I have a question regarding a current consumption of a current transducer (LEM LA 55-P). In the current transducer datasheet it says that the current consumption at \$ \pm 15~\text{V} \$ is \$ I_c = 10 + I_s~\text{mA} \$, where \$ I_s \$ is the secondary current (in my case, up to \$50~\text{mA}\$). This means that the current consumption is \$ I_s = 60~\text{mA} \$.

This current transducer has 2 supply pins:

  • \$ V_{sup} \$ (in my case \$ + 15~\text{V} \$)
  • \$ -V_{sup} \$ (in my case \$ - 15~\text{V} \$)

I was wondering, what is the total power consumption of this transducer:

  • \$ P_c = 15 \cdot 60 + 15 \cdot 60 ~\text{mW} = 1800~\text{mW} \$, or
  • \$ P_c = 15 \cdot 60 ~\text{mW} = 900~\text{mW} \$

In other words, is this transducer going to draw up to \$ 60 ~\text{mA} \$ for every supply pin, or in total. If this is total consumption (for both supply pins), can I except that the transducer is going to draw equally on each supply pin, i.e., \$ 30 ~\text{mA} \$ per supply pin?

Thanks, M.


enter image description here enter image description here

Figure 1. The LEM LA 55-P internals.

All of these devices work using Hall-effect sensors. This is represented in the schematic by the X-box.

enter image description here

Figure 2. Source: Power Guru.

The amplifier runs a current through the coil winding so that the flux in the core sums to zero. This is rather clever as it removes all non-linearities of the core from the performance of the device. It is never driven into saturation (if operated within spec) and never leaves zero so there are no hysteresis effects either. The flux cancelling current is fed to the output and will be directly proportional to the primary current.

It should be clear then that for the device to source current the top transistor of the push-pull output stage will turn on and you could have quiescent + 50 mA at full scale drawn from + supply. Similarly, when current is reversed you could have quiscent + 50 mA returned to the negative supply.

At zero primary current there will be zero output current so the quiescent current must flow from V+ to V-.

What is the total power consumption of the transducer?

You might want to re-phrase the question. The power will be consumed by the transducer and the load RM. Worst case for the transducer will be with maximum current and voltage drop into minimum RM. RM minimum is listed as 10 \$ \Omega \$. At 50 mA this will dissipate \$ I^2R = (50m)^2 \cdot 10 = 25~\mathrm mW \$ and drop \$ IR = 50m \cdot 10 = 500~ \mathrm mV \$. The power dissipated in the output transistor will be \$ (V_+ - V_{RM})I = 14.5 \cdot 50m = 725 ~\mathrm mW \$.

Add to this the quiescent power = \$ 30 \times 10m = 300 ~ \mathrm mW \$.

  • \$\begingroup\$ No matter what \$ R_M \$ do I use, the power dissipated can be calculated as \$ I_S \cdot V_+ \$, with an addition to the quiescent power. The resistor \$ R_M \$ will only determine how much power will be dissipated in the output transistor: the smaller \$ R_M \$ is, the more dissipation occurs at the output transistor? If I got this right, the positive supply will supply current to \$ 0~\text{V} \$ for one direction of \$ I_P \$, and the negative supply will draw current from \$ 0~\text{V} \$ for other direction of \$ I_P \$? \$\endgroup\$ – Marko Gulin Oct 7 '16 at 8:32
  • 1
    \$\begingroup\$ \$ R_M \$ will determine the proportion dissipated in the device and in the resistor. I suspect you only care about your power budget in which case your analysis is correct. \$\endgroup\$ – Transistor Oct 7 '16 at 9:08
  • \$\begingroup\$ Yes, I only care about the power budget since I need to size a power supply. Thank you for clarifications, this is really helpful! \$\endgroup\$ – Marko Gulin Oct 7 '16 at 9:22

Probably not. The datasheet is not very comprehensive so I am guessing a bit. If it's important: measure!

The device itself needs 10mA (continuously) from both supply rails to work.

And next to that it will draw Is from the supply "in general". If Is flows out of the device it will be drawn from the positive supply (and vv).

So the maximum power drawn will consist of two components: 10mA @ 30V (300mW) and Ismax @ 15V (750mW). Together 1050mW.

Note that this is the worst case possible, where you continuously measure 50A DC. If you measure sine-shaped (or small) currents, Is will on average be smaller, and so the average power will drop. In that case you will need to find the rms value of the current you measure to determine the average power you will need for driving Is.


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