You are attempting to draw more current from your supply than it can provide and still remain in regulation.
Fixing your supply might fix the problem. The circuit you show is a mess:
- The two bottom diodes in each of the (intended, apparently) full wave bridges do nothing. Stop and actually think about it. They are two back to back diodes in series. One of them will always be reverse biased, and no current will flow thru them. You would get the same effect by simply removing them.
- The third diode in each bridge that is connected between the DC output and ground also does nothing. This is the top right one in both bridges. Again, think about it. It is directly connected between the DC output and ground in reverse. It will always be reverse biased, so it won't conduct. You would get the same effect by simply removing it.
- The remaining diode, the top left in both cases, makes a half-wave rectifier.
- Two separate unregulated DC outputs from two separate rectifiers makes no sense in this case. You could use the two diodes that are actually doing something, together with the center tapped secondary to make a single full-wave rectified unregulated supply. That can feed as many 7805 regulators as its current can support.
You don't get more current by using multiple rectifiers. In this case, you get less because each of your unregulated supplies is only half-wave, and therefore has lower current capability.
- You don't show the values of the caps around the 7805s, but a single cap on the input can't be right. You need a decent amount to store significant energy to ride out the half power line cycle between peaks (in the fixed design). That would take a electrolytic with current technology.
However, the regulators need their inputs to be low impedance at high frequencies for stability. That would take a not-electrolytic with current technology. Something like a 10 µF 20 V ceramic would do.
In other words, what you show obviously isn't right because there would need to be two caps on the input of the 7805.
- If you want to make 3.3 V, it's simpler to do it directly from the unregulated supply. Using a 7805 intermediary does spread out the power dissipation, but there are other simpler ways to do that.
Even better would be to replace the bottom 5 V and 3.3 V regulators with a single buck switcher. That gets around the problem of getting rid of the heat.
Do some math. You say each end of the secondary puts out 6 V. That is presumably RMS, so 8.5 V peaks. Using just the two diodes to make a full wave bridge looses one diode drop in the rectifier. Figure 700 mV for that, so now you're left with 7.8 V peak out of the diodes.
Read the 7805 datasheet. See the minimum dropout figure, or maybe its shows as the minimum required input voltage. Some variants need up to 2.5 V headroom, so require a minimum of 7.5 V in to be able to keep the output regulated at 5 V.
There is very little room between the 7.8 V coming out of the rectifiers and the 7.5 V the 7805 might require. Just for laughs, let's see how large a capacitor it takes to provide 1 A while not dipping below 7.5 V. (1 A)(8.3 ms)/(300 mV) = 28 mF. That's a lot, but it could be done.
However, that is assuming the transformer still puts out 6 V under 1 A load, and that the diodes still only drop 700 mV.
A much better answer is to replace the 7805 regulator with a buck switcher. Those can work with less headroom, and can also work with higher input voltage without extra power dissipation. Now you could use the 12 V out of the ends of the secondary with a real 4-diode full wave bridge. That would cause too high of a voltage drop in the 7805s for them to dissipate as heat. The buck switcher, however, will be much more efficient, so it can use the extra voltage to give you a lot of headroom.