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interfacing of msp430f235 with op-amp

Other Parts Discussed in Thread: MSP430F235, TLC071

1. I want to interface output of op-amp  tlc071 with adc12 of msp430f235. the maximum current at any pin of msp430 could be  2mA. but  tlc071 supplies current more than 2mA. In order to prevent microcontroller from damage what should be done?

2. If i do current limitation using a resistor than can it load to that particular input pin's voltage?

3. What is the input impedance of any pin of msp430?

  • Hi,

    akhilesh sati said:
    1. I want to interface output of op-amp  tlc071 with adc12 of msp430f235. the maximum current at any pin of msp430 could be  2mA. but  tlc071 supplies current more than 2mA. In order to prevent microcontroller from damage what should be done?

    I don’t see any problem. The Amp-Op only will provide the necessary current to charge the input capacitance. For example, you can connect a voltage source ( < 2,5V) to the ADC also without any trouble. Why do you select a high-output-drive Amp-Op? Have you noticed that the minimum supply voltage is 4.5V?

    akhilesh sati said:
    2. If i do current limitation using a resistor than can it load to that particular input pin's voltage?

    This situation will affect the sampling time. If the ADC input sees an high impedance value, then you must provide an adequate sampling time. You can found information on this subject here (23.2.4.3 Sample Timing Considerations), on page 571.

    akhilesh sati said:
    3. What is the input impedance of any pin of msp430?

    From the device's datasheet (12-bit ADC, Power Supply and Input Range Conditions - pag 61)  is 2000 Ohm. Not production tested, limits verified by design.

    Best Regards,

    AES

  • but sir the problem is that msp430 can sustain the max current that is 2mA. Now what is the guarantee that output current of the op amp will be lesser than 2mA OR which factor will decide the output current of op-amp in this situation?

  •  Hi, why use an operational in front of adc? Need some gain to adapt level? OK

     ADC has from model 2KOhm in series to a 27pF capacitor (sampling circuits) so when you drive at Vcc max current is limited by input resistor @ 3.6V maximum peak current is 1.8mA decreasing with exponential rate driven by values.

     Adding an  OPAMP in front simply the output impedance is so low to be neglected in series of MSP model so max current is from previous but no sense drive so high value and @Vref=2.5V max PEAK current is less than 1.25mA.

     I hope you need complete a basic course to understand how thing work.\

     Regards

     Roberto

  • akhilesh sati said:
    the maximum current at any pin of msp430 could be  2mA. but  tlc071 supplies current more than 2mA.

    A normal OpAmp outputs a voltage, not a current (well, there are some specific current transducers etc. but that's not the common OpAmp).

    The specified output current is the maximum current at which the OpAmp can still maintain the output voltage. Since the MSP ADC input is high impedance, no significant current is required to apply the output voltage of the ADC to the MSP.
    Things are different if the OpAmp output voltage is above the MSP VCC. In this case, the overvoltage creates a current through the MSP pins clamp diodes to VCC, which mus tnot exceed 2mA pe rpin and must not exceed the current consumption of the MSP in total (else VCC begins to rise, eventually exceeding the maximum value allowed).

    To prevent this from happening, a simple series resistor can be used. Take the difference between the OpAmp supply voltage and VCC and fdivide it by 2mA. The result is the minimum resistance that will keep the current below 2mA even if the OpAmp outputs its supply voltage.

    So if Vcc is 3V and the OpAmp is povered by 5V, then 2V/2mA=1kOhm are enough between OpAmp output and MSP input. The 1k input series resistance of the ADC doesn't count here, as it is after the clamp diodes that protect the input.

    akhilesh sati said:
    If i do current limitation using a resistor than can it load to that particular input pin's voltage?

    ???
    Teh series resistor will influence the voltage that appears on the ADC input pin. The voltage at the ADC input drops with the relation of the series resistance and the ADC input impedance. Assuming a static input impedance of >500kOhm for the ADC, it has an influence of <0.2%. It does, however add to the ADC series input resistance (which in conjuntion with the samplign capacitor forms a dynamic input impedance that is much lower than 500k when the sampling gate opens), so it will increase (in case of 1k double) the required S&H time.
    But since the OpAmp itself will reduce the source impedance from high to ~0Ohm, you can still sample much, much faster than without the OpAmp.

    akhilesh sati said:
    What is the input impedance of any pin of msp430?

    It depends. The digital circuitry defines a maximum leakage current of 50pA which converts to a virtual input impedance of 20MOhm. This is, however, for digital input voltages near VCC or GND and does not necessarily apply to voltages near the threshold level between low or high (which also will increase the digital current consumption). Any active pullup/pulldown will lower the input impedance, and so does the analog circuitry, especially if the S&H gate is open on this pin.
    There might be differences on other pins, like the XTAL pins, RST, TEST adn the JTAG connectors.

  • If I use such a circuit configuration before ADC12 of MSP430f235.

    1. Would it affect the performance of ADC12 ?

    2. What would be  the maximum value of load resistor that I can connect instead if 1k resistor (1k resistor I have connected in my schematic) to retain the performance of ADC12 ?

  • One more question I would like to add in addition to previous questions. What value should I take, of the load resistor, (I have taken 1k in my schematic) so that it does not affect the performance of ADC12 in anyway ?

