Part Number: TRS3232
Our customer would like to ask the value of absolute maximum rating for terminals C1 +, C1-, C2 +, C2-.
However, I could not confirm the these value in datasheet.
Could you tell me the maximum rating values for these four terminals?
In reply to Y.Ottey:
I'm sorry to hear the issue is still there. Is it possible to capture the waveforms of C1+ and other pins (like C1-, V+, etc) with a specific loading condition? How does the voltage change under different load? Can you share the model of the capacitor? Does it make difference with different batch of IC? I'm more worried if the charge pump control loop functions.
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In reply to Hao L:
Thank you for your reply.
I have confirmed with the customer the model number of the capacitor and the measurement results for other batches.
When Vcc = 5V, C1 = 0.047μF, C2 = C3 = C4 = 0.33μF, the waveform of C1 + is as follows.
The rising edge of the waveform exceeds 7V.
Please check this waveform and give us some advice.
Thanks for your waveform. How was the load condition? How many channels are turned on? What's the data rate? Can you zoom out a bit to show some periods?
Here is a waveform I took with no load for TRS3232. Green is C1+ and purple is C1-. You can see that since there is not much discharge, the charge pump works with a low duty cycle.
Thank you for your reply and send your waveform.
I would like to ask a question.
If the C1 + rating is exceeded as in the customer's waveform, what kind of problems will occur in the operation of the IC?We think that the movement of C1 + is due to the movement of the IC itself, and there is no problem with the original input / output state of the IC.
It depends on the percentage of the life time when the voltage exceeds the abs max value. For example, if the overshoot only shows in the power up or it has a very low duty cycle (like 0.01%), 200mV over the limit might not be a big risk. After all the reliability is a statistical measure. With higher voltage and longer duration, more IC could be damaged in life time.
You could also try powering up the IC after the supply is established and also check if there is any current limit on the supply. Again, C1+ voltage might be different with different loading condition.
I would like to ask question about your answer
It may be a basic question, but please let us know as it is necessary to confirm loading conditon1. Where does the loading condition you say?2. Is there any recommended loading condition for the above question?
What I meant was the bus load, like cable capacitor, any resistance to ground. I was trying to say since the charge pump might behave differently with different loading condition, please focus on the most common case in your application.
I have confirmed to you the confirmation that you were saying before.
The confirmed results are as follows.
①The load condition was changed, but the waveform did not change.
②They replaced it with another IC on same board, but the waveform did not change.
③The following capacitor models are used
C1:GRM40B473K50(MURATA)C2:C1608JB1A334K(TDK)C3:C1608JB1A334K (TDK)C4:C1608JB1A334K (TDK)
④We had you measure the zoomed out C1 + and C1- waveforms with no load, but they differed from the waveforms you posted earlier.
※Green is C1+ and purple is C1-
Could you get advice from the above?
I only found the information about C1608JB1A334K online. It looks like they're 10V capacitor. Can you try with the capacitor with higher voltage rating , like 16V? I'm not sure why the C1 voltage drops the comes back after each charging. The waveforms may look similar. With heavy load, the charging time might be longer and discharging time (idle time) shorter.
I tell our customers to try with the 16V rated capacitor you said.
I have more questions.Is it correct to recognize that the voltage of C1 + changes in the following procedure?1. When the switch is in the state ①, the current flows from Vcc to GND, so the C1 + voltage is Vcc. (C1 is charged.)2. When the state changes to switch ②, current flows from Vcc to V +.At this time, when a charge discharged from C1 is added, the voltage of C1 + becomes Vcc × 2.
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