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I see that some power supplies provide a wide range of output voltages meanwhile maintaining the current.

For example: 2.7 A, 20-36 V output.

I understand that the voltage rate will depend on the end-unit consuming the power (eg., an LED chip).

What happens if the output from the PSU is divided into two parallel connections, one lead connected to an end-unit needing to use 27v, and another using 30v?

What happens internally? What's the reason why for the end result?

Example:

[[80W 2.7 A PSU with 18-30 V]] |--> (21-24) V, 1 A Diode 
                               |--> (27-31) V, 2 A Diode

How would it regulate the voltage in each case?

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  • \$\begingroup\$ Phil, if the supply is in fact a constant current supply, it is the current that is regulated (made regular), not the voltage. Also, by definition, two parallel connected circuit elements (diodes in this case) have identical voltage across. As written, your question is ambiguous. I recommend using the schematic editor to produce a circuit diagram of what you're proposing. \$\endgroup\$ – Alfred Centauri Dec 13 '14 at 14:44
  • \$\begingroup\$ Yes. That's the whole question. Current is regulated, voltage is varying. So how would it deliver the voltage to each node? I will take your advice onboard and try better next time, thanks. It is a simple parallel circuit though as you can see and read I hope. \$\endgroup\$ – Phil Dec 13 '14 at 14:46
  • \$\begingroup\$ Since, I assume, we're talking about a single power supply, it's not possible for there to be two different voltages; there is just one voltage across the two terminals of the power supply. Also, if the two diodes form a "simple parallel circuit", surely you know that there cannot be a different voltage across each? \$\endgroup\$ – Alfred Centauri Dec 13 '14 at 14:56
  • \$\begingroup\$ That's exactly the whole point dear @AlfredCentauri. I'm wondering what would be the voltage at each end? If there is single LED attach, the V gets to somewhere between MIN and MAX of the diode. What if on the parallel there is another LED with different MIN, MAX and TYP. values? \$\endgroup\$ – Phil Dec 13 '14 at 15:27
  • \$\begingroup\$ Phil, what does "voltage at each end" mean? There is just one voltage, not two. Regardless, if the two diodes are indeed connected in parallel across the power supply terminals, and if one diode has a significantly lower operating voltage rating, almost all of the power supply current (2.7A) will be through that diode. Thus, the voltage across will essentially be entirely determined by that diode. To find that voltage, consult the IV curve for that diode (assuming that much current has not destroyed the diode). \$\endgroup\$ – Alfred Centauri Dec 13 '14 at 15:55
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I understand that the voltage rate will depend on the end-unit consuming the power (eg., an LED chip).

I don't believe there's such thing as a "voltage rate" in the context you're using and, in the same context, the power supply will generally be indifferent as to what the load looks like.

What happens if the output from the PSU is divided into two parallel connections, one lead connected to an end-unit needing to use 27v, and another using 30v?

The voltage across the loads will be the single voltage the supply generates in order to force a total of 2.7 amperes through the parallel loads.

What happens internally? What's the reason why for the end result?

The supply senses the current in its output lead and adjusts its output voltage so that its output current will remain constant at 2.7 amperes, regardless of the load resistance.

As an aside, your:

I see that some power supplies provide a wide range of output voltages meanwhile maintaining the current.

For example: 2.7 A, 20-36 V output.

Indicates that since - from Ohm's law - R = E/I, with 20 volts across a load and 2.7 amperes through it, the load's resistance must be 7.41 ohms.

Similarly, at the high end of the supply's compliance, with 36 volts across a load and 2.7 amperes through it, the load's resistance must be 13.33 ohms.

Note that the range of load resistances the supply can accommodate is from 7.41 ohms through 13.33 ohms, regardless of whether the load is a single resistor, a couple of parallel diodes, or whatever, as long as the compliance of the supply is maintained within its limits.

Next, re. your diode example:

Example:

[[80W 2.7 A PSU with 18-30 V]] |--> (21-24) V, 1 A Diode |--> (27-31) V, 2 A Diode How would it regulate the voltage in each case?

It CANNOT regulate the voltage, it can only increase it until it either pushes 2.7 amperes through the load or until it runs out of headroom.

In the first instance, assume a diode with a 4 volt drop when there's 1 ampere through it.

If it's running from a 23 volt constant-voltage supply it'll need a ballast resistor to drop the remaining 19 volts, that resistor being: Rs = (Vsupply - Vled)/Iled = (23V - 4V)/1A = 19 ohms

Notice though, that since it's not a constant voltage supply, but a constant current supply it'll raise its output voltage in an effort to get 2.7 amperes through the load. In this case its output voltage can only rise to 30 volts, so when it gets there the current into the load will be: Iled = (Vsupply - Vled)/Rs = (30V - 4V)/19R = 1.53 amperes.

