Measuring an IC's input resistance

[BlackMelon]

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".

 

[davenn]

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".

Have you read the datasheet for this IC ?
The input is going to be a little more involved that the way you have shown
This is because the input + and - are the pins to an op-amp ( comparator) and as such, the input impedance is likely to be quite high.
This is typical so that the IC doesn't load down ( affect) the incoming signal or its source

there's lots of other good info in the datasheet that you may or may not be aware of


Dave

 

[BlackMelon]

I have read the datasheet several times. For IN+ and IN-, they clearly belong to a comparator, having a high input resistance, of course. But for the HEN, how can you know that it belongs to an op-amp (or a comparator)?

Thank You

 

[(*steve*)]

I have never read the resistance of a chicken. However for high resistance inputs you could use something like an electrometer to measure the input current.

If the current is as high as several pA, a more simple picoammeter may suffice.

However these instruments are finicky beasts, and you probably don't have one hanging around.

A simpler option is to add a large series resistance then increase the input voltage to produce an output the same as that of the original. If you add 100MΩ and need to increase the voltage by 20% to get the same output, then 100MΩ is 20% of the input impedance (which thus must therefore be 500MΩ).

 

[BlackMelon]

A simpler option is to add a large series resistance then increase the input voltage to produce an output the same as that of the original. If you add 100MΩ and need to increase the voltage by 20% to get the same output, then 100MΩ is 20% of the input impedance (which thus must therefore be 500MΩ).

That's a good idea! Thank you. But I am not sure this method will always work. Thinking of a half-wave rectifier (a diode in series with a resistor), you can get a lower resistance reading just only in one direction of hooking. So, to measure an input impedance, do we need to input the exact waveform that drives the input pin when the IC operates?

 

[(*steve*)]

That's a good idea! Thank you. But I am not sure this method will always work. Thinking of a half-wave rectifier (a diode in series with a resistor), you can get a lower resistance reading just only in one direction of hooking. So, to measure an input impedance, do we need to input the exact waveform that drives the input pin when the IC operates?

If the input impedance is asymetrical, you can handle the positive and negative cycles separately.

If the resistance is non-linear, you can start with different initial voltages.

With really high input impedances, the additional input resistance can cause a significant increase in noise. It pays to have a signal significantly greater in amplitude than the noise floor.

 

[Alec_t]

The HEN input is a logic-level Schmitt-trigger input. It typically sources 200uA for logic low and sinks a maximum 1uA for logic high. It can be driven directly by an MCU.

 

[BlackMelon]

The HEN input is a logic-level Schmitt-trigger input. It typically sources 200uA for logic low and sinks a maximum 1uA for logic high. It can be driven directly by an MCU.

Thank You, my MCU is capable of sinking/sourcing those values.

After reading the datasheet of HIP4080A, I have some doubts about the offset voltage (Vos) of IN+ and IN-. Here I have the picture, showing my understanding based on https://en.wikipedia.org/wiki/Input_offset_voltage.

Now I see a contradiction between the application note AN9324.4, page 2 and the information from an e-mail from the company. The app note says the input offset voltage must be kept lower than 5mV. But the chip's company says Vin- and Vin+ can be 100mV different from each other.

So, do I understand their terms right?

 

[Alec_t]

Over the full operating temperature range the input offset voltage (a random-among-devices error voltage) might be anywhere between -15mV and +15mV according to the datasheet. The App Note's reference to 5mV looks wrong to me.
I think the company was trying to tell you that it is preferred to keep the differential voltage between the two inputs below 100mV, though I don't understand why since, as per the datasheet, the common mode input range can go from +1V up to Vdd-1.5V.

 

[(*steve*)]

I believe the offset voltage listed in the application note is the maximum offset of the op amp (although as noted above, other specs say 15mV), i.e. the differential voltage required to make the output 0V.

Without seeing the email, I can only suspect what it means, however you typically don't want to drive the inputs with too disimilar voltages as it can cause the op-amp to be slow to respond to changes in the input. Perhaps the 100mV is the recommendation in this area.

 

[BlackMelon]

Here is the pdf file of my e-mails to and from Intersil.
https://www.mediafire.com/?8k71t53771cb8pd

EDIT: The information about the comparator is at the bottom of the page 1
They have just apologized for the wrong circuit on HEN!

 

[BlackMelon]

Here is the reply I have just got from Intersil about the "offset"

"The offset should not matter. The offset is the functional difference between the two inputs when both are tied to the same voltage. Note that the OUT pin is usually used for generating hysteresis but is not needed if you overdrive the input by a 100 mV or so.
When you bias both inputs with the 35.3K and 4.99 K the inputs will be about 2 V (1.98 if you do the calculations) .
If you then drive one of the inputs from the 49.9 K which is being driven by the 5V logic signal its value will change about +/- 150 mV when the logic signal goes high and low. This is enough voltage to drive the comparator input to be higher or lower than the voltage on the input that is not driven . This will overcome any offset of 5 or 10 mV or even more."

Both IN+ and IN- are tied to the same voltage so why the functional difference is not zero?

 

[Alec_t]

Both IN+ and IN- are tied to the same voltage so why the functional difference is not zero?

Because of slight mis-matching of transistors within the opamp in the IC. No manufacturing process is perfect.

 

[BlackMelon]

Because of slight mis-matching of transistors within the opamp in the IC. No manufacturing process is perfect.

Thank you Alec_t for making me realize that issue. May be I have to review the differential amplifier topics for more information again.



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.".

 

[BlackMelon]

In Pspice's parametric sweep, does anybody know how to display a sweeping parameter for each trace?

(For example, like in the attachment, how can I know (or display) a particular resistance value for each V1 curve? I got so many traces now but I don't know their parameter values)

 

[Alec_t]

Try a DC Operating Point analysis instead of a Transient analysis.

 

[BlackMelon]

Hi y'all! I have just figured out how to find the sweeping parameter (Vcc in my case)! Just right click on each trace > Trace Information. The first time I did; I was unable to view the value of Vcc for each particular trace because I did sweeping with too many Vcc values, so the traces are packed too tight.

 

[BlackMelon]

Alect_t, your circuit works excellently!

In the simulation, the upper waveform is the input of the circuit, the lower one is that of Q2 collector. I sweep the value of Vcc from 0 to 30V (so I got many traces).

What does cause the undershoot in the collector voltage when the input is at the rising edge?

 

[Alec_t]

What does cause the undershoot in the collector voltage when the input is at the rising edge?

Probably junction capacitance.
Vcc=30V

 

[BlackMelon]

Thank you! I'll try it! Thank you for warning me about 30V too. I'm not gonna input that to the Vdd. I just want to see the extreme trend of the circuit.

I have a question about driving the IN+ pin of HIP4080A. I took into account the battery (Vcc) drain. The IN- voltage is fixed at Vcc/2. The TTL logic of an MCU is passed to VIN+ via a voltage divider, having 3 resistors. I did a Vcc sweep from 16V down to 7.6V, and saw that the DC level of IN+ is higher than Vcc/2 when Vcc = 7.6V. And, the opposite situation happened when Vcc = 16V.

I was trying to solve this problem, so I made a BJT circuit with 2 base resistors, Rc, and Re. I picked two Vcc values to be 30V and 7.6V (again, I am not going to input 30V to the driver's VDD). I hope that lowering of the battery will lower the Vc bias point. But what I have got is the VIN+ always greater than VIN-.

So, is there any way to address this problem without using switching power supply to fix the Vcc?

EDIT: The voltage divider task is to confine the VIN+ to be in the common mode voltage range of the "IN" comparator.