Hi,
pls clarify my calculation corrent or not?
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Hi Prema,
Thanks for reaching out and providing details related to your question.
I see that you have used a straight-line (constant slope) relationship to estimate the input current for a load of 12.64mA. Unfortunately, the device might not exhibit such a relationship with constant-slope. Please refer to the efficiency curve provided in datasheet which shows the relationship (or slope) between output and input currents is not going to be a constant. i.e., The slope (also efficiency) varies as the load decreases.
Due to the above stated reason, you would have to use empirical data from datasheet to find out input current consumption for given load conditions. I understand that IDD is not provided for ILOAD of 12.64mA but it does have IDD value for ILOAD of 30mA as 112mA max. This is the closest and safest value that I recommend you to use. It also gives you the assurance that the device has sufficient margin to the actual load.
Let me know if this helps, thanks.
Regards,
Koteshwar Rao
Hi Koteshwar,
Thank you for spending your valuable time .
I am using the efficiency curve to calculate the supply current, I want to make sure my calculation is correct or not.
VCC | 3.3 V |
VISOut | 3.3 V |
Iload | 12.6 mA |
Efficiency | 0.36 |
IIO | 0.03512 A |
IIO = (VISOUT*Iload)/(Vcc*Efficiency)
Hi Prema,
Thanks for following up with additional questions related to the topic.
Although your calculations are correct, please note that the efficiency curve provided is a typical efficiency curve and thus the current calculated is also a typical input current. There will be variation in efficiency which could make the input current higher than what you have caculated.
I would also like to mention that the currents you have considered only refer to the load on VISOOUT while the I/O channels of ISOW7741 might also consume some current based on their operating conditions and that current at Side1 and Side2 needs to be accounted for to estimate a more accurate value.
Let me know if this helps provide you with more clarification, thanks.
Regards,
Koteshwar Rao
Hi Koteshwar
As per your suggestion I take the current from both side and I add these sum of the current to IIO. Kindly verify it and give your valuable comments.
VCC | 3.3 V |
VISOut | 3.3 V |
Iload | 12.6 mA |
Efficiency | 0.36 |
IIO | 0.04812 A |
IIO = [(VISOUT*Iload)/(Vcc*Efficiency)]+0.0103
Hi Prema,
Thanks for considering to include the current consumption from the I/O channels as well.
Please note that the current consumption of I/O channels is going to exist on both sides of device. i.e., VISOOUT/VISOIN will supply current to the I/O channels on Side2 while VDD will supply current to I/O channels on Side1. Hence, the equation to include this current can be expressed as below.
"I_{DD} + I_{IO}" = [ ( V_{ISOOUT} * {I_{LOAD} + I_{ISO}} ) / ( V_{DD} * Efficiency) ] + 0.0103
where IDD is the current into VDD pin, IIO is the current into VIO pin (Side1 I/O channel current), IISO is the current consumed by the I/O channels on Side2. The sum of IDD and IIO will form the total current consumed by device. I am not sure about your application I/O channel operating conditions but assuming "IIO = IISO = 10.3mA", "Iload = 12.6mA" and "Efficiency = 0.4 @ 22.9mA total load", the total current can be estimated to be as,
"I_{DD} + I_{IO}" = [ ( V_{ISOOUT} * {I_{LOAD} + I_{ISO}} ) / ( V_{DD} * Efficiency) ] + 0.0103 = 68mA
Please note that this is a typical current while when measured on a typical sample at room temperature. The maximum current across devices and across temperature can be higher. Thanks.
Regards,
Koteshwar Rao
Hi Koteshwar,
Thanks for your effort to clarify my queries.
1. Here I'm using ISOW7741DFMR this isolator, for this part number which current I consider for this max current calculation?
2. As per your comments I changed the some points, but I need some clarification for that formula. Here you took IISO value is 10.3mA and also add the 0.0103 A, here both current is same, is there any specific reason to add this current or you took any reference formula. If you referred any formula means kindly send to that or clarify why took this current for twice.
here I'm attached my calculation, kindly referred through it and give your valuable comments.
Hello Prema,
Please allow another day for us to get back to you on this thread.
Best,
Michael
Hi Prema,
Thanks for sharing additional details.
1. Here I'm using ISOW7741DFMR this isolator, for this part number which current I consider for this max current calculation?
The current listed in the table that you have highlighted is given for two test conditions. One test condition is when VI = VCCI which means INx = HIGH and the other is VI = 0V and this means INx = LOW. Please choose the suitable current based on whether your inputs are HIGH or LOW. If your inputs are switching with equal HIGH and LOW bits then you can consider average of two currents or alternatively, you can consider the one that is highest between the two to account for worst-case operating conditions.
Here you took IISO value is 10.3mA and also add the 0.0103 A, here both current is same, is there any specific reason to add this current or you took any reference formula.
The current IDD can be represented as below,
I_{DD} = [ ( V_{ISOOUT} * {I_{LOAD} + I_{ISO}} ) / ( V_{DD} * Efficiency) ]
But the total device current is not just IDD instead it is sum of IDD and IIO. Hence, total device current can be expressed as below,
Total device current ITOTAL = "I_{DD} + I_{IO}" = [ ( V_{ISOOUT} * {I_{LOAD} + I_{ISO}} ) / ( V_{DD} * Efficiency) ] + I_{IO}
I used IISO as 10.3mA as only an example but the actual current depends on your test conditions. I hope this clarifies that the equation has both IISO and IIO currents considered. Let me know if you have any other questions, thanks.
Regards,
Koteshwar Rao