Computing stuff tied to the physical world

Collecting energy into a reservoir

A simple way to collect energy from the Current Transformer and turn it into a low-power voltage source, is to use a half-wave rectifier plus reservoir capacitor (from Wikipedia):


But this throws away half the cycles, and hence half the energy we could benefit from.

Meet the Delon bridge voltage doubler, which is essentially twice the above circuit:

499px Bridge voltage doubler svg

To understand what is going on, we must look at each phase of the incoming AC signal:

  • assume the secondary coil produces 5 Vpp, i.e. swings between +5V and -5V
  • assume also that the diodes have the usual 0.6 V forward “drop”, i.e. voltage loss
  • when the voltage is zero, nothing happens, the caps start out empty
  • when the upper side of the transformer rises to +5V, the top diode will conduct
  • the top capacitor will start to get charged up via the top “loop” in the circuit
  • the bottom side will do nothing, since the bottom diode blocks, it’s reverse biased
  • when the voltage drops below the top capacitor’s current level, both diodes block
  • the voltage drops back to zero, and no further current is flowing anywhere
  • for the negative-going cycle, the same things happens with the bottom loop
  • at the end, the top and bottom capacitors will each be charged up to 4.4V
  • so the total output voltage over the caps is 8.8V, minus some resistive losses
  • as long as the caps are kept fully charged, no further current will flow
  • when a cap drops below 4.4V, it gets charged up again in one of the two cycles

The effect is that a voltage transformer rated at 3.5 VAC will produce up to 8.8V on the caps – hence the name “voltage doubler” (by approximation). That’s because a 3.5 VAC transformer rating is usually specified as RMS (Root Mean Square), which is less than the peak voltage by a factor of ≈ 0.7 (for sine waves). This is equivalent to 5V peak, and since it’s alternating current, these swings are in both directions, for a total of 10V peak-to-peak.

The Delon bridge is just a clever trick to take advantage of both sides of the voltage swing. Unlike a four-diode “full bridge” rectifier, its diode loss is only 0.6V instead of 1.2V.

In the case of our Current Transformer, there is no sine wave, and the CT’s output voltage is in fact related to the current in the wire passing through it. We don’t really care much about the signal’s shape, we just need to extract as much energy as we can from it, at a voltage level which is suitable for the LPC810, i.e. anything between 1.8 and 3.6 V.

Here is an example of the CT input- and output- voltages (no load on the output side):


The input (yellow) is from a 6 VAC transformer, with a 10 Ω resistor as load. Ideally, it should be a pure sine wave, but the tops are flat, as you can see (AC mains in is probably similar). The output from the CT is a very complex waveform, but that’s because there are magnetic flux reversals involved, and ringing from the unloaded output.

On the output side of the Delon bridge, there is about 5.8V without load.

Loading the bridge with two red LEDs in series makes them light up, which shows that there is definitely some energy coming out at about 3.4..3.6V. With LEDs connected, the bridge output voltage fluctuates about 40 mV up and down. Note that this is now a continuous DC current, even though power comes from that 50 Hz signal shown above.

The current through the 6V test circuit is roughly 0.6A – this corresponds to a 140 Watt load on 230 VAC mains if the CT were directly clipped onto (one side of) a mains power line. This setup might provide just enough energy to power up a µC!

Note: several of these findings and observations are the result of numerous fruitful discussions with Martyn Judd, who helped explain the quirks of CT’s. Many thanks!

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