having a resistor in the loop:

The HART communicator is communicating with the transmitter through FSK protocol. Understanding this protocol, it is a communication in analog platform. Digitalization of the signal happens within the HART devices and not along the transmission line.

The resistor's position in the loop is between the transmitter and the DC power supply. As we connect the communicator it should be tap between the resistor and the transmitter or directly parallel to the resistor. Why? Because the power supply has a LOW PASS CAPACITOR filter that basically squelches analog signals like ripples. Thus, if there is no resistance in between the PS and the TX that analog signal coming from the HART devices will only be filtered out.

Explaining the value 250:

The resistance value is not necessarily be 250 ohms. Maybe because of the idea that the controlling current of 20mA will give us 5 volt calibration if there is a loop resistance of 250 ohms. Maybe not a good theory.

Actually you can use a number of possible resistance values that will allow communication between HART devices. This value depends on the transmitter minimum operating voltage requirement. Simple OHM's LAW. For example, if the voltage requirement for a certain transmitter is 12-24 volts use a resistor that will not give you a drop of below 12 volts or else you will not deliver power to your transmitter. Most transmitters are supplied with 24 volts and as per computation 250 ohms to 500 ohms gives a drop of almost 18 v which is somewhat in between the operating voltage.

1 out of 1 members thought this post was helpful...

You don't.
The HART signal is a high frequency superimposed on the analog signal. In order to receive the signal you need a means of creating a some impedance in the loop, a resistor does this.

If you were to connect the transmitter directly across a 24 VDC power supply, the analog signal would be just fine but the HART signal would be damped right out. If you add some resistance in the loop you will get some DC voltage drop but also across the resistor you will get the HART signal.

There's nothing magic about the value 250, it will work with less e.g. 200. It would also work if instead of a resistor you added a choke.

Do us all a favour - If you have a communicator and power supply try a range of resistors and let us know the minimum value that you can get to work.

The literature says 250 because that's a common value.


0 out of 1 members thought this post was helpful...

Sorry to disagree again with you Roy, but the 250 ohm resistor in the HART circuit is there for the ANALOG signal, not the digital signal. With 20 mA DC in the loop, typically full scale, the voltage drop across the precision 250 ohm resistor is exactly 5 volts. The A/D converter in the Analog Input card expects a range of 1-5 volts to interpret the analog transmission of a HART instrument. The resistor has nothing to do with the LOW Frequency digital signal (1200 bps)
of the HART digital transmission.

Changing the resistor to something else would certainly not effect the digital HART signal, but it would kill the calibration of the analog HART signal.

Dick Caro
Richard H. Caro, Certified Automation Professional, CEO, CMC Associates,
2 Beth Circle, Acton, MA 01720
Tel: +1.978.635.9449 Mobile: +.978.764.4728
Fax: +1.978.246.1270

Page 1-4 of the Rosemount 275 communicator manual

shows the 250 ohm resistor adapter plugged directly into the 275 (and in series with the loop), using the pair of banana jacks on 3/4" center with the supplied banana plug adapter across which a ~250 ohm resistor is connected.

When using the 250 ohm resistor in this fashion, the voltage drop across the 250 ohm resistor is placed outside of the receiver device's analog input, if there is one in the loop. The illustration cited shows no receiver device in the loop, only the DC power supply, the HART device and the 275, not untypical of a bench 'cal' for commissioning.

In my experience, the 275 communicator works just as well if the 250 ohms is across a receiver device's AI, but bench testing or field commissioning might not have a complete loop circuit through an AI with its dropping resistance, which is what I think the illustration is trying to show.

Carl Ellis

Carl, you have it right. The HART instrument expects to operate in a full loop with an analog input board where the termination has a 250 ohm precision resistor to provide the 1-5 volt drop for translating the 4-20 mA current signal to a voltage. If you attempt to operate the 275 without the termination of the AI circuit, you must provide another 250 ohm resistor, as for bench calibration.

If the 274 is connected across the 4-20 mA line of an operating instrument, then the bench test 250 ohm resistor is not needed. The 275 draws its power from the operating 4-20 mA loop so must be placed in series with that loop, but it can do this when the AI board with a 250 ohm resistor is already installed, by connecting across the line. The 275 only needs to detect the digital HART signal.

Dick Caro
Richard H. Caro, Certified Automation Professional, CEO, CMC Associates,
2 Beth Circle, Acton, MA 01720
Tel: +1.978.635.9449 Mobile: +.978.764.4728
Fax: +1.978.246.1270

By Jonas Berge on 22 December, 2008 - 7:06 am

You need the resistance. It can be as low as 230 ohm. It can the wire resistance, it does not matter, as long as there is a resistance.

