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TLV767-Q1: Help confirming ripple spec in Datasheet?

Jim Smith1
Prodigy 160 points
Part Number: TLV767-Q1

Hello, I'm having trouble discerning something in a datasheet for the linear regulator TLV767-Q1. 

Here is the datasheet: 

https://www.ti.com/lit/ds/symlink/tlv767-q1.pdf?ts=1646337144566&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FTLV767-Q1

The line I am confused about is on page 6. 

In the Electrical Characteristics, it is the Line Regulation (ripple) information at the far right, stating 0.02 %/V

Here is the whole line copy/pasted:

ΔVOUT(ΔVIN) Line regulation(1) VOUT(NOM) +1.5 V ≤ VIN ≤ 16 V, IOUT = 10 mA 0.02 %/V

What exactly does 0.02 %/V result in, if we can state it in mV?

The Output voltage in my application is 9V. The input is 12V, give or take 10mV. 

I am looking for a regulator with a ripple spec that exceeds [ 1.6mV TYPICAL / 32mV MAX ] since that seems to be the best standard achievable for XX7809's of all brands/variations. 

So, does 0.02 %/V mean 9 X 1.0002? If so, that gives us a high point in AC ripple of 9.0018. That is 1.8mV MAX, which would be quite good.

Or, does 0.02 %/V mean 9 X 1.02? If so, that gives us a high point in AC ripple of 9.18. That's 180mV MAX, which would be quite poor. 

So i'm thinking it must be 1.8mV MAX. Does that seem correct?

Thanks to anyone who can help to confirm this! 

  • 0
    TI__Mastermind 38115 points

    Hi Jim,

    First, it is likely that the test condition VOUT(NOM) +1.5 V ≤ VIN ≤ 16 V for line regulation has a lower test limit for VIN as VOUT(NOM) +1.5 V because the max dropout spec is 1.5V so that for every case the LDO is not in dropout. From that VIN point, increasing VIN will change the output by .02% of VOUT (9V for your application) for each volt. So for your application, the VIN differential from the spec you see is 1.5V (VOUT + 1.5V = 10.5V to 12V), and so converting this to volts you have .02% * 9V / 1.5V = 1.2mV max. Alternatively, some people like to leave it as a percentage and calculate the additional output voltage error due to line regulation (and load regulation) and add it to the overall output accuracy spec. In this case, you would add .02%/V * 1.5 V = .03% to the output accuracy spec for line regulation. 

    I would also like to clarify that the line regulation spec is not a measure of AC ripple. It is a measure of a DC change in the output when the input is increased at DC. 

    Regards,

    Nick