Measuring an IC's input resistance

Discussion in 'Analog & Mixed-Signal Design' started by BlackMelon, Jun 25, 2017.

  1. BlackMelon

    Thread Starter Member

    Mar 19, 2015
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    Hello,

    My IC takes in a square wave input as written on the attachment. I want to apply that input and measure the current, so I will be able to determine the input resistance value. However, my multimeter is not a true RMS one. As what I know, the RMS current is accurate only if it is in a sinusoidal waveform. So, can I apply the ripple waveform shown on the attachment?

    Thank You

    Ps. My IC is HIP4080A, a mosfet driver. The input pin is "HEN".
     
  2. wayneh

    Expert

    Sep 9, 2010
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    Did you read what the datasheet says the input impedance should be? The current may be too low to measure with your meter.
     
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  3. Alec_t

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  4. BlackMelon

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    Mar 19, 2015
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    http://www.communica.co.za/Content/Catalog/Documents/D0199184670.pdf
    My digital multimeter has a micro-amps range with a resolution of 0.1uA. Anyways, thank you for warning me. :)

    Thank you. I have just read that section. But, I sent an e-mail to the IC's company, Intersil, asking for how to make the IC be able to interact with the MCU. They replied me back to use the circuit in the attachment. The thing is that the voltage swing on HEN pin is not going to be 0 and 5V like in the datasheet. It is going to be 16V and something near 0V instead.

    So, on page 5 of the datasheet, can I just find the resistance by, when HEN is high, 5V/10microamps = 0.5Mohms? How about when HEN pin is low?
     
  5. Alec_t

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    Sep 17, 2013
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    That 16V is perilously close to the Absolute Maximum rating. If you want the chip to have a long life it is not good practice to run it at/near its max.
    The component values in your circuit don't allow HEN to switch both high and low. Consider using either a different configuration of the NPN transistor (with appropriate logic inversion), or perhaps use two NPNs?
    Why aren't you switching HEN at normal logic levels, e.g. with 5V as used by the MCU?
     
    Last edited: Jun 25, 2017
  6. BlackMelon

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    Mar 19, 2015
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    I am just trying to find what's good in the company's suggestion about driving HEN. May be they just want to make things more complicated?
    I was trying to find how low can the HEN voltage go when the input logic from MCU is low. I got two paradoxical values shown in my new attachment. I have no idea why they suggested a common-base amplifier like that.
     
  7. Alec_t

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    That circuit simply won't work reliably, if at all. Vbe varies from part to part and with temperature. Why not just drive HEN with the micro output directly?

    Edit:
    Here's a simulation of your circuit
    HEN-drive.PNG
     
    Last edited: Jun 25, 2017
  8. BlackMelon

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    Mar 19, 2015
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    Thank you. I intended to drive it with MCU logic, but the HIP4080A's engineer suggested that darn circuit. That made a lot of confusion to me, and now your simulation clarify them all. So, I will drive the HEN with the micro output then.

    Thank You again :)
     
  9. BlackMelon

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    Mar 19, 2015
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    I have just recognized that the company suggested me a lot of funny things. For the IN- pin, we can just use a regulated 5V, and half it down by a voltage divider for IN-. Using +16V from a battery will cause problems when the battery is running low.

    The IN+ pin can be driven by MCU signal via resistors in that manner. But the resistors must make the IN+ pin swing in the common mode voltage range, which is between Vdd-1.5 and 1V. They said, for the chip to consider IN+ greater/lower than the IN-, the voltage difference between the IN+ and IN- should be more than 100mV. This value is low enough to fit in 5V to 1V range. So, we don't need that high voltage supply to control the IN+ and IN- pins.

    Do you agree with me?
     
  10. Alec_t

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    From my reading of the datasheet, pins 15 and 16 of the IC need the same supply voltage, in the 9.5V-15V range, but the IN+, IN- and HEN input pins will work ok in the 0V-5V range (but will tolerate voltages up to the level on pins 15 and 16). In particular, HEN <0.8V = logic 0; HEN > 2.7V = logic 1.
     
