ogulcan123
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My power stage produces 3 voltages which are 3.3V, 3.8V and 5V, but I'm not using 3.8V yet, I'll use it later for GSM. You can check it out below.Does your power source produce more than one voltage and if so, do the supplies rise in the same order as before?
In fact, everything has changed except for the voltages. You can check it out. I've really upgraded my power design to endure in even truck environment.What change was done in the power stage in new version ?
In this power design, I've actually used some large filter capacitors which is cumulated as 702uF at my 3.3V net.Two likely reason:
1. Reset applied too early;
2. Reset voltage changes slowly.
Both can be related to the power stage: too large filter capacitors can worsen the case.
Great - thank you.
Almost certainly what is happening is the 3.3V supply to the MCU is rising slower than it used to. In order for the MCU to reset properly, the RST pin has to be held low until the 3.3V has stabilized and the clock has had time to start and perform some internal actions to get the MCU ready to run. The delay between power being applied and the reset being released is created by the time constant of R15 and C7. The voltage across C7 has to stay below the reset pin threshold for a while after the 3.3V is present. If the supply rises slowly, C7 charges up at the same time as the 3.3V line rises so the delay is reduced or not there at all.
You can try fixing it by increasing the value of C7 to say 10uF so the time constant is 100 times slower. A better solution is to use a dedicated reset IC that monitors the 3.3V then waits a few mS before producing the reset signal.
If you increase the value of C7 you should also make two other changes to ensure safety:
1. add a resistor of around 100 Ohms in series with the reset switch to prevent sparking as C7 is discharged through it.
2. add a small Schottky diode (BAT85 or similar) across R15 with it's cathode to the 3.3V line. This will ensure that if the 3.3V turns off quickly, the remaining charge on C7 goes back to supply instead of discharging into the MCU.
Brian.
the problem is not the reset button. My reset button works perfectly. The problem is that the MCU does not boot when I directly plug the power connector.
That's not what I meant. The fact that the reset button work is justifying what I said. Think of it this way, if you hold the reset button down while you apply the power then release it, does it start properly. If it does, you need simulate that electronically by increasing the time constant on the reset components.Thanks Brian. However, the problem is not the reset button. My reset button works perfectly
Okay, I get it, so you mean the MCU is actually resetting every time it boots and if I increase the time constant in reset button, it will work.That's not what I meant. The fact that the reset button work is justifying what I said. Think of it this way, if you hold the reset button down while you apply the power then release it, does it start properly. If it does, you need simulate that electronically by increasing the time constant on the reset components.
Brian.
The MCU may not being restarted because it is likely not being reset. It suggests that the power supply stage is somehow retaining some charge at the 3.3v power rail, preventing the reset capacitor from being fully discharged during the power turn-off. You could check this, and once confirmed, just adding a diode reversely to the power input could ensure an instantaneous discharge in the power off event.
Yes. You may omit the big electrolytics completely. I see no benefit.and I will also try to lower the filter capacitors a bit.
Hi,
Please confirm:
C42 and C54 are ceramic capacitors and placed very close to the Vin pins of the switching regulator.
Yes. You may omit the big electrolytics completely. I see no benefit.
Especially the combination of small input capacitors (144uF at Vin_D) and large output capacitors (2650uF) is no good combination.
I´d rather place the big ones to the input side (Vin_D), this gives a large energy reservoir.
Some calculations:
* your switching frequency is about 600kHz
* this generates a soft-start time of 1.6ms
* assuming you charge the output capacitors within 1.6ms to the (mean) value of 4.5V then you need a current of about 7A!!!
*****
Another aspect:
* your output capacitor is 1450uF at 3.8V. This is a stored energy of 10mJ. (this determines the size of the capacitor)
* if now the voltage drops by 0.4V to 3.4V ... you are close that your system refuses to operate safely. The remaining energy in the capacitor then is about 8.4J .. but useless.
You can just use about 16% (1.6J) of the stored energy.
The same energy can be used from a 55uF capacitor (at input) when you allow it to discharge from 11.5V (12V - diode drop) to 8.5V (ENABLE threshold).
But this means you need to specify the minimum input voltage to 12V.
The problem now is that your input voltage range is huge. It is 6.5 : 1. Does it need to be that huge?
Especially the lower limit of 8.5V is problematic.
Klaus
It has to be big, because my circuit has to survive in an automotive environment such as truck, where there are high voltage spikes. But I can lower the value of my TVS diode and narrow this range a bit.The problem now is that your input voltage range is huge. It is 6.5 : 1. Does it need to be that huge?
Especially the lower limit of 8.5V is problematic.
Maybe I was not clear enough:That's because the bigger the capacitor the more my circuit handles short electric cuts and abnormalities from the grid.
It discharges at a slow rate but it discharges. I've tried after waiting hours and it is not fixed. I'll try some solutions though
just out of curiosity, how slow is this discharge rate? Regardless of whether or not it is the cause of the problem reported above, the voltage of Reset capacitor should be discharged faster than the MCU's VCC voltage.
Here are the transient responses where yellow is the input voltage and blue is the reset voltage
I mean 3.3V voltage, the one that feeds the MCU.Anyway, what exactly do you mean by 'input' voltage ?
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