See also page 10 of the HART application guide. Found a copy here:
"A minimum loop impedance of 230 is required for communication."

Transmission is by current, reception by voltage, so you need the shunt.


By Juan Pinzon on 22 December, 2008 - 9:56 pm


Maybe this will help a little:

Page 39: *Positioner or Controller with HART Communication*



Your statement "The 275 draws its power from the operating 4-20 mA loop" is not quite correct. It has its own internal battery. If you read the original post, Shankar asks, "Why we need to use 250 Ohms While using HART"? From that I assumed he was using HART without analog, that's why I said "You don't."

If you connect a transmitter directly across the power supply, that's an operating instrument and the HART signal will still be there, but you need some form of reactance or resistance so the HART communicator can pick up the change or modulate the current. When you use the communicator in the field the resistance is supplied by the wiring and any input resistance.

I did a project with a Modicon (50 Ohm 0.2-1V) I found the communicator would not work at the panel but worked OK at the transmitter because of the additional wire reactance. I guess what I am trying to get across, there's nothing magic about the number 250, 220 or 570 will work just as well.

Regards & Merry Christmas,



I don't disagree with you that the 250 is used for analog in a 1-5 volt input, but not all input modules are 1-5. I have used Modicons where the input was 50 Ohms.

If you are testing a Hart device in the shop you might have a power supply, a multimeter and a HART communicator. You hook the transmitter, the multimeter and power supply in series. Now where do you hook the HART communicator? Obviously you need some loop resistance, All I am saying it doesn't have to be a precision 250 Ohm resistor, any old thing will do, e.g. 147 Ohm, 220 Ohm, 570 Ohm, an RTD, a transformer primary, a light bulb... anything that will give you some loop impedance.

You Don't Have to Have 250.000 Ohms


By sujit ezhuthassan on 20 March, 2010 - 1:18 pm

here why 250 ohms!

there's no magic behind 250 ohm..yeah! could give a drop of volts from 1v to 5v, but get this..

the dc power supply we use in the loop has low pass capacitor filters, and in absence of a resistor, the filter is in parallel with the Tx, thereby bypassing the digital information , and adding a resistor just provides an impedance btn the Tx and filter circuit!

and yeah! can go up to 1000 ohms too!..
so chill out!!!..yo!!

By Roy Matson on 21 April, 2012 - 12:43 pm

You are correct of course,

What I was disputing is the need for 250 Ohms in the HART communications.

For instance you have a transmitter on the bench for calibration.
If you connect it directly across a 24Volt power supply it will be quite happy but you will not be able to communicate with it. You need some impedance so the HART can modulate the lines.
It doesn't have to be 250 Ohms for that.

Sometimes you strike inputs that use much lower voltage levels e.g. 0.2 - 1 Volt (50 Ohms), I'm not sure if HART will work with that in which case you can add another resistor in the loop.
250 Ohms will work of course but so will 100, 330, 270 or any other value thereabouts.

the HART communicator to function properly a minimum of 250 ohs resistance must be present in the loop. The HART communicator does not measure loop current directly.

Hi there,

HART Communication technology has a very low power requirement typically working on less than 3.8 mA of current. HART utilizes a frequency shift keying (FSK) modem to communicate over the primary variable analog signal wires. By imposing the digital signal on top of the 4-20mA analog loop, HART allows for a simultaneous analog signal with a continuous digital communication signal that has no effect on the analog signal.

HART circuits or networks can be configured in one of three ways, through a conventional connection method (point-to-point), multi-drop (point to multiple points) or through a multiplexer (HART to RS-485). HART Communication is designed so that there must be minimum resistance in the current loop before communication will occur. The HART specification accommodates loop resistances of 230 to 1100 ohms, typically a 250-ohm resistor is recommended.

Power supply impedance is minimal as opposed to current source being infinite.

To wire 24V right across a transmitter and using hart communicator connectors across the transmitter would be the same as connecting them on the same side.

> Why do we need to include 250 ohms resistor in the loop while using HART protocol?

This video explains why you need a 250 ohms resistor. Please watch it.

By Jonas Berge on 3 March, 2011 - 9:59 am

Great that somebody made a video to prove the theory. Finally we can lay this topic to rest. You must have a resistance for the HART transmitter.

Note that for HART positioners you do not need the resistor

I did not watch the clip to the end, there is a minor detail in the beginning "T" in "HART" is transducer, not transmitter


By Roy Matson on 2 March, 2011 - 6:10 am

Simple answer - you don't.

But you need some means of developing a voltage drop that the HART communicator can see e.g. a reactor or resistor of high enough value.
Without the resistor the load is too low in impedance and it shunts the signal.

