Power bricks in series?

[budm]

Re: Power bricks in series?

That rectifier output is typical of SMPS output, so hooking two power supply up in series that is what it looks like.
How about your theory about the solenoid and when the switch is open?

[momaka]

Re: Power bricks in series?

@momaka
i really want to know this, looking forward to the results, i really do not want to kill a pair of perfectly good adapters

You won't kill them by hooking them up in series (for a higher voltage output.) Just look at the do's and don'ts in the PDFs that budm attached, and you'll be fine.

However, hooking PSUs in "inverse-series" is (i.e. grounds hooked together and load connected between two positive voltage rails)... not exactly a good idea, as I found out from my experiment. Actually, I kind of knew that ever since I thought of the idea of hooking PSUs in inverse-series a long time ago. I just didn't know how good/bad it was and how far it can be pushed.

In the case of the 5V and 12V power adapters mentioned earlier...
Unloaded, my 5V adapter was outputting 5.14V and the 12V adapter was outputting 11.95V.

First I tried a 12V 70 mm PC fan that draws about 0.2 Amps @ 7V. With that load, the 12V PS dropped slightly to 11.93V, but the voltage on the 5V adapter, unsurprisingly, went up (!) to 5.49V.

Next, I tested a 12V incandescent bulb that draws about 0.4 Amps at 7V. With this one, the 12V power adapter pretty much stayed the same - dropped to around 11.92V. But the 5V adapter voltage raised even more - almost touching 6V!!

This shouldn't be surprising because there is technically no DC current path between the positive 12V output and positive 5V output terminals when the entire circuit of the two adapters is considered. Any current flowing out of the 12V terminal is going -into- the 5V terminal, thus raising the voltage. This is because at DC steady state, you have a charged cap on both outputs connected across a series connection of a rectifying diode and a transformer winding. Even if you treated the transformer windings as a "short-circuit" to ground in both power supplies, the rectifying diode will block any DC current going back into the transformer winding - both for the 12V and 5V power supplies. Since the 12V supply has a higher voltage, there will be current going in the forward directions through the rectifier (or just charge flowing from the output cap, since it is at a higher voltage than the 5V adapter). But for the 5V supply, current can only go into the output capacitor.

I suspect the only reason why my above experiment works is because those adapters probably have a small minimum load resistors inside that loads the outputs down for stability purposes at low power. Also, there may be something to do with the high frequency current of the SMPS charging the output caps that allows this kind of loading (I imagine something similar to DC-blocking caps on the output of a single-rail single-ended [non-BTL] audio amplifiers.) I haven't done an in-depth analysis for that yet, though.

Gosh. All these people being taught EE 101 and KCL/KVL and they don't teach you this stuff - what happens with "real" devices ... heh.

Exactly!
College books really don't tell you this and instead treat power supplies as zero impedance sources, which is actually done to simplify the math calculations and make learning/analyzing a circuit easier. But even when more advanced courses are thought with Thevenin-equivalent power supplies, that's still not an entirely accurate representation... again, because the real-world power supplies we show here have diodes on the output, thus allowing current to flow only one way.

With that being said, I think the verdict on hooking power supplies in inverse-series is:

1) It should generally work, but only for small loads relative to the total rated output. Just BE CAREFUL! Clearly not all PSUs will allow inverse-series loading. On PC ATX PSUs, of course, it generally can be done when the PSU is attached to a motherboard, as the motherboard will provide a big enough load on all rails so that small currents between positive rails (such as fans and lightning) can be connected without issue. Just make sure two positive supplies do not short-out to each other, as that may cause damage to the PSU and/or mobo (use fuses or other means to limit current.)

2) Bigger loads may be used if there is a larger load connected between the positive and negative output of each power supply (in this case, the larger load connected across each PS will provide a low impedance to ground, so the voltage on the lower-rated supply may not rise then.)

[eccerr0r]

Re: Power bricks in series?

Okay...now that the test has been done, thanks momaka:

Yep, as expected. Now the solution that doesn't involve hacking up/redesigning the PSUs:

momaka hit the right nail. The minload resistor is indeed what's causing it to work. On motherboards, the motherboard is also providing the load that we need. Note that the voltage is going up on the lower voltage PSU and exceeding the cap voltage or backfeeding the regulator can cause ICs to blow because you're now using a weak path (like the b-e junction of a bjt), but yes, it depends on how it's designed.

The solution however is not quite right. Yes, the solution is to add a load. The load needs only to be on the lower voltage supply - the 5V one in this case.

How to calculate the load? A resistor will do just fine. The trick is that the current through the resistor on the 5V supply must equal or exceed the current that you want to pass through that 7V load. No load on the 12V is necessary.

Yes, this still wastes energy, but it's about all we can do without redesigning the 5V PSU to be a regenerative PSU (heh, an isolated, two tandem motor "power supply" would actually kind of work in this case!)

[budm]

Re: Power bricks in series?

