Is power rating of transformer determined by input or load?

Question

Scenario

I have current carrying conductor with a current transformer around it. If the load is rated at 5V, 2A and only requires 2W of power, but the primary side conductor is carrying 100V 12A.

Question

In this scenario would the power rating of the transformer be determined by the load or the primary side rating. AKA would I need a transformer core rated at 1200VA or 10VA.

Background

I am designing an energy harvesting unit to power my Raspberry Pi

I am aware that regulatory circuitry will be needed, but let's omit that for now. I plan to use a switch mode power converter for regulation, shown in diagram above.

Some proof would be great as I have heard conflicting answers to this question.

Answer 1

I have current carrying conductor with a current transformer around it. If the load is rated at 5V, 2A and only requires 2W of power, but the primary side conductor is carrying 100V 12A.

The first problem is that the CT primary current will vary from 0 at no load to a maximum depending on the load on the line. This makes it very difficult to extract even power from such a scheme and it is the reason that constant voltage is the preferred method of power distribution.

I understand that constant current power is used for runway lighting with CTs driving lights along the sides of the runway. The advantage is that there is no drop in lamp power at the far end of the runway as the current is constant throughout. Note that the current is constant and that changes everything!

Back to your question:

You are proposing to draw \$ 5 V \times 2 A = 10 W \$. The nearest standard type might be a 50:5 A unit such as an INSTRUMENT TRANSFORMER 15RL-500 Current Transformer, 50:5, 50 A available from Farnell / Newark. A quick look through the datasheet shows:

Figure 1. Note the maximum burden is 2.5 VA on this CT.

That's not enough power so we need to go bigger.

Figure 2. This one can handle a 12.5 VA burden.

Let's use this one. Note that you've now spent $50 to $100. (I can't find a price.)

To get 5 A out of the secondary we'll need 200 A in the primary but we only have 12 A in our circuit. Let's wind the primary through the core to increase the ampere-turns closer to 200. \$ \frac {200 Aturns}{12 A} = 16.66~turns \$. Since you can only have integer turns let's go for 16. Now 12 A on the primary will give us \$ \frac {5}{200} 12 \cdot 16 = 4.8~A \$ on the secondary.

The maximum secondary voltage we can get is given by \$ \frac {VA~rating}{max~current} = \frac {12.5}{4.8} = 2.6~V \$. Now we can see that this is getting awkward. 2.6 V RMS (it's AC, remember) isn't a great starting point for an efficient low-voltage PSU.

Add to this the problems of variable current and ensuring that, should the current increase for some reason, that you don't exceed the VA rating of the transformer and you can see that the exercise becomes far from trivial and far from cheap.

By the way, your secondary impedance, \$ \frac {2.6V}{4.8A} = 0.54~\Omega \$ will be reflected back to the primary as \$ R_P = R_S (\frac {N_P}{N_S})^2 = 0.54 (\frac {16}{40})^2 = 0.086~\Omega \$ and this will cause a \$ V = IR = 12 \cdot 0.086 = 1.04~V \$ drop in voltage to your mains load.

^{simulate this circuit – Schematic created using CircuitLab}

Background: I am designing an energy harvesting unit to power my Raspberry Pi.

No you're not. Harvesting suggests that you're getting energy that would otherwise be wasted. This is a complex way of connecting a load to a mains supply. As you can see, power is drawn from the mains and you (or someone else, if you're stealing it) will have to pay for it.

Answer 2

When you clamp a current transformer around a current carrying wire it would be totally expected that the current through the wire to the load wouldn't drop hardly at all i.e. the current in the wire would remain at least 99.9% what is was before the CT was clamped around it.

This means that the extra volt drop in the wire is going to be micro volts or a milli volt at worst case. So you have a wire carrying 12A passing through a CT and the CT primary (yes it's a still a primary winding) sees maybe 1 mV.

That's a power input to the CT of about 12 mW. That would normally be dissipated in the CTs burden but now you have a circuit instead of a burden BUT it will still only receive 12 mW of power.

Is that enough for a raspberry Pi?

No.

Answer 3

Your challenges are going to be having a regular supply of power to your 'regulator' circuit or whatever follows the current transformer.

You cannot use a regular transformer for this application (it would turn out massive and inefficient).

If your main current was always between 10 and 100A you could then be sure you will always have enough to power your device through a suitable current transformer.

You need to size your current transformer based on the maximum current it will pass in practice, including inrush and such but not more than your circuit fuses.

You need to make sure you have the correct turns ratio so that the output current is close to your device current when the main load is consuming the minimum expected current.

You would them locate or develop a shunt voltage regulator (with credible safety features) that will prevent excessive voltage or current excursions in your regulator circuit or in the current transformer. A pair of anti-parallel high wattage Zener diodes greater than your device requirements in series with a resistor that will limit the current transformer secondary voltage if your shunt regulator fails or the protection fuse opens. Your shunt regulator should be able to handle the maximum expected current transformer secondary current.

This is not a very common application but perhaps there are circuits out there that you could copy or work from.

Here are a couple of pages that may give ideas. I found them with a google image search just now but have not tried to establish the direct relevance.

https://sound-au.com/project79.htm

http://www.electronicproducts.com/Power_Products/Batteries_and_Fuel_Cells/Energy_harvesting_from_an_ambient_electric_field.aspx

http://www.discovercircuits.com/DJ-Circuits/energy-harvesting.htm