  • What is the purpose of this circuit? The diode will prevent negative output voltage from the OpAmp (latchign the OpAmp output to the negative rail on the smallest negative input voltage. However, the input cannot be negative as the input voltage, as you have drawn it, is positive only. So the diode is superfluous.
    The R1 load is a load, but makes no sense here too. It jus tdraws a current, while the OpAmp keeps the output voltage stable.
    To limit any current into the MSP pin, there must be a protection resistor in series to the MSP input, not parallel to it to GND.

  • Jens-Michael Gross said:
    What is the purpose of this circuit? The diode will prevent negative output voltage from the OpAmp (latchign the OpAmp output to the negative rail on the smallest negative input voltage. However, the input cannot be negative as the input voltage, as you have drawn it, is positive only. So the diode is superfluous.

     HI Jens, why input cannot be negative? Pulse generator can be any voltage. Circuits appear as a strange variation of precision rectifier, I don't understand why to power from negative supply when with a rail to rail you can obtain same result with less component less wasted power and more precision.

     Regards

     Roberto

  • Roberto Romano said:
     HI Jens, why input cannot be negative? Pulse generator can be any voltage.

    Not the one drawn in this schematics. It generates GND-based positive pulses. :)

    Roberto Romano said:
    Circuits appear as a strange variation of precision rectifier

    That explains the diode. As soon as the input goes negative, the output will be pulled to exactly GND by the load resistor. So rectification does not have the diode voltage as offset. The circuit exactly cuts off the negative part of a signal.

    The negative supply is required to allow negative input voltages. Without it, you'd probably destroy the OpAmp input when the input goes negative. Sicne the negative OpAmp output is blocked by the diode, the negative supply is not required to produce a negative output. Here a real r-r OpAmp would be okay (but then, R-R outputs usually never reach exactly 0V. There is almost always a border of some mV, even if no current is drawn at all.

  •  Hi Jean, again I don't know program used by our friend, I suppose is a spice front end from TI, reading I see a menu named TI Tools. Where can I download?

     Found:  http://www.ti.com/lit/sw/symlink/tina-ti_v-9-3_english-b.zip

     About used operational it is a rail to rail output and  can support input voltage range of +-16V from input and +-8 from power rails. So negative voltage are well accepted by this operational without pain, exceeding that limit result in component gate punch trough. Removing negative supply load resistor and diode with a simple buffer you can obtain a more precise precision rectifier from ground to VCC.

     Chosen operational has a min power supply voltage of 5V, choosing a low voltage opamp still this problem be removed and can be supply'd from MSP VCC rail too, TLV822 from NS for example can be powered with as low as 2.7V but no negative voltage can be applied to input. No negative tolerant choice is possible from Opamp selector page.

     Regards

     Roberto

  • Roberto Romano said:
    About used operational it is a rail to rail output and  can support input voltage range of +-16V from input and +-8 from power rails. So negative voltage are well accepted by this operational without pain, exceeding that limit result in component gate punch trough.

    It's not a question of negative input beign accepted or not. The signal source shows that there ar eonly positive pulses coming (maybe jsu thte wrong symbol has been schosen).

    And well, r-r output does not really mean 'down to exactly the rail'. Depending on load, the rails cannot be reached. The difference is still much, much less than with non r-r- output OpAmps, which couldn't get nearer than 0.45V to the rails (one C-E voltage)

    Also, r-r outputs does not mean r-r inputs. There are many OpAmps which cannot bear inputs closer than 2V to the positive rail.

    Roberto Romano said:
    emoving negative supply load resistor and diode with a simple buffer you can obtain a more precise precision rectifier from ground to VCC.

    No. The purpose of the OpAmp/Diode is that teh diode will prevent any negative offset while the OpAmp willauto-compensate for its own input offset and the diode voltage. The load resistor ensures that the blocking is instantaneous and no underrun will occur due to the diode recovery time or the OpAmp propagation delay. It really keeps the output between 0.000V and somewhere near the positive rail (at least less by the diode voltage).

    A buffer wouldn't by far be less precise and fast.

    Actually, this circuit will act as an ideal diode. No recovery time, no diode voltage. Just a small delay.

  • Jens-Michael Gross said:

    A buffer wouldn't by far be less precise and fast.

    Actually, this circuit will act as an ideal diode. No recovery time, no diode voltage. Just a small delay.

     Hi Jens, see here for linear, I am not able to find equivalent from National Semiconductor or TI, I am sure NS has a similar solution.

     Regards

     Roberto

  • Roberto Romano said:
    Hi Jens, see here for linear,

    Yes, this one is a good choice, but "V0min=6mV". 6mV, not 0.0mV. And only if the OpAmp does not have to sink any current (even not if it has to discharge line capacitance ot the input capaictance of an ADC S&H circuit). The diode/pulldown circuit has no problems, the minimum output voltage is determined by the sink current and the pulldown resistor. It is exactly zero for no current, and very, very low for any normal current that needs to be swallowed.

    I agree, that for most cases, teh fe wmV offset won't hurt, but for a precision rectifier that shall, say, measure oltages in the mV or µV range, this is inacceptable.