Single loop circuit and Kirchhoff's law.
https://www.youtube.com/watch?v=GOYLaD_4zRw
http://www.engineeringvideos.org/

[momaka]

Re: Power bricks in series?

The solution however is not quite right. Yes, the solution is to add a load. The load needs only to be on the lower voltage supply - the 5V one in this case.

How to calculate the load? A resistor will do just fine. The trick is that the current through the resistor on the 5V supply must equal or exceed the current that you want to pass through that 7V load. No load on the 12V is necessary.

Right on.

Note that the voltage is going up on the lower voltage PSU and exceeding the cap voltage or backfeeding the regulator can cause ICs to blow because you're now using a weak path (like the b-e junction of a bjt), but yes, it depends on how it's designed.

Exactly.
Design does matter.

Since you mentioned it, it crossed my mind that my 5V power adapter could have a Zener diode on the output to protect from over-voltage. I probably could find out if I opened the adapter, but I didn't feel like it. So instead, I decided to perform another test. This time, I decided to injected a DC voltage on the 5V power adapter while it is disconnected from the wall. This was done to see how much current the 5V adapter will sink in and if there is a Zener or some other active/non-linear path to ground.

So first I tried a 5V regulated output (measuring 5.08V) from the 5VSB of an ATX PSU feeding into the 5V power adapter.
- Result: about 0.007 Amps (7 mA) of power draw and no change in the output voltage from the ATX PSU.

Next, I tried injecting a slightly higher voltage. I cobbled together a "supply" from 4 weak AA batteries in series, measuring about 5.6V unloaded. Once connected to the 5V adapter's output, the voltage dropped to 5.15V (close to the no-load voltage when the adapter is plugged in, which is around 5.13-5.14V). The interesting part is that the current going into the adapter nearly doubled at 0.013 Amps (13 mA).

Of course, I didn't settle for this, so I performed yet another test by switching out 2 out of the 4 AA batteries in my supply bank for stronger (nearly new) ones. This brought the unloaded voltage of the battery supply to ~5.85V. Connecting that across the 5V power adapter revealed even more interesting results: the voltage across the batteries dropped to about 5.5V, but the current going into the 5V adapter increased to around 0.065 Amps (65 mA) - a near 5-fold increase for just a slight voltage increase of ~0.3V.

So what does that mean? Without opening my 5V adapter, I can't say for sure. However, there is indeed a circuit on the output that is providing extra loading when the voltage goes higher. And like you said, that may be through a weak "active" path, possibly from the feedback circuit. Or it could also be from some kind of a self-loading circuit (series Zener-resistor or similar.) In any case, it's probably not a good idea to abuse it and force extra current through it, as it may not have been designed for it.

Thus, on that note, design can indeed matter and without knowing your adapters 100% inside, running them in inverse-series without a large enough load on the lower-voltage PSU may not be good for it.

Also, I gave it a bit more thinking, and there is another "curve-ball" path when connecting power adapters in inverse series. Gonna have to draw a circuit diagram for that to make the explanation easier, so stay tuned...

[eccerr0r]

Re: Power bricks in series?

Oh wow yeah you brought up another good point. if it has a crowbar circuit and you go OVP on that, it shorts out and then you get 12V on your circuit. Granted probably you're using a 12V device you want to use at 7V, this is something else to consider.

If it has just a zener shunt, it actually is the best case scenario, may well just shut off the 5V supply...and hope the zener doesn't fry on you

[evilkitty]

Re: Power bricks in series?

does the stuff about inverse series apply when using a ATX PSU, i know 7v adapters are common for a cheap way to control fan speed
i know voltages lower than 12v are make via dc to dc conversions, but does that difference matter?

I have a 5v and a 12v power supply - https://www.meanwell.com/productPdf.aspx?i=691#1
while i have not had a reason to use 7v from this, i just assumed i could, they have both a common earth ground and a common dc ground

[eccerr0r]

Re: Power bricks in series?

Initially I thought it doesn't apply, but after thinking about it more carefully and considering the schematic of ATX PSUs... again,

YES IT DOES APPLY!

Unfortunately there's no free lunch, though fortunately if you're using the ATX PSU for powering an ATX MB, then you're fine as long as you don't drain too much current... and yes, as long as you're within specs, it is a free lunch...

[evilkitty]

Re: Power bricks in series?

so how do we know/calc how much current is safe for 7v?

[eccerr0r]

Re: Power bricks in series?

The current that the 5V supply is passing to GND is the upper limit. Note that if the 5V supply current changes, then your 7V supply changes, and there's nothing you can do other than increase/waste more current on the 5V supply.

[momaka]

Re: Power bricks in series?

Initially I thought it doesn't apply, but after thinking about it more carefully and considering the schematic of ATX PSUs... again,

YES IT DOES APPLY!

