• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Answers 2 answers
  • Subscribers 46 subscribers
  • Views 243 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Original question:

DAC7750: Alarm detection

DAC7750: Working of ALARM feature in DAC7750

Swapniil Kanwar
Prodigy 160 points
Part Number: DAC7750
Other Parts Discussed in Thread: DAC8750

(In the following text, ALARM refers to ALARM_BAR pin of DAC7750)

I was experimenting with DAC7750 by noticing the behavior of ALARM pin. As per datasheet it should go low when there is an open circuit. That's what happens in practical too.

The thing which intrigued and confused me about how ALARM feature is that when I gave command to set 0mA current to DAC7750, and IOUT pin is floating in air, ALARM pin is LOW indicating open circuit. But when I shorted IOUT pin to ground pin, IOUT became HIGH. Then I attached an LED on IOUT pin (the current is still set to 0mA) thinking that LED will create a barrier of atleast 1.7V for it to switch on so it will be open circuit at 0mA. But the moment I connected LED, ALARM pin became logic HIGH. Instead of LED I even tried connecting a Voltmeter to see the IOUT voltage wrt gnd thinking that Violtmeter has significantly high input impedance and ALARM will remain LOW, but after connecting multimeter too, ALARM pin became HIGH indicating 'No Open Circuit'. !!!! 

The thing I want to ask is that how does ALARM circuit inside is detecting even in 0mA current that if IOUT pin is open circuit or not. Why didn't it treat a Voltmeter or and LED to be an open circuit eben at 0mA current output/ In the datasheet, ALARM circuit is not given. 

The registers I have set for the experiment are: 
Reset= 0x0001                                            // Reset registers to default
DAC_NOP= 0x0000                                    // NOP operation
CONTROL= 0b0001000000000110           // o/p enable, o/p->0-20mA, current setting res enable
CONFIG= 0b0000000000100000              // disable HART, watchdog, error-check, calibration
GAIN  =  0x8000                                        // Gain of 1, default is 0.5
DATA=  0x0000  

  • 0
    TI__Guru* 93215 points

    Swapniil,


    The DAC7750 (and DAC8750) does implement an open circuit detect, but you may not be able to see it when the DAC output is set to 0000h.

    The non-detection that you see comes from the nature of the detection circuit in the device. The open circuit is detected with the output amplifier in the device. When the DAC tries to output a current, and there is no load, the output voltage of the amplifier goes to the positive supply. An internal circuit detects that the amplifier output has been pulled to the positive supply and issues the alarm.

    However in your case, the DAC code is 0000h and there is no resistive load. Because there is no output current, the output buffer may not pull to positive supply. Depending on any parasitic leakages on the IOUT pin, a DAC code of 0001h or 0002h may not pull all the way up. It may take several codes to pull the output up.

    You can also measure the output voltage of the IOUT pin and see that the voltage goes to the positive supply. Note that multimeter has an input impedance. Unless the multimeter has a very high input impedance setting, it will increase the number of codes that are required to trip the alarm.

    Joseph Wu

  • 0 in reply to Joseph Wu
    Prodigy 160 points

    I am using a KEITHLEY DMM7510 digital multimeter, on which I am able to see that on output code 0000h (for 0 to 20mA mode), the open circuit voltage at IOUT pin is +2.5V. Since multimeter is connected between IOUT and Ground, ALARM is high in this stage, the moment I disconnect multimeter probes and let IOUT floating in air, ALARM goes low. 

  • +1 in reply to Swapniil Kanwar
    TI__Guru* 93215 points

    Swapniil,

    So your results aren't unexpected. The DMM7510 has a input impedance of about 10MΩ in the standard configuration. When you have the meter attached, the DAC sees the meter as a resistive load. The voltage at IOUT is at about +2.5V, so there may be some small amount of leakage current coming out of the IOUT pin. Regardless, the output is within the DAC's output buffer operating range.

    As soon as you remove the meter, that 10MΩ load is removed and the output voltage at IOUT goes to positive supply and triggers the ALARM. The voltage of the IOUT pin goes to the positive supply.

    You could also repeat the test with the meter configured to very high impedance (>1GΩ) and run the same test gain (just to see if the alarm triggers immediately) Or you could run the test with very high load resistances (say 100kΩ) and increase DAC code for IOUT until the ALARM trips. That way, you could check the voltage on the output directly without the input impedance being much of a factor.

    Joseph Wu