Capacitor Burning issue due to unusual reasons

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You didn't yet answer the question if the burned capacitor was observed in the single side post #1 or double side post #15 PCB. You say "We will plan for 2 Layer", sounds like the #15 PCB exist only as a CAD design yet.
 


ULN2803 must be grounded or it will not work.
What I meant was that each solenoid gets its own free wheeling diode in addition to whatever the ULN2903 does.
that way, one solenoid, one diode. I recommend at least 50V and 1 A, preferably 100V and 3A.

since you do not add a series resistor (for example) to limit the solenoid current, you could draw a lot of current though the solenoid,
perhaps more than the ULN2803 specification, but that would mean the driver part of ULN280 is also carrying more current than specified

what is the nature of the 15V source to your circuit?
 

What I meant was that each solenoid gets its own free wheeling diode in addition to whatever the ULN2903 does.
that way, one solenoid, one diode. I recommend at least 50V and 1 A, preferably 100V and 3A.
Click to expand...
To achieve what, particularly? 2803 diodes are just fine for regular solenoid load, no reason to add anything. If you fear shorted solenoids however, a series resistor would either have no protections effect for 200 mA rated solenoid or dissipate huge power.
 

When posting schematic, try to post correct one.

Your capacitor is, according to schematic, is connected to 5V, on PCB is across 15V
If the supply is 15V, your cap should be for at least 25V working voltage.

Also, missing decoupling capacitor of cca 0.1 uF on HC595 supply pin.
LEDs should be connected to 15V supply. No need for additional supply just for LEDs.

Second, PCB is designed by amateur.
 

My point about LED PIV is this:

If the solenoid is ON, the output pins of the 2803 will be close to zero and about (VCC -Vf)/470k = 7uA of LED current will flow. (not really enough to be visible)
If the solenoid is OFF, the output of the 2803 will be at +15V and the anode of the LEDs will be at +5V so the reverse voltage will be 10V which almost certainly exceeds the LED rating. It will not light up but could be permanently damaged. It would be more sensible to wire the LEDs and thier series resistors across the solenoids and use a resistor of say 12K instead of 470K.

Brian.
 

Is it - or was it - a tantalum capacitor? these are famous for failures

else a high capacitance SMT part with an unknown dielectric - these can overheat due to ripple current - or volt spikes ...

how is it obvious that it was the upper left corner - please explain ?
 



ceramic capacitors blow up when they see a serious overvoltage.
since the current does not flow through the capacitor, that is not likely the issue.
all the other things being discussed, LEDs, solonoids, ULN2803... etc are not likely the culprit causing your fire.

If you replace the capacitor, does the board work?
or is something else broken also?

you posted the make and part number of the power supply
have you looked at its output voltage transients when the solenoids go on and off?
specifically when there is a large current change from the power supply?
especially when the current drops significantly - say all of the solenoids turn off at the same time?
 

since the current does not flow through the capacitor
Click to expand...
sorry, wwfeldman, this is not quite correct - capacitors carry AC and spike currents - some times quite large currents, a 10uF cap going from 12V to 15V in 10uS must carry (import ) 3A while the change is taking place, and 3 amps out for a similar discharge ( I/C = dv/dt)
 

@Easy peasy
you're right
caps see transient currents
but this particular cap does not see the current from the supply to the circuit

is your example likely to blow up the the cap?
or does one need a much larger deltaV and hence a much larger transient current?

what I'm thinkng is that the power supply has a large voltage transient when the current drops significantly.
such as when a lot of the solenoids turn off
not 3V, more like 10 or 20 V transient
 

I agree the source power supply is likely spiking up when some or all of the relays turn off, a 5W 18V zener should cure it ...
 

Your pcb design (post #1) seems to be flawed.
For example - the island marked "GND" is actually isolated. Also I see at least one unrouted connection.
 

correct - in both pcb's the gnd copper appears to have no connection to the pcb push on connector ...
 

Both ground nets are apparently connected by a SMD jumper next to the 10µF input capacitor.

The ground net isn't exposed at the solenoid connector, only at the input connector.
 

FvM said:
Both ground nets are apparently connected by a SMD jumper next to the 10µF input capacitor.

The ground net isn't exposed at the solenoid connector, only at the input connector.
Click to expand...

That's possible, since it's shown as an airwire. However the keyword is "apparently" and worth crosschecking.
 


Dear!
Thanks for your suggestions.. Sure We will change the resistor Value. If so is there any chance to getting heat from LED? Since it is closed environment( Metal Case) there will no air circulation or very less. If We increase the brightness temperature is moderately increasing( Device will work 20 Hours Per a day). More than 200 PCBs will be used in a sngle application. 200 x 8 = 1600 Leds.. Please consider

- - - Updated - - -

FvM said:
You didn't yet answer the question if the burned capacitor was observed in the single side post #1 or double side post #15 PCB. You say "We will plan for 2 Layer", sounds like the #15 PCB exist only as a CAD design yet.
Click to expand...

It is observed in 2 Layer PCB.

- - - Updated - - -


So We need to add 100V/3A Diode extra?? As fly wheeling diode

Shall We add any minimum resistance for Current limiting purpose..? Say Less than 3oHM( 1206 Package). We will separate ULN2803 Power from main input power

We using Meanswell SMPS for powering purpose.. Power directly goes to Slave board. From Slave board to this PCB We will connect Ribbon cable(14 way).. data and power lines are traveling through 14 way cable
 

(15V - 1.6V)/12000 = 0.0011A so all 8 LEDs turned on at once will pass ~9mA and combined power dissipation of all LEDs and their series resistors will be about 0.13W, barely warm enough to notice. If you use 470K resistors the LEDs the LEDs will be so dim that you won't see them anyway.

Brian.
 


Hi..
Well said.. After reading all suggestions in this thread I am like floating wire and expecting a good current flow..... Just joking..

Practically there is no problem with ULN2803 part. No issues We faced yet.. We have seen one Chinese board a week ago for same application. They have not added any diodes after the ULN output.. No series resistors too.. Since they have used there own (Customized) components We could not reverse engineering all other inputs.. Anyhow 1oHm series resistor is good I think so
 

If the chinese board is based on similar components as yours, it's easy to reverse engineer
You know the pinout of voltage regulator, you can alwaus trace the supply for shift register and so on, and so on.

BTW, which LEDs are you using? Maker, part number? Color?
If you realy need LEDs, ordinary LEDs will work with 20mA nominally but will be visible even if 5mA is flowing through them.

For example, green Kingbright 0603 LED has Vf of arround 2.1V. If you supply them from 15V, for 5mA current the series resistor should be 2580 Ohm, first standard value 2k7 If you need all 20mA through LED then the series resistor should be 645 Ohms, first safe standard value 680 Ohms.

Also, check the datasheet for ULN2803, Page 5/8, Figures 11 and 12 (Peak Collector current in function of duty cycle)

View attachment ULN2803.pdf
 

Was it ever disclosed if the original offending cap was a tantalum type?
 

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