Some seem fixated on 250 Ohms but in reality 100 might be enough and 500 not too much.


By Jonas Berge on 3 March, 2011 - 9:57 am

100 ohm is too little. You need 230 ohm or more


By Carl Ellis on 3 March, 2011 - 10:01 am

For the record

The HART Communication Application Guide,
HCF_LIT-039 Rev. 1.0 Preliminary
Date of Publication: March 25, 2010, on page 9, states

"A minimum loop impedance of 230 Ω is required for communication."

The character after the value 230 (above) is the Greek letter Omega for Ohm. Whether that character actually makes it to the forum's web page is iffy.

Document is publicly available here:

2 out of 2 members thought this post was helpful...

the 250 ohm is required in loop to do the following:

hart communication is established by superimposing fsk signal of two tones 1200 and 2200 over the 4-20 ma. the amplitude of these signals is + or - 0.5 ma so the resistor is required to make voltage drop sufficient to be sensed by the hart master and slave. if we remove this resistor what happens is, the voltage across the terminals of the transmitter will remain constant 24 v. the hart and transmitter are connected in parallel so the only way to initiate or receive signals is to vary the voltage across the terminals of each others.

> Why do we need to include 250 ohms resistor in the loop while using HART protocol?

because of ohms law..

V = I * i correct?
so what is the resistance?
then r=1/4=0.25#250ohms

Yes, you need a 250 ohm resistor to get 1-5Vdc signal for an analog input, but that's not why HART requires 250 ohms.

When a transmitter is fired up on the workshop bench, the transmitter will light up and do its functions while connected directly to a suitable DC power supply with zero load in the loop [no resistor, 24Vdc across the xmtr (+) and (-)], that is, it's functional except for HART communications. HART comm will not function without at least 230 ohms in the loop.

The reason for this was explained in posts above and has to do with an impedance across which the FSK HART signal can develop and be detected, because the power supply acts as a low pass filter.

A workbench setup might or might not require a 1-5Vdc for something like calibration but HART definitely requires a minimum loop resistance.

>because of ohms law..
>V = I * i correct?
>so what is the resistance?
>then r=1/4=0.25#250ohms

4ma = 0.004Amps not take 4 in the divisor value, so R = 1/0.004 = 250 ohms. Current take in terms of Ampere only not in milliamps.


There are two things

1. 4- 20 ma signal

2. HART protocl

you need to look separately above two

HART protocol uses conventional 4-20 mA signal with digital information.

1. To achieve 4-20 mA, (DCS standard 1-5 VDC) you need 250 Ohm resistor irrespective of you are using HART protocol or not.

2. HART protocol based on 4-20 ma hence of course you need 250 ohm resistor to achieve this current.

so in first place, the basic purpose of 250 OHMS to achieve 4-20ma, then HART comes.


By Roy Matson on 21 April, 2012 - 2:03 am

You don't, but you do need something to create some impedance so the HART communications can pull the Voltage up and down.

Without some impedance the FSK signal would be shorted out. I'm sure 220 Ohms works just as well, a small choke would probably work better.

> Why do we need to include 250 ohms resistor in the loop while using HART protocol?

After reading all the posts and contributions for the reply of this question,i think the answer can be summed up into this,

The 250 ohms resistance in the loop serves two purpose :-
1) The HART Communicator communicates to the transmitter by FSK method.The capacitor in the power supply for the transmitter distorts this HART signal.Hence the resistor serves as an aid to get the signal back,for the communicator.

2) DCS,PLC needs voltage(1 to 5V) signal to understand the signal from the transmitter. The resistor again serves to convert this current signal into voltage signal for the DCS/PLC.

And last, but not least, the value of the resistor can be anything as long as it doesn't come in the way for supplying power to the transmitter and filtering out the HART signal.

By Roy Matson on 24 April, 2012 - 12:48 am

You summed it up pretty well but don't worry some people are hung up on the magic value of 250 Ohms.

I guess if they can't find a 250 Ohm resistor they just throw up their hands and give up. LOL


By Dick Caro on 24 April, 2012 - 4:20 pm

If I understand this thread, a current sink resistor of about 200 ohms is needed to allow the HART chip to send a digital signal on the wire pair. As it happens with a HART circuit, a precision 250 ohm resistor is typically required at the analog input terminations in order to get the 4-20 mA analog signal from the transmitter that is being sent to have a voltage drop of precisely 1-5 volts. That 250 ohm resistor, then is also functioning as the load resistor for the HART digital signal. As Roy says, if you don't care about the 4-20 mA signal, then you do not need a precision load resistor, but it is unusual to find a HART installation in which the 4-20 mA signal is not used.

