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Full Wave Rectifier

In my last post, I showed the most basic use case of a diode. The problem was though, it was pretty inefficient - only giving under half the wave. If the input voltage was less than the diode turn on voltage, we just sat there. This loses half the wave, and has pretty terrible efficiency. We need to design some kind of circuit that can handle both positive and negative voltages, and give us the correct orientation no matter which case we’re in.

Let’s think about the design of this circuit. Let’s imagine we had two normal half-wave rectifiers in parallel. One could see how that would with the proper orientation, we should always have the correct orientation to our load. The problem would be making sure we had no accidental shorts, where we would just skip the load entirely. As we’ve already learned, we can use diodes to “steer” currents, as no current goes in a reverse biased diode. The final design we end up at is the full wave rectifier shown below.

* Simple Full Wave Rectifier

Vin top 0 SIN(0 3 5 0 0) 
d1 top right DIODE1
d2 left top DIODE1
d3 left 0 DIODE1
d4 0 right DIODE1
r1 left right 5

.tran 2m 1
.print tran v(top) v(left) v(right) i(Vin)

full-wave-rect-sch full-wave-rect-out

So let’s work out what’s going on here. We really have two cases: a positive input voltage, and a negative input voltage. When we’re at positive voltage, D1 will conduct current. D2 will not, due to reverse biasing, so we go RIGHT to LEFT across the load. Following that we could go to D3 or D4, but D4 won’t make a current loop, so that option is removed. We go through D3 and finish our loop.

In the opposite case, we apply a negative voltage, so we start at the bottom. We conduct through D2 (since voltage is all about magnitudes!), and then again, RIGHT to LEFT across the load. We then have a similar D4/D3 choice, but here the loop is up to D4. In both cases, we get the same voltage orientation across our load. The only loss here is that now for each path, we have to turn on two diodes, so the minimum voltage is much higher.

In my test circuit, the input is the blue wave, and the output is the green wave. It takes about 1.6V to turn the circuit on, which is just double the turn on for one diode. Whether or not that is more effective than only getting the half the wave obviously depends on your input function!