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  11. BlackMelon

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    Mar 19, 2015
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    Hi all,

    Thank you everyone. I think my doubts are all cleared out because of you all. I have just received a reply back from Intersil. They apologized for the wrong circuit for driving HEN. Also, they clarify terms in the datasheet. I attached my e-mails back and forth to the company for you all.

    So, to conclude everything, the input resistance of the HEN is very high because it's a Schmitt trigger. The HEN pin and DIS pin both use TTL, according to AN9324.4 app note. The maximum voltage difference between IN+ and IN- to set the output of a comparator to 0V is +/-15mV (Vos, offset voltage). So the company recommend the IN+ to be driven higher/lower than the reference, IN-, +/-100mV. And, we cannot feed TTL logic into the comparator because the "low" of TTL is lower than the lower limit of the common mode voltage range (+1V). Also, the bootstrap cap of the circuit can be determined by the formula at the end of the second paragraph on page 6. The higher micro-farad of the bootstrap cap is the lower the frequency you can use.


    https://www.mediafire.com/?e8j818z8as5rzw2

    I think that's pretty much of it

    Thank you again :)
     
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  12. BlackMelon

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    Mar 19, 2015
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    I still have one more doubt about the low level input current of HEN (check my attachments). The company put the current (270uA) in that way in the schematic. Why does the current have to be connected in that way?

    EDIT: Will the current be added up with the current through 40.2k, inputing more current than 270uA into the HEN pin?


    PS. This is a message from Intersil about the reason that we need transistors at DIS and HEN pins: "The reason for the transistors is to allow the pull up resistors to go to the supply voltage on the HIP4080A. The problem with driving with logic alone is that the proper power up conditions can not be guaranteed if the logic supply and the HIP4080A supply do not have exactly the same start-up conditions. For example, how can the logic output exactly follow the supply rise of the HIP4080A without over voltaging the pin on the HIP4080A when power up occurs.".
     
    Last edited: Jun 28, 2017
  13. Alec_t

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    Point taken about power-up conditions.
    In that case I'd probably use one of these drive circuits for HEN :
    HEN-drive2.PNG
    The first requires logic inversion; the second doesn't. The Q1 or Q3 transistor can sink at least 1mA, which easily satisfies the input current requirement.

    Edit:
    A diode added in series between R2 and Q1 collector (similarly between R5 and Q3 collector) might be advisable to guard agains current flowing from the MCU via the collector junction to HEN in the event the MCU 5V rail comes up before the '4080's supply rail does.
     
    Last edited: Jun 28, 2017
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  14. BlackMelon

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    Mar 19, 2015
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    Thank you for the circuit. They said in their FAQ that driving the input pins during the power-up period may bias the internal ESD circuitry. Ok, I can remember this as a rule. But could you draw a simple ESD circuitry? I just want to know more about the nature of this power up problem, so I can know how the transistor circuit protects the HIP4080A from the over voltage.
     
  15. Alec_t

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  16. BlackMelon

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    Mar 19, 2015
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    If these circuits are the case in HIP4080A, can we just hold the HEN pin at 0V during the supply start-up time without the transistor circuits?
     
  17. Alec_t

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    That seems logical, but would you risk it? Safest to go with what the manufacturer recommends.
     
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  18. BlackMelon

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    Mar 19, 2015
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    I am not that adventurous. LOL

    I am doing a PSPICE simulation on your BJT circuits, plotting the MCU waveform and the HEN2. Because I have done a Vcc sweep, I got so many traces of HEN2. How can I know the Vcc value for a particular trace on the simulation result?
     
  19. Alec_t

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    I don't use Pspice, so can't say.
     
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  20. MrChips

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    Oct 2, 2009
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    Put a resistor in series between the signal source and the input pin.
    Measure the voltage at the source.
    Measure the voltage at the input pin.
    Increase the value of the series resistor until the voltage at the input pin is one-half the voltage of the source.
    The value of the input resistance is the same value as that of the series resistor..
     
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