A bridge design is odd for this use, however it'll work if you use a single resistor as the load of the bridge. That resistor will determine the current through the LEDs, and thus the brightness.
Flogging the FREDs
Voltage fed converters with transformer isolation will exhibit ringing in the secondary. Ringing is caused by parasitic inductances and capacitances in the circuit, with the dominant elements will being the transformer leakage inductance (\$ L_ {\text {Lk}}\$) and junction capacitance ( \$ C_j\$)of the bridge diodes. The diode data sheet shows \$ C_j\$ of 32pF. I'm going to make a naive guess at \$ L_ {\text {Lk}}\$ of 500nH, but it will have to be measured to really know. So, an LC of 500nH and 32pF is what must be snubbed.
Spike amplitude without snubbing will be \$ 2 n V_ {\text {in}}\$, where \$ n \$ is transformer turns ratio and the factor of 2 is what you get for a high Q resonance.
There are different types of voltage snubbers; Clamping, Energy transfer resonant, and Dissipative. The clamping and resonant types require more parts and some involvement of active switches which I think make them impractical for this case. So, I am only going to cover dissipative snubbers because they are the most simple and work well with passive switches (like diodes or synchronous rectifiers).
The form of dissipative snubber that I will cover is a series RC placed in parallel with each bridge diode.
Some facts about RC dampening snubbers:
- They are all about impedance matching. You don't get to choose the snubber resistor value \$ R_d\$. The parasitic LC determines that for you by characteristic impedance Zo.
- You do get to choose the value of the snubber cap \$ C_d\$. That's important since the cap value sets the snubber loss (\$ P_ {\text {Rd}}\$)as \$ C_d F V^2\$ . Where V is the pedestal voltage and F is switching frequency. The snubber cap must provide a low impedance at the LC resonance of the parasitics, so it needs to be several times \$ C_j\$.
Some guidelines, and what to expect with RC dampening snubbers:
For \$ L_ {\text {Lk}}\$ of 500nH and \$ C_j\$ of 32pF, Zo will be 125Ohms. So, \$ R_d\$ would be 125 to match Zo. You may have to fine tune this a little since \$ C_j\$ is non-linear and falls off with reverse voltage.
Choosing the snubber cap \$ C_d\$ : Choose \$ 3 C_j\leq C_d\leq 10 C_j \$ . Higher values in the range do provide better dampening. For example, \$
C_d\$ of \$ 3 C_j\$ will result in a peak diode voltage of \$ 1.5 n V_ {\text
{in}}\$, while \$ C_d\$ of \$ 10 C_j\$ will result in a peak diode voltage of
\$ 1.2 n V_ {\text {in}}\$.
Dissipative snubber performance will not improve for \$ C_d\$ values
greater than \$ 10 C_j\$.
Power loss \$ P_ {\text {Rd}}\$, with a pedestal voltage of 1250V and F of 50KHz.
- If \$ C_d\$ is \$ 3 C_j\$ or 100pF, \$ P_ {\text {Rd}}\$ = \$ C_d F V^2\$ or 7.8W.
- If \$ C_d\$ is \$ 10 C_j\$ or 330pF, \$ P_ {\text {Rd}}\$ = \$ C_d F V^2\$ or 25.8W.
\$ C_d\$ of \$ 10 C_j\$ gives the best dampening with peak voltage of 1.2 time the pedestal voltage, but you can save some power with smaller snubbing caps if you can stand the higher peak voltage.
Best Answer
Yes, you can connect the output of the regulator to as many USB ports as you want, but the total current drawn by all ports can't exceed what the regulator can supply. 5 ports is really pushing it unless you know that the devices that you will be charging don't take much current. Normal USB ports only guarantee 100 mA, and then up to 500 mA after negotiation. However, the pure charging interface doesn't require any negotiation and can supply significantly more. Note that the LM2596 only guarantees 3.6 A.
Yes, you need a capacitor on the output of a full wave bridge. It sounds like you plan to feed the LM2596 from the full wave rectified output of a transformer secondary. That's fine, but the reservoir cap is still important. Without the cap, the voltage will drop to 0 twice per line cycle. You need a big enough cap so that the minimum input voltage of the LM2596 is maintained between power line peaks when the cap will be recharged.
For example, assuming 60 Hz power, there is 8.3 ms between line cycle peaks where the cap gets charged. With a 3.6 A load, a 10 mF cap would drop 3.0 V. If you have at least that much headroom above the regulators minimum input voltage, then that's fine. One nice characteristic of such a switching regulator is that it can handle a wide input voltage range. This one can handle up to 40 V. If you can arrange the transformer output after the rectifiers to provide nearly that much under no load, you have a lot of room to allow the voltage to sag between peaks and can use a smaller cap than 10 mF.
If you really want to drive a large number of ports, you could use multiple regulators, in which case the total current required out of the transformer will be higher and the cap will need to be larger to achieve the same voltage drop.