How can I eliminate the voltage sinking?

[ay001]

I build a pcb.
When it plug in the socket of the back-board,the vlotage will sink to 4.35V for 20us.
But I use the +5V power. So the boards that is sharing the power with it will reset due to the power sinking.
How can I deal with it?

Thanks.

 

[lkm]

It sounds you are building a hot swap PCB and found the PCB will reset. Also, you PCB has a micro processor inside.

I suggest you check the peak current of your PCB. It seem when you put the board to the slot, the capacitor draw a lot of current make the voltage drop. The way you can do is limited the current go into your PCB and reset the CPU after the voltage supply stable.

Remember, you must reset you microprocessor after you supply voltage. It can avoid your processor working wrong.

I hope it help.

 

[jbartos]

Suggestions:
1. The 5 V power supply internal impedance is too high causing the voltage dip to be too low. Try some 5 V power supply with lower internal impedance.
2. Also, try to connect two 5 V in parallel to find out the difference.

 

[nbucska]

I disagree with jbartos:
1.) The when you connect your board which has decoupling caps on it, its impedance = 0 i.e. pulls too much current.
Reducing the source impedance will INCREASE the current spike.
2.) Paralleling two 5 V-s possible only is they are exactly
the same value ELSE one will load the other

I suggest adding a series inductor BEFORE any decoupling cap
on the new board -- this will reduce the peak current and slow down the current spike. May want to add parallel
diode or resistor to prevent sparking on disconnect.

<nbucska@pcperipherals.com>

 

[nbucska]

How many microF decoupling does the new board have ? <nbucska@pcperipherals.com>

 

[jbartos]

Suggestion: My posting is based on the following:
1. Consider a voltage divider circuit applicable to the posted problem, namely:
E4= 4.35V = 5V x Z4 / (Z4 + Z5).....Eq1
where
Z4 is impedance of the board that exhibits 4.35V across its terminals
Z5 is internal impedance of 5V source
E4 is given terminal voltage that sinks.
Now, from Eq1
Z5/Z4 = (5 - 4.35)/4.35..........Eq2
Next, by paralleling the 5V source Zsourceinternal is
Z5 x Z5 / (Z5 + Z5) = Z5 / 2 ....Eq3
Then, Eq1 becomes
E = 5V x (Z4)/(Z4 + Z5 / 2) =
= (10)/(2 + Z5/Z4)..............Eq4
Substitute Eq3 into Eq4
E4' = 4.65V (93% of 5V) that is higher than E4 = 4.35V (87% of 5V)
2. Now, when it comes to transients over capacitor C and resistor Rc in series, the following holds:
Io = E4/Rc.......................Eq5
I4 = Io x e**(-t/CRc)............Eq6
Ec = E4 [1 - e**(-t/CRc)]........Eq7
Therefore, even if the current I4 is spiking more across the capacitor since
E4' > E4 or 4.65V > 4.35V
The capacitor voltage is greater for E4' than for E4 since
Eq7 holds, and voltage E4 or E4' governs the transient process related to Rc and C in the series connection.

Obviously, if the Zsourceinternal < Z5 /2 then the E4' will be even higher, which I implied in my previous posting.

Question remains, if the recommended inductor addition, as posted in other posting, is feasible.