Dick Caro

1 out of 1 members thought this post was helpful...

The issue of resistance in a HART loop typically arises when a new or replacement transmitter is fired up on the bench, not in the field where there's already a shunt/load resistor on the analog input.

On the bench, the tech connects power supply leads to the transmitter signal terminals and the transmitter fires up, regulating its output like it should.

Then the tech connects the leads from a HART communicator. The transmitter won't talk HART because the master's commands and the transmitters response, are filtered out by the power supply, effectively a low pass filter. There isn't enough resistance in the loop to develop a 'seeable' FSK HART signal.

The HART Foundation says a loop needs a minimum of 230 ohms. 240, 270, 300 and 333 ohms are standard resistance values. 250 ohm precision resistors abound in the process world. But zero ohms in a HART loop means no 'seeable' HART signal.

The other less frequent case is devices whose analog inputs use less than 250 ohms. I ran into a Precision Digital panel meter indicating flow from a DP transmitter the other day. The PD's input resistance was only 50 or 100 ohms. There was no HART signal until I broke the loop and inserted a resistor.

Thank you all for giving me clear information about the use of 250ohm resister for the HART transmitter.

But I have one Doubt. If I connect 3 numbers of 9 volt battery in series with the HART transmitter, Does the transmitter sent HART signals to the communicator?

I have already done that as soon as this doubt came into my mind. The result is the communicator didn't receive HART signals, Why this is happening? its just a battery, Does the battery eat up the HART signals?

> Does the battery eat up the HART signals?

Yes, the 9 Volt battery is low impedance shorting out the FSK signal, you need some resistance to develop some signal Voltage, try a few different values in series e.g 220, 330, 470 Ohms and it will work.
With 3 x 9V batteries you have lots of options. You may find you can cut down to 2 x 9V Most transmitters give you a formula for calculating the maximum burden or loop resistance.

I have been trying to get people here not to fixate on 250 Ohms, that's only required if you need to pick off 1 - 5 Volts

hai friends
I was already search the result of this question and after i got the sufficient answer this is the answer that resistor is using only with the power supply but no need for the field because in that that power supply is contain the resistance. in our power supply's have capacitance that capacitance is kills the ac current from the HART communicator that's why we are putting the resistor.

By control.student on 5 December, 2014 - 10:41 am

Hi all,

Can we summarise this (old) discussion as:

i. The resistance is necessary for Hart comm only and its value is approx 230 to 1000 ohms.

ii. The Tx calibration is for 4-20mA to reflect the measurement span and not in terms of voltage. Hence the bench calibration is unaffected by the resistance value.

iii. In service, the receiving instrument will most likely have a 250 ohm precision resistor. Hence we do not need to use a precision resistor during bench calibration, unless we are simultaneously bench calibrating the receiving instrument also.

Dick Caro's clarification about the need for a precision resistance was confusing. Hence I have tried to summarize it here.

By Mujahid Chougle on 29 January, 2015 - 10:17 pm
2 out of 3 members thought this post was helpful...

> Why do we need to include 250 ohms resistor in the loop while using HART protocol?

- Control Systems (i.e.PLC & DCS) Communicates only with Digital Signals.

4-20mA converted into voltage by using 250 ohm Resistor as DCS & PLC are accepting only voltage signal 1-5V.

In short,for setting the Communication between Field Instrument & Control System,need to convert the Analog Signal into Digital Signal & it is only possible via using 250ÔŽ resistor in HART System.
Which will use to convert the Analog Signal (4-20mA) into Digital Signal.

For your understanding how it is works, Please find below Calculation.

For 4mA Analog Signal,
For 20mA Analog Signal,

So in this Ways 4-20mA Analog Signal converting into +-1-5VDC Digital Signal by using 250ÔŽ resistor in HART System.

By FICE - Instrumentation & Control Engg on 26 January, 2017 - 8:37 pm
1 out of 1 members thought this post was helpful...

Hello Guys,

Let us focus to original question,

"Why do we need to include 250 ohms resistor in the loop while using HART protocol?"

Simple Answer: To maintain the Loop impedance (230 ohms minimum) in order the HART communicator can communicate with Smart Devices.

Some answer, they said to convert the 4-20 mA into 1-5 VDC?

I agree on this, however is not the correct answer of the original question. 250 ohms resistor is already inside Analog Input Card of Host Controller to convert 1-5 V in which the processor inside AI card may read this signal. (Not part of the loop as I said it is inside the card)

I have a question?
If I put the HART communicator parallel to Smart devices, do you think the the HART reading is only 1-5V? I DON'T THINK SO....