Author Topic: Input Power Peak Requirement  (Read 944 times)

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Offline PseudobyteTopic starter

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Input Power Peak Requirement
« on: July 24, 2019, 07:51:13 pm »
I have a design project I am working on and I am having a bit of confusion with a specification. The specification is:

The input power peaks shall meet the following requirements:

Current time product <= 10 A.s
Peak time <= 0.5 s

I am not quite sure how to interpret the requirement. It is just confusing me. Say the input voltage is 12V. 12 V * 10 A.s = 120 J, 120 J / 0.5s = 240W.

To me it seems like the peak time could be tiny and you could have a huge current briefly and the requirement would be satisfied. I guess what I am trying to get help with is the intent of a specification like this.
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Offline bob91343

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Re: Input Power Peak Requirement
« Reply #1 on: July 24, 2019, 07:53:36 pm »
Yes the input isn't completely specified and does allow for large currents of short duration.

Your computation of 240 W is only the peak power.
 

Offline T3sl4co1l

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Re: Input Power Peak Requirement
« Reply #2 on: July 24, 2019, 08:23:55 pm »
A.s == C (coulombs).  It's a charge figure.

They shouldn't call it a power figure, since it has nothing to do with power, in and of itself.  Can you ask the customer to clarify/correct, or is this something you have little recourse about?

If the supply voltage is 12V, you can use a maximum of Q/V = C (that's capacitance) or (10 A.s) / (12V) = 0.83 F (farad).

Seems like quite a lot, and even a modest motor would have trouble drawing quite that much inrush I think?

The time frame needs a definition (peak above what threshold?) but it further implies that the input current should be limited to, roughly, something around 20A.  Depending on how that is defined (instantaneous, average over the duration, etc.?), it may be less stringent than that.

The energy associated with this charge, is also unclear; but if it is a linear capacitance being charged, then the energy is not the product again, but half.  Consider E = 0.5 C V^2, then sub in C = Q/V.  Mechanically speaking, it's the same relationship between momentum (kg.m/s) and energy (kg.m/s^2). :)

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Offline PseudobyteTopic starter

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Re: Input Power Peak Requirement
« Reply #3 on: July 25, 2019, 02:31:11 pm »
Thanks Tim. Yeah I sent an email yesterday asking them to clarify the requirement, and its intent. I will let everyone know what exactly that is.
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Offline bigamps

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Re: Input Power Peak Requirement
« Reply #4 on: July 25, 2019, 03:53:14 pm »
I guess I'd interpret it as 20A would be OK for 0,5s, but 40A would also be OK as long as they don't stay for more than 0,25s, or even 50A for 0,2s, and so ok for even shorter times. That is, the current * time product must be <=10As. Like somebody said it is like limiting the input capacitance.

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Offline PseudobyteTopic starter

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Re: Input Power Peak Requirement
« Reply #5 on: August 05, 2019, 06:28:11 pm »
The requirement was changed to:

"The input circuit must be able to withstand its own maximum inrush current."

Much less confusing but still pretty vague.
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