VCA2615 Vclamp function

Takumi Suzuki1

Mastermind 8610 points

Hi,

I have one question about the Vclamp function.

I think that the description of "Fig 48" and the explanation of datasheet "page 22" seem to be different.

Though page 22 explains that "Vclmp is equal to the peak differential signal" at H*/L=0, the output is about x2 value of Vclmp on Fig 48.

Probably I misunderstood Vclmp function.

Could you please tell me what I mistook?

Best Regards,

over 9 years ago

0 Takumi Suzuki1 over 9 years ago

Mastermind 8610 points

Sorry, as I failed to insert the grafic image, let me show them as below.

Thank you for your support.

Best Regards,

0 Amy Schnoor over 9 years ago in reply to Takumi Suzuki1

TI__Mastermind 18285 points

Takumi-san,

I am assigning your post to the correct applications engineer. He should respond soon.

0 Takumi Suzuki1 over 9 years ago in reply to Amy Schnoor

Mastermind 8610 points

Hi Amy-san,

Thank you for your support!

Let me wait for the reply.

Regards,

0 Takumi Suzuki1 over 9 years ago in reply to Takumi Suzuki1

Mastermind 8610 points

Hi,

Can I ask you the situation going?

I would like your idea as soon as possible.

Thank you for your support.

0 Amy Schnoor over 9 years ago in reply to Takumi Suzuki1

TI__Mastermind 18285 points

Takumi-san,

I apologize for the delay the team has been on travel the last few weeks and will return shortly.

0 Takumi Suzuki1 over 9 years ago in reply to Amy Schnoor

Mastermind 8610 points

Amy-san,

Thank you for your kindness, I understood.

0 Amy Schnoor over 9 years ago in reply to Takumi Suzuki1

TI__Mastermind 18285 points

Takumi-san,

I will try to answer your question. Based on the datasheet, if a vclmp voltage is applied the output will be that given by fig 48. Page 22 explains the HIGH/LOW gain function. HIGH gain is active low, so the output is 2x, LOW gain is active high so the output is equal to vclmp voltage. Hope this helps.

0 Takumi Suzuki1 over 9 years ago in reply to Amy Schnoor

Mastermind 8610 points

Amy-san,

Thank you so much for your kindness.

Let me ask you again to confirm if my understanding is correct.

1) About Fig48

Vout(pp) is "differential" signal, that is to say, Vout(pp) = (VCAOUT+) - (VCAOUT-).

Is this correct?

2) Datasheet P22 description

There is a description "Vclmp signal is equal to the peak differential signal developed ..."

I would like to confirm the meaning of "equal" word.

Does this mean " Vclmp = (+Vout) - (-Vout)", right?

Does "equal" mean "2x" (equal to the subtraction result) not "1x" (equal to Vclmp voltage), correct?

Sorry, my question maybe the question to ask the meaning of "equal".

Thanks for your support.

0 Amy Schnoor over 9 years ago in reply to Takumi Suzuki1

TI__Mastermind 18285 points

Takumi-san,

It is all based on the High/Low gain function. HIGH gain is active low, so the output is 2x, LOW gain is active high so the output is equal to vclmp voltage. The output is based on whether High or low gain is set if you get the 2x vs. 1x vclamp.

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VCA2615 Vclamp function