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I've been asked to build a circuit to switch a constant power (varying current & voltage) supply between two loads, for a research experiment the customer is carrying out. The loads are unknown to me, but apparently they can be approximated by a linear Varistor. As such, the supply varies from 1Vdc @ 10A, to 150Vdc @ 0.06A.

The switching will be initiated by a pulse from a computer, possibly as part of an automated routine. I already have the switching part of the circuit sorted.

The customer wants the switching to take place when the power supply is off, but I'd like to check the supply is definitely off before my circuit switches.

In the past, with CC or CV supplies and a fixed load, I've used a simple high or low-side current/voltage sense circuit. But I'm struggling here, tying myself in knots trying to figure out how to do this without affecting the varying load.

  • I've made the assumption that my circuit needs to measure the supply all the time, so must work for all circuit conditions. In reality, it could be argued that the software requesting the switch means I only need to do a single measurement at the point of request, and that changing the load is not a problem at that point.

I'd appreciate any suggestions, as the problem genuinely interests me. Thanks!

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  • \$\begingroup\$ This is the classic measurement problem in physics. You simply cannot learn the state of a system without influencing it. Any measurement you perform must necessarily inject or extract energy from the system under test. The best you can do is minimise the influence, by keeping that energy to a minimum, which you do with the smallest sense resistors you can get away with, for instance. \$\endgroup\$ Jan 31 at 15:45
  • \$\begingroup\$ @SimonFitch Yes indeed! I should have clarified: minimal effect on the load. So absolutely I could use a small sense resistor, but the problem then becomes how small. At 1Vdc, the load would be 0.1 Ohms, and the sense resistor would need to be unrealistically small. \$\endgroup\$
    – Joe Mills
    Jan 31 at 16:36
  • \$\begingroup\$ Can you clarify if your question is about how to check when the supply is off, or about implementing a constant power source over that large range of conditions? \$\endgroup\$
    – DamienD
    Jan 31 at 20:09

1 Answer 1

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Here are two suggestions, one for current sensing, and one for voltage.

First the current:

schematic

simulate this circuit – Schematic created using CircuitLab

This is a shunt regulator. The closed loop tries to keep voltage across current sense resistor R1 at 20mV. It can lower that voltage, but it can't raise it, because the MOSFETs can only sink current. Therefore the voltage across R1, which is the potential at node A, is capped to a maximum of 20mV, at which point the MOSFETS begin to shunt current around R1. Here's a plot of \$V_A\$ as \$I_1\$ is swept up to 100mA:

enter image description here

R1 is never permitted to develop a voltage comparable to the 1V...150V supply, and yet for currents up to 20mA it will act like any regular current sense resistor.

The op-amp output goes high only when the threshold of 20mV (when \$I_1=20mA\$) is reached, which switches on the MOSFETs. It can therefore be used to signal that 20mA or more is flowing. Since your application only needs to know if current is flowing or not, perhaps \$V_{OUT}\$ is your signal of choice:

enter image description here

Now for the voltage sensing. This is a similar problem, in that we require accurate voltage sensing at low voltages, but we don't want to use a voltage divider which attenuates the signal so much that this becomes impossible.

We can use a similar technique, to cap the voltage across the lower divider resistor to some maximum, and somehow have the upper element take up the rest:

schematic

simulate this circuit

This is a just a voltage regulator. It employs a MOSFET as the "pass element", and a closed loop to make it more resistive as the potential \$V_B\$ at node B exceeds 100mV. This allows \$V_B\$ to follow \$V_{IN}\$ for all inputs below 100mV, but that's where it stops rising. This is \$V_B\$, as \$V_{IN}\$ is swept from 0V to 200mV:

enter image description here

This arrangement will only draw a tiny current from the 1V..150V supply:

$$ I = \frac{100mV}{R_1} = 100\mu A $$

For input voltages between 100mV and 150V, the MOSFET takes up all the remaining potential difference, so it needs to be able to handle that voltage. The maximum power it will dissipate is:

$$ P = IV = 100\mu A \times 150V = 15mW $$

Just like the current sense circuit, the op-amp output is a good indicator of when input voltage has exceeded 100mv. This is \$V_{OUT}\$ for the same sweep of \$V_{IN}\$:

enter image description here

These circuits are just preliminary ideas. They need work to keep them stable, and to protect against problems I have yet to identify. They also need to be made fail-safe.

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  • \$\begingroup\$ I like that current sense circuit, that's cool! \$\endgroup\$
    – Aaron
    Jan 31 at 18:37
  • \$\begingroup\$ Apparently it's not acceptable to thank people here, so I will just say that those are both really interesting and useful circuit ideas. \$\endgroup\$
    – Joe Mills
    Feb 1 at 11:19
  • \$\begingroup\$ @JoeMills It's also not OK to say "you're welcome", but you are, and I consider myself thanked. \$\endgroup\$ Feb 1 at 12:05

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