I was asked by a Client to provide calculations for a typical IS circuit of their project. I found different Hart certified DP transmitter vendors like Yokogawa, Emerson and Invensys stating a minimum of 16.4-17 V at the transmitter terminals in order for Hart to operate (despite of their transmitter can work with 10.5-11 volts without Hart).

However to power the transmitter from an isolator type Safety Barriers from P+F; MTL; (and others Hart certified) a minimum of 15 or 16 volts is provided to power the transmitters. So evidently if both requirements were true they cannot work together in the same circuit which we know that is not the case.

So trying to find the truth I found the Hart Physical Layer specification that in page 50/51 list the power supply requirements for Hart and there is no a single mention to a minimum voltage for Hart to operate.

I called both; IS barrier and TX vendors and none of them wants to change or admit their statement is incorrect so based on what I've learned about Hart I am pretty sure there should be no minimum (at least below 10.5 v) voltage requirement for Hart to operate and I am planning to move on with my calculations discarding this voltage limitation.

Am I correct on this?

 

The vendor assertions about minimum voltage are indeed correct. What you've missed is the minimum resistance requirement.

It isn't up to the HART Foundation spec to require a minimum loop voltage because the minimum voltage required is a function of the device, dependent on its circuitry, and the loop resistance. That's up to the vendor to spec.

The HART spec does, however, require a minimum loop resistance (not a voltage, a resistance) of 230 ohms, because that's what's needed for the HART signal to develop and be read by commercial HART devices.

The lower loop voltage range you cited, 10.5-11V, is the voltage needed when a power supply is directly connected across a transmitter with virtually no (copper wire only) loop resistance. The transmitter operates its 4-20mA output just fine with that minimum voltage. But you won't get a HART signal when it's connected across a power supply like this, because the power supply acts as a low pass filter. This can be proven in a bench test. Try it and see. You'll get a "no communication" failure message.

Only when the minimum required 230 ohms is put into the loop will the HART signal be sensed. The 5-6V increase in required voltage over the bare minimum is needed to drive the loop current through the additional 230 ohms of loop resistance needed to get a valid HART signal.

The transmitter will operate its 4-20mA output at zero ohms loop resistance, but neither you nor your client will be able to read a HART signal.

Only when there's sufficient minimum loop resistance will the HART signal be readable.

All the vendors you cited have a formula or a graph of loop resistance (starting at zero ohms) vs voltage. Some of the charts I have show a minimum voltage point on the graph for HART or "PC communications", crossing at about 240 ohms The requirement for minimum loop resistance for HART is sometimes stated there as well.

 

> The vendor assertions about minimum voltage are indeed correct. What you've
> missed is the minimum resistance requirement.

David thank so much for replaying.

Well, I understand the minimum loop resistance for Hart is 230/250 ohms but what I am questioning is the 16 volts min due to Hart stated by most transmitters vendors even though their product state 10/11 volts a minimum.

> It isn't up to the HART Foundation spec to require a minimum loop voltage because the minimum voltage required is
> a function of the device, dependent on its circuitry, and the loop resistance. That's up to the vendor to spec.

I understand that, but again what is in question is the minimum voltage requirement by Hart they are proclaming to be 16 volts.

> The HART spec does, however, require a minimum loop resistance (not a voltage, a resistance) of 230 ohms, because
> that's what's needed for the HART signal to develop and be read by commercial HART devices.

I agree; that is clear to me.

> The lower loop voltage range you cited, 10.5-11V, is the voltage needed when a power supply is directly connected
> across a transmitter with virtually no (copper wire only) loop resistance. The transmitter operates its 4-20mA output
> just fine with that minimum voltage. But you won't get a HART signal when it's connected across a power supply like
> this, because the power supply acts as a low pass filter. This can be proven in a bench test. Try it and see. You'll get a
> "no communication" failure message.

Thank you. I understand that as well.

> Only when the minimum required 230 ohms is put into the loop will the HART signal be sensed. The 5-6V increase in
> required voltage over the bare minimum is needed to drive the loop current through the additional 230 ohms of loop
> resistance needed to get a valid HART signal.

I do not understand why that 5-6 volt increase. Where did you get that number from? If 5/6 volts over the 10/11 volts is required then the barrier can not meet that requirement since it provides only 10.5/16 volts at barrier terminals!!!

> The transmitter will operate its 4-20mA output at zero ohms loop resistance, but neither you nor your client will be able
> to read a HART signal.

> Only when there's sufficient minimum loop resistance will the HART signal be readable.

I am Ok whith this but still not clear

> All the vendors you cited have a formula or a graph of loop resistance (starting at zero ohms) vs voltage. Some
> of the charts I have show a minimum voltage point on the graph for HART or "PC communications", crossing at about
> 240 ohms The requirement for minimum loop resistance for HART is sometimes stated there as well.

David; that is the chart I am talking about but why the 16.6 volts is coming from? Again; if true then how the barrier would work providing only 15.5 volts?

Have anybody else can help?

 

It's Ohms Law.

To drive the same current through greater resistance requires a higher voltage:

With a loop load resistance of zero, the 'internal resistance' of the transmitter circuitry is
R = E/I,
R = 11/0.023 = 478 ohms (0.023A = 23mA, NAMUR upscale failure mode)

Add 230 ohms for a HART resistor and the loop's total resistance becomes 478 ohms + 230 ohms = 708 ohms

E = I*R E = 0.023A*708ohms = 16.28V

If the HART barrier provides a minimum voltage of 15.5V to the transmitter, then that minimum is lower than the minimum voltage needed by the transmitter. Being a minimum voltage from the HART barrier, one assumes that the barrier can pass more than that, assuming the power supply is up to the task. I don't see the problem.

 

Thanks David for your detailed explanation. I am into commissioning and i was wondering about the same doubt. Thanks for the rectification

 

> Thanks David for your detailed explanation. I am into commissioning and
> i was wondering about the same doubt. Thanks for the rectification

David; thank you for clarifying this unclear statement.

I believe your interpretation and calculation of the 16 volt statement from transmitter manufacturer is correct.

However for the last part regarding the IS barrier I think the 15 volt minimum voltage already include the voltage drop by the current to voltage converter resistor inside the barrier (isolator type)therefore all other voltage drops in the loop can be by the wires and remote current indicator if any. Furthermore 15 volts is conservative number just in case you are using a 19 volt Power Supply being 24 volts the most common case.

 

The minimum voltage on the transmitter terminals specified by the device manufacturer is sufficient for HART communication. If manufacturer says 10.5 V, HART communication will work. The additional 5 V is the minimum voltage you will need from the power source (DC power supply or AI card). The 10.5+5=15.5V is not required on the device terminals.

Also note that many devices require more than 10.5 V on the terminals to operate. Some require 12 V, others require 19 V. Always refer to the product data sheet when designing loops or replacing devices.

Don't forget other voltage drops in the loop. If the cable is long, its resistance is large, and there could be several volts drop along the cable. If you have loop powered indicators or wireless adapters in the loop, there will be additional voltage drop across those too. Same goes for shunt resistors in recorders and signal conditioners etc. Make sure to take all into account when you do the design.

Cheers,
Jonas