Multiple USB Vcc

Two different computers both connect to one board via USB Vcc, and ground.

Because both are ~5V and your sharing a common ground there, any slight differences in voltage will cause a slightly greater strain on one of the computers regulator when it gets pulled up to the other.

Combined with the fact that this happens all the time when you connect any cable from one computer to another that share a Vcc/ground.

Any reason why this wouldn't be true?

Two different computers both connect to one board via USB Vcc, and ground.

How is that going to happen?

I would warn against connecting two 5V from two computers together, it will lead to trouble.

harddrive123:
Because both are ~5V and your sharing a common ground there, any slight differences in voltage will cause a slightly greater strain on one of the computers regulator when it gets pulled up to the other.

Correct.

Although the strain might not be "slight", it could destroy a PSU.

harddrive123:
Combined with the fact that this happens all the time when you connect any cable from one computer to another that share a Vcc/ground.

Ground is necessary but I doubt they connect their Vcc to the cable.

harddrive123:
Any reason why this wouldn't be true?

Yes, there many reasons why it's a BAD idea.

So basically I need to put some isolation resistors in to burn off any difference.

Its difficult to determine exactly how much the total board circuitry equivalent is but 22ohms should be small enough that I don’t need more then 1/4 watt.

Or perhaps only have one USB side with the isolation, its primarily for data, but its very realistic that its the sole power supply for the board.

Untitled.png

Typically if one can't prevent a system from connecting two voltage sources of the same nominal voltage value together, one effective protection method is to have each voltage source wire through a series diode (a low Vf diode is best) and then the module well draw current from the common cathode wired together (assuming a positive volt is being used) such that neither one can force current through the other and the device will just draw current from whichever voltage source is slightly higher then the other. There are more complex methods of isolating (like the Uno uses) but simple diode ORing is effective.

harddrive123:
So basically I need to put some isolation resistors in to burn off any difference.

No not at all, it will not burn off the difference, you should get more familiar with ohms law.

If I use a diode I will lose line voltage ~0.6-0.7V, thats cutting it really close for operational voltage.

I realize there is some current feedback when using a resistor but 10mA or less is very small, and the USB ports are almost certainly already feedback protected.

No not at all, it will not burn off the difference, you should get more familiar with ohms law.

:roll_eyes:
ex. one is 5.1V and the other is 4.9V, we have one resistor value of lets say 20ohms isolating them. That's a difference of 0.2V across the resistor, with 0.2V/20ohm = 10mA current flow, which puts us at 2mW power dissipation on the resistor.

But please tell me, what is wrong with this ohms law math? Do you know how voltage and resistance works in a circuit?

Do you know how voltage and resistance works in a circuit?

My god I don't know, after 40 years in electronics I finally see I have been doing it wrong all my life.

Ya… but tell me, whats wrong with that math or concept?

What I’m suggesting is this attached circuit as a likely worse case scenario, your not going to get more then 0.2V difference on sources that are suppose to be 5V on USB ports made in the last decade.

Untitled.png

harddrive123:
Two different computers both connect to one board via USB Vcc, and ground.

Because both are ~5V and your sharing a common ground there, any slight differences in voltage will cause a slightly greater strain on one of the computers regulator when it gets pulled up to the other.

Combined with the fact that this happens all the time when you connect any cable from one computer to another that share a Vcc/ground.

Any reason why this wouldn't be true?

Never connect two voltage regulators in parallel, this can lead to instability and
turn the combination into a high power oscillator (worst case). Each regulator
has gain and uses feedback, the combination may have too much gain at some
frequencies - the resultant oscillation can put the full input voltage on the output
rail.

USB hosts do have quite a lot of extra protection facilities so this particular
failure mode may not happen, but I don't expect it to work.

I would put a P-channel MOSFET on the Vcc of one side. It should switch off if the other side has power.

Is there a reason why a isolating resistor won't work?

After thinking about it I would probably want closer to 10ohms to prevent current limiting from the port to the board (~500mA).

If there is a difference between the 2 sources of voltage the resistor will drop the difference, the load will only see the lower of the 2, and the drop on the resistor through the load/board is negligible, far better then any diode would give you. The feedback current on the resistor is also going to be very small % in relation to the regulator capabilities.

I was never that good/interested in thevenin theorem, but this seems like a basic application of it, so if I'm wrong, how so?

P.S 40 years doing what with electronic's? If its 40 years of hobby I could see you doing it wrong. At this point 40 years of actual design in a job seems very unlikely.

harddrive123:
Is there a reason why a isolating resistor won't work?

After thinking about it I would probably want closer to 10ohms to prevent current limiting from the port to the board (~500mA).