And here is that diagram of two PSUs connected in inverse-series (continuing with the example of the 12V and 5V adapters), just to help with this:


A few important notes about the above diagram:
** It is meant only for steady-state DC analysis.
** C1 and C2 are the filtering capacitors inside the 12V and 5V adapters, respectively. Likewise, D1 and D2 are the internal rectifying diodes.
** V1 and V2 were drawn as ideal voltage sources (which they aren't at all) for model analysis simplicity. These would actually correspond to the transformer output winding(s) of each SMPS. I also set them as 12.7 and 5.7V instead of 12V and 5V, just to (non-ideally) account for the voltage drop from the diodes, D1 and D2.
** The 12V PSU is drawn "taller" on the left side than the 5V PSU on the right side - this was done to help a bit with visual representation of the voltages.
** R_load is the load that can be placed between the two positive outputs to get "7V output".

So before R_load is placed in the circuit, C1 is sitting at 12V and C2 is sitting at 5V (assuming 0.7V drop through each diode and a very tiny and small discharge current provided through each cap.)

Once R_load is placed as shown above, there will be an instantaneous current flowing through it, because there is a difference in potential between capacitors C1 and C2. Eventually, the current should settle down to zero as cap C2 charges to the voltage of cap C1 (i.e. 12V). This is because on the above diagram, there is no DC path to ground. However, we saw that this is not the case when I performed the test with my 12V and 5V power adapters. The 5V adapter had a non-linear DC current path to ground that quickly started to draw more current as the voltage increased above 5.5V.

But that alone couldn't provide the current that I was getting when connecting my fan and light bulbs between the 12V and 5V output, which were much higher than the current I meansured when trying to "force" the 5V adapter output voltage to a higher one.

So what gives?

Well, that can be answered with an AC analysis diagram.

Here is the same circuit above, but drawn for AC analysis:


Note the caps for the 12V and 5V adapters were replaced with impedances, Z1 and Z2. (In AC circuits, the impedance of a component is the analogous to resistance in a DC circuit.) For capacitors, Z = 1 / jwC where w = 2Pi x f, in which f is the frequency in the circuit and C is the capacitance. And for resistors Z = R. The voltage sources V1 and V2 were also replaced, now more properly, with pulse voltage supplies, to reflect the output of an SMPS.

Comparing the above AC analysis circuit to the DC analysis one, we now see that with R_load (or in this case, Z_load, as the load could be non-linear, which both my fan and incandescent bulb are) there is actually a current path between the 12V adapter's output and the 5V adapter's: part of the current from V1 will go through the impedance Z1 of capacitor C1. But, another part of the AC current can go through Z_load and then through the impedance Z2 of capacitor C2. How much, of course, will depend on the capacitances of the output capacitors, along with the frequencies of the switching in the two adapters and also the impedance/resistance of the load.

And that folks, is what I think provided at least some of the "extra" / "mysterious" current that supplied my fan and light bulb without letting the voltage on the 5V adapter raise too much.

Unfortunately, I don't have an oscilloscope to measure the AC current or voltage across R_load / Z_load, but I imagine we would actually see a rather noisy voltage, despite both adapters outputting supposedly smooth DC voltage.

Now, if a large resistive load (larger than the R_load across the 12V and 5V outputs) is placed across the output of the 5V adapter (i.e. across C2), then much or all of the DC current through R_load could go through that and reduce the noise.

So going back to what eccerr0r said: the best way to safely use PSUs in inverse-series is to place a large-enough load across the lower-voltage supply that can "carry" the current from the load placed between the outputs of the two PSUs.

To give some numbers to work with for the above example with the 12V and 5V adapters...
Let's say you want to to draw up to 1 Amp at 7V (maybe connecting a bunch of fans like I did?) To do that, first you need to make sure both your 12V and 5V power supplies are rated for at least 1 Amps of output. Next, place a 5-Ohm load resistor across the 5V adapter. This will draw 1 Amp from the 5V adapter (i.e. 5 Watts, so your resistor should be rated for at least 5W dissipation) when there is nothing connected on the "7V" combined output of the two adapters. Once you connect whatever device you wanted between the 12V and 5V outputs, that device will draw a current from the 12V supply, and return it through the 5-Ohm resistor back to the 12V supply. The "7V" output will regulate properly as long as the current draw of your device is under 1 Amp (so you can have a fan that only needs 0.15 Amps at 7V, or you could have a bunch of them, but their total combined draw should not exceed 1 Amp.)

[eccerr0r]

Re: Power bricks in series?

Actually there's no AC component, the analysis can/must all be done at DC. i(R_load) will be DC, i(C1 or Z1) and i(C2 or Z2) only makes sense from the charging/discharge from the power supply whether it be a SMPS or linear or even battery. In fact the AC component of R_load is meaningless because as shown the steady state will make the lower voltage capacitor charged to the higher one and there will be zero voltage across R_load.

As mentioned, the minload resistors that aren't shown are the main contributor to apparently unloaded PSUs working in this fashion. In fact if you draw these resistors, you can apply KCL/KVL and things just "work" (with diode models). Without the resistors, you can't really do any calculations other than the steady state.