Hi Team,
There is a good new that we DIN TPS7B8601-Q1
And there have one thing need your support!
Please help check this SCH, THX
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Hi Team,
There is a good new that we DIN TPS7B8601-Q1
And there have one thing need your support!
Please help check this SCH, THX
Hi Kygo,
What is the purpose of the inductor L9 at the output of the PTS7B86-Q1? There is also an inductor at the input (L10) but I can't find the part from the part number. What is the purpose of that inductor?
Also, what is the expected load current?
Regards,
Nick
Hi Nick
After discuss with customer , we have updated SCH
And Io= 0.4A
Please help check this SCH, THX
Hi Kygo,
It looks to me like they just cropped it because I can still see the inductor. With 0.4A and a 47uH inductor there is a high potential for large voltage spikes that can cause an overvoltage on the output and destroy the device.
Other than that, everything looks fine, but I do have one thing to note. 220uF on the output is fine for stability, but remember that with such a large output capacitor and with a device that does not have a soft-start, there will be a lot of inrush current that can drag down the 12V line during startup, and the TPS7B86-Q1 will certainly start up in current limit which is a leading cause of overshoot on startup.
Regards,
Nick
Hi Nick
Got it , I will discuss with RD!
I'm calculating the startup time, can you help me confirm whether my calculation is correct?
TPS7B86-Q1 after get 12V then will need 100uS to give Vcc9V
Hi Kygo,
The spec that you mentioned is not for startup time; this spec sets a max time that the output will take to fully respond to a load transient.
We don't have a clear startup plot without entering current limit (as in Figure 6-34 in the datasheet), but this plot can still give a decent approximation to startup time because the device is not in current limit until the reference voltage has come up all the way. From the plot, the output takes about 200ms to come up to its final value.
Regards,
Nick
Hi Nick
After discuss with customer,
Customer feedback if the load change to 0.1A and a 47uH inductor, is there still will have a high potential for large voltage spikes ?
How could we confirm whether the spike would cause an overvoltage on the output? Is the fomular is "V=Ldi/dt" ? THX
Hi Kygo,
Yes, using V = L*dI/dt is a good starting point. The caveat here is that this formula describes instantaneous behavior of the circuit, and it is often misused because people want to approximate the rise of fall as linear, but the actual peak voltage spike you would see would be where the slope of the current is steepest. However, doing a quick calculation can be useful because if the calculated voltage spike is much smaller than the abs max rating of downstream components then that can help to build confidence that this won't be an issue.
Here's a quick example. With 100mA load current, if for some reason the load abruptly stops drawing current and takes, say, 1us for its current to go to zero, there would be a 47uH * 100mA / 1us = 4.7V spike in the direction indicated in the following image. Load currents could collapse faster than this, which would also increase the peak spike. This is a rough approximation, but you can see that the magnitude of the spike already looks like it could be problematic.
Can you please explain why you want an 47uH inductor at the output of the LDO?
Regards,
Nick