At 500mA, how much voltage drop will a 10 Ohm resistor give you? (Hint: Use Ohm's law)

Even at 100mA, how much voltage drop will a 10 Ohm resistor give you? (Hint: Use Ohm's law)

the load will only see the lower of the 2,

No that is not how it will work. Everything that MarkT said, plus what happens is that you start to back feed current into the lower supply. The normal current that the load will take will also cause a voltage drop across the resistor to add to the differential voltage. The source impedance of the power supply will be higher and could also cause oscillations.
If you want better than a diode you could use a Schottky diode as these have a much smaller forward voltage drop.

At this point 40 years of actual design in a job seems very unlikely.

Well left school → two years in industrial electronics → degree in Physical Electronics (4 years) → 3 years grad school → 21 years University Lecturer ( U.S. read Professor ) in Physics / micro computer instrumentation → 4 Years designing Digital Set top boxes → 5 years designing RFID systems → 3 years designing a Trans-modulation system → 4 years freelance consultant → supposed to be retired but now writing books. It adds up. :slight_smile:

Back to the topic at hand, retrolefty gave you the solution if what you describe has to happen. In many ways that solution is similar to what you do when you have a battery connected as a back-up power supply. The two USB would get connected like the regulator and batter in the following circuit with a pair of low Vf diodes.

Which ever of the two has the higher output voltage ‘shuts off’ the diode of the other and the device with the higher voltage provide ALL of the power to the circuit in question.

Lets put it this way, how much resistance do you think the entire arduino board is? (the one being powered from this source) 10, 100, 1000 times larger then 10ohms?

If we put that in a voltage divider (1000/1010)*5V = 4.95V. I think its safe to say that a 0.05V drop on that resistor is the largest drop you could expect, I would think its going to be far less.

However that's not the only possible drop on that resistor if you have 2 USB like my schematic, its also going to drop the difference between the 2 sources across it. This is where thevenin theorem comes in.

The 10 ohms is to both drop the voltage difference from 2 sources and not create current limiting in the event that your only plugged into 1 USB.

I already stated that I was aware of the current feedback but that level of current shouldn't cause any issues. (10-20mA)
The resistor is there as an energy absorber for the voltage difference. I see no way that oscillations could start; that only happens when 2 regulators are truly in parallel with no resistance matching.

I'm aware of germanium diodes for smaller voltage drops, but I want 5V not 4.3V or 4.7V.

Aside from the current feedback which I think we all know isn't enough to cause issues on the regulators rated for currents 100+ times larger. Especially on USB ports that have there own protection and very like have there own diodes built in.

harddrive123:
The resistor is there as an energy absorber for the voltage difference. I see no way that oscillations could start; that only happens when 2 regulators are truly in parallel with no resistance matching.

In real circuits there is ALWAYS resistance in the connections (granted usually much less than 10-20 ohms), but never-the-less such oscillations do statrt. Further, while what you propose may work most of the time, what is the end result if your circuit is connected to a source that was produced by some shady third world company who cut all possible costs and removed every thing in a normal USB design that you are relying on to allow your approach to possibly work?

harddrive123:
I'm aware of germanium diodes for smaller voltage drops, but I want 5V not 4.3V or 4.7V.

Getting schotky diodes with voltage drops of 0.15V is not difficult to find.

harddrive123:
Lets put it this way, how much resistance do you think the entire arduino board is? (the one being powered from this source) 10, 100, 1000 times larger then 10ohms?

If we put that in a voltage divider (1000/1010)*5V = 4.95V. I think its safe to say that a 0.05V drop on that resistor is the largest drop you could expect, I would think its going to be far less.

However that's not the only possible drop on that resistor if you have 2 USB like my schematic, its also going to drop the difference between the 2 sources across it. This is where thevenin theorem comes in.

The 10 ohms is to both drop the voltage difference from 2 sources and not create current limiting in the event that your only plugged into 1 USB.

I already stated that I was aware of the current feedback but that level of current shouldn't cause any issues. (10-20mA)
The resistor is there as an energy absorber for the voltage difference. I see no way that oscillations could start; that only happens when 2 regulators are truly in parallel with no resistance matching.

I'm aware of germanium diodes for smaller voltage drops, but I want 5V not 4.3V or 4.7V.

Aside from the current feedback which I think we all know isn't enough to cause issues on the regulators rated for currents 100+ times larger. Especially on USB ports that have there own protection and very like have there own diodes built in.

Well as you are the one that started the topic, concluding with:

Any reason why this wouldn't be true?

And I don't see a single reply from anyone saying, "yes, go for it", I guess you didn't really need any advice or conformation in the first place. So rather then continuing this ego battle just go ahead with your solution, it most likely won't result in smoke but know your 'solution' is unique in my 50 years of electronics also.

@ retrolefty +1
Just came back to my computer and that is exactly what I was going to say.

Grumpy_Mike:
@ retrolefty +1
Just came back to my computer and that is exactly what I was going to say.

That just might mean we are both old and fixed in our way. :wink: