# Wiring a voltage specific device onto a constant current circuit?

I've got a bunch of LEDs wired in series to a constant current driver. The driver runs the circuit at 500mA to 700mA (I can vary it up or down a little) and can supply 50 to 100 volts, maximum 60 watts. I'm using about 63 volts and 53 watts.

I found a temperature probe that calls for 4 to 28VDC. No mention of a current rating. Can I wire the probe into the LED circuit or do I need to find a separate power supply?

Why I'm confused:

From reading other threads, I believe that I'm okay supplying half an amp to the device even if the device would normally only require a fraction of that. However, if the probe consumes anything close to its maximum 28 volts, the circuit's wattage will exceed the driver's rating. How much voltage will the device "ask for?"

• poof <- that's the magic smoke leaking out... May 7, 2018 at 23:12
• electronics.stackexchange.com/questions/34745/… should tell you why. May 7, 2018 at 23:13
• When it's constant current it's actually the other way around from usual: the current most match, the voltage must be equal or greater. I wouldn't recommend connecting a constant voltage load to a constant current supply, it seems like it's asking for trouble. May 7, 2018 at 23:25
• Where is the datasheet? If it is only a few mA , no sweat May 7, 2018 at 23:32
• Please provide a part number and datasheet for the probe, or at least a link to the product sales site if you can't find a datasheet. May 8, 2018 at 2:58

Can I wire the probe into the LED circuit or do I need to find a separate power supply?

Maybe you could, but it is not a good idea.

In all of this, I am assuming the probe requires much less than 53 watts of power. I am also assuming that it contains an internal DC-DC converter (a.k.a. switchmode power supply), given that it is a complex device with such a large specified input voltage range. Such a converter has an important characteristic: it draws as much power from its input as is required by its output — therefore as the input voltage goes up, the current it draws goes down (power = voltage × current). (This is opposite to a resistor or LED.)

Remember, a "constant current driver" cannot control current directly — it can only adjust the voltage across its output terminals to attempt to hit the current target. Let's look at what happens when it tries.

1. If you hooked up the probe in parallel with the LED string, it would see the 63 volts required to drive the LED string, which exceeds its input specification, so no good.

2. If you hooked it up in series, the constant-current driver would turn up the voltage until 500 mA runs through the entire series string, which either again produces over-voltage or just results in the driver's control circuit fighting with the probe's power supply. Again, no go.

3. We can try to reduce the voltage. See below.

The 63 volts across your LED string is "dropped" along the length of the string, so if you were to measure between two points separated by exactly a third of the length of the LED string, you would find a region which has 21 volts across it. So, you could use that as a source of 21 volts, which is in the specified supply voltage for your probe.

However, the probe will take some current. This means that, compared to the probe not being in the circuit, somewhat more current will be drawn at the same voltage — so the constant-current driver will reduce its output voltage to compensate. The probe will react to the lowered voltage by drawing more current — in approximately inverse proportion, to obtain the required power for the actual probe circuit.

This won't likely cause runaway positive feedback driving the supply all the way down to zero volts — the actual outcome will depend on the probe's power requirements and the I-V curve of the LEDs — but it will mean that the LEDs in the segment you've connected the probe across will be dimmer, and might mean that the voltage drops too low to actually supply the probe with its minimum 4 volts (and compensating for this by connecting it across a wider span puts it out of range in the startup condition).

And that's not even considering inrush current or changing power requirements over the probe's operation (e.g. if it has a radio transmitter to report its measurements).

1. We could also consider connecting in parallel and reducing the voltage without involving the LED string. This could be done using another DC-DC converter rated for 100 volt input.

This would work just fine, for powering the probe (given my initial assumptions) — but it would mean the LEDs flicker if the probe's power consumption varies at all (because the constant-current driver is keeping the current the same but it doesn't distinguish between the two loads).

And if you are going to put in another DC-DC converter, you might as well not hook it up to the LED circuit at all and have a completely independent, ordinary, stable power source for the probe.

So, I would not recommend doing this. You might be able to make it work, but it won't be predictable and well-behaved.

• Thanks for the response, Kevin. I'm a little confused by your remark about the driver dropping its voltage to maintain a fixed current. If the driver can supply up to 100 volts at the fixed current, wouldn't adding additional load be okay so long as the maximum voltage and wattage were not exceeded? May 7, 2018 at 23:28
• @rpeinhardt I've edited my answer to put in some more words about what's going on which I hope clarify things, but to directly address your comment: The problem isn't the additional load. The power ratings here are (probably) fine. The problem is that the driver only has two output terminals. They can't supply 63 volts to the LEDs and less-than-28 volts to the LED driver separately; you have to wire them up as some sort of combined circuit, series, parallel, or what-have-you, and the outcome depends on which circuit you choose but none of them are great. May 8, 2018 at 0:49
• I see. The point about constant current drivers manipulating voltage to maintain current is obvious now that you point it out. That's what I'd been overlooking. Thanks for taking the time to educate me! May 8, 2018 at 2:23