In the instance where you have two LEDs in parallel, the output voltage of the supply will rise until the total current out of the supply is 2.7 amperes, or until its output voltage rises as high as it can if it can't output 2.7 amperes.

With a 1 ampere LED in parallel with a 2 ampere LED and each LED's current limited by a ballast resistor, the voltage of the supply will rise until the current into the parallel combination equals 2.7 amperes.

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  • \$\begingroup\$ Cheers @EMFields, this really made sense! If I may, I'd like to understand one more thing: In my example, there are two diodes connected in parallel to the PSU. Normally, when I use a 30-36v rated LED, I read the voltage around 32-33v range. I can't test this because I don't have a second LED but, what if there was another, with a typical voltage level of ~27v. What would be the voltage levels I would read using a multimeter then? Is there a way to calculate this using the ohms law and power supply and diode specs? \$\endgroup\$ – Phil Dec 13 '14 at 18:16
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    \$\begingroup\$ Let's assume, just for now, that your diodes have a 1 volt drop at 2.7 amps, and that both the 27-volt LED and the 32-volt LED both draw 2.7 amps at full load. Now let's think about what would happen if you gradually increase the voltage. At 28 volts, the 27-volt LED will draw 2.7 amps while the 32-volt LED will draw (approximately) nothing. At this point the power supply is in full CC mode, and the output voltage cannot go higher, or the 27-volt LED would draw more current, and that cannot happen. \$\endgroup\$ – WhatRoughBeast Dec 13 '14 at 18:54
  • \$\begingroup\$ @Phil: I'll edit my answer to include your comment. \$\endgroup\$ – EM Fields Dec 13 '14 at 18:54
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    \$\begingroup\$ @EMFields thank you for your thorough, complete (and most importantly) kind and patient answer :) Very much appreciated. \$\endgroup\$ – Phil Dec 13 '14 at 23:17
  • \$\begingroup\$ Thank you @WhatRoughBeast that makes sense! Great example. \$\endgroup\$ – Phil Dec 13 '14 at 23:17
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With a purely diode load like that, the voltage will be clamped at the forward voltage drop of the lowest voltage diode. The majority of the current will flow through that diode, and the other diode will get little or nothing (depending on it's I/V curve).

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  • \$\begingroup\$ Ok. So if I get you straight, the output V is the lowest of the lowest end-component (in our case 21v)? How does it decide on that though, could you elaborate your answer? This is very curious :) \$\endgroup\$ – Phil Dec 13 '14 at 14:39
  • \$\begingroup\$ It doesn't "decide" on it, it's just a property of the diodes. There is a fixed voltage dropped across a diode (or chain of diodes), and that is all there is to it. \$\endgroup\$ – Majenko Dec 13 '14 at 14:45
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    \$\begingroup\$ If you had a constant voltage supply, yes. But not with a constant current supply. The current is constant, the voltage is whatever is needed, and with diodes the lowest voltage one wins. \$\endgroup\$ – Majenko Dec 13 '14 at 14:49
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    \$\begingroup\$ But none of this has any bearing on the original question, so has no place in the answer. To answer the "Why" of the lower limit - show me a schematic for the power supply and I'll explain it to you. The PSU is designed for a specific voltage range. If it's switching, maybe it can't generate a duty cycle suitable for a lower voltage. If linear, it would definitely dissipate massive amounts of extra heat at lower voltages. \$\endgroup\$ – Majenko Dec 13 '14 at 15:34
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    \$\begingroup\$ I don't think you edited the question in a way that even remotely encompasses what you're asking here. The simple answer to what you are asking about the power supply is: Because they designed it that way. Pure and simple. Yes, you can make a CC PSU that would cover 0-30V, but why would you go to that expense when it is meant to be used with LED lamps that operate within a specific voltage range? It'd be pointless and the end product would be overpriced and unsellable. \$\endgroup\$ – Majenko Dec 13 '14 at 15:42
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It cannot. A single output power supply will be adjustable to a single voltage level. Parallel loads can split the the current capacity of the supply as required. For example a 5A load and a 3A load could be sourced by a single supply that had a capability to supply 8A or more. But the voltage delivered to the two parallel loads would be the same.

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  • \$\begingroup\$ So how is it going to regulate and decide on how much voltage to send given that the power supply in question is a constant-current one? \$\endgroup\$ – Phil Dec 13 '14 at 14:38
  • \$\begingroup\$ A constant-current power supply will adjust its output voltage such that it delivers (or the load demands) the supply's rated current. If you connect two or more loads in parallel, the power supply's current will split between the loads depending on the characteristics of the loads. \$\endgroup\$ – Peter Bennett Dec 13 '14 at 17:13

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