Circuit question

Sorry. It's DC. I should have labelled "AC adaptor output wires"!

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You didn't say if the output of this adapter is DC? I can only assume it is AC ? If so, shouldn't you be using a rectifier, cap and a 7805 Post reg? Or something close to that?

Jamie

the you should be passing that into a REG... use a 7805 or some LDO type

Something like those.

Jamie

Why do I need to use a regulator?? It's 6.2V DC reduced to 5V for the USB hub.

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Ripple removal, stability.. etc..

suit yourself.. It's your hub..

Jamie

--- Here's your circuit: (View using a fixed-pitch font)

. 6.2V E1 . | . [10k] R1 . | . +----> E2 . | . [40k] R2 . | . GND

With the resistors you've shown:

E1 * R2 2.48e5 E2 = --------- = -------- = 4.96 volts R1 + R2 5.0e4

and the current in the string will be:

E1 6.2V I = --------- = ------ = 124 microamperes R1 + R2 50kR

Now, let's say you have something connected to E2 that wants to take

1mA.

Then you'll have this:

. 6.2V E1 . | . [10k] R1 . | . +---->>----+ . | | . [40k] R2 [5k] R3 . | | . GND GND

But, since R3 is in parallel with R2 the combination will look like:

R2 * R3 40kR * 5kR Rt = --------- = ------------ ~ 4400 ohms R2 + R3 40kR + 5kR

Your circuit will look like:

. 6.2V E1 . | . [10k] R1 . | . +----> E2 . | . [4k4] R2 . | . GND

and the voltage at E2 will fall to about 1.89 volts.

For that reason you need to interpose a regulator between the 6.2V source and the load so that 5V will be supplied to the load regardless of the current it draws, but within the limits of the source and the regulator, of course.

Unfortunately, since it has a dropout voltage spec of 2 volts, the common 7805 won't work, so you'll need to find, most likely, an LDO regulator which can do the job.

-- JF

To expand this slightly, you (Man-wai) proposed: "I am going to use a

10k and a 40K resistor to make a potential divider to reduce 6.2V to about 5V." That gives you 80% of the input voltage and is fine when nothing is connected to the common point but what happens when the hub chip draws some current and you have four high current devices connected each drawing up to 500mA so the hub tries to draw about 2A from the junction of the resistors?

Also what actual voltage does the power supply output? You said it is a 6.2V one, but did you measure that or just read the label? If you measured it, did you also check what the voltage is under load? Can it even supply enough current? For a hub you need 500mA per port + a small margin for the controller chip so a 4 port hub need a 2.1A PSU and a 7 port hub needs a 3.6A PSU.

You don't even know if your hub needs 5V or somewhat more. If it has an internal regulator it may need between 7.5 and 9V. It is *MUCH* easier to design a circuit to switch properly between Self powered (external supply) and Bus powered (from the 5V on the USB cable)if the external supply is a little more than 5V. It is absolutely forbidden by the USB spec to 'back-feed' power to the port that goes to the host and simply connecting another supply to the USB bus WILL cause problems and may damage the host so it MUST have some sort of switching circuit.

Your idea isn't going to work the way you expect. It may do nothing, it may be worse than useless i.e. it prevents the hub working from bus power, but with those resistor values you are unlikely to cause damage if you get the polarity right.

and your PC, hub, USB devices, adaptor etc. As long as you can afford to replace them and dont leave them unattended till you know they aren't going to catch fire by all means experiment, but if you don't have lots of money, use a load that doesn't matter like a collection of

6V torch bulbs wired in parallel to test your power supply modification and be certain it works not expensive computer peripherals possibly with valuable data on. . . .

p.s. If you are trolling, you are good at it and have got me!

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Because I was not a trained professional in electronics, so I asked. I thought it's just a simple dc circuit.

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Was the hub originally unpowered?

That won't work. The resistors will fatally reduce the ability to supply current to the loads.

Can you buy a powered USB hub? They are fairly cheap.

John

USB supplies power to the devices plugged into the hub. If you connect four devices, it might change the current requirement by 4 x 500 mA, and a regulator would keep the hub voltage nearly constant when the current requirement changes by up to two amps.

Some hubs have internal regulators. Some expect that function of their power supply module.

If I had thought there was more than a 20% chance you were trolling, I wouldn't have bothered giving a detailed reply. We all started with the basics and a genuine question from anyone willing to learn (at least a little) is always welcome.

It *IS* a simple DC circuit, but unfortunately it seems you initially misunderstood the likely requirements of the hub. That stuff I mentioned about a switching circuit would all be inside the hub - unless it is a dirt cheap one from the dollar store that ignores the USB standard.

IF the power supply is rated for a couple of amps output current and is REGULATED (i.e. outputs the same 6.2V no matter whether you measure it unloaded or with a 1A load), and you can confirm the hub needs 5V, it may be worth taking this project further, otherwise 5V regulated SMPSUs with a suitable current rating are readily available from many suppliers, although buying a powered hub *WITH* a PSU may be simpler and cheaper!

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You can put two diodes in series, it will drop 2 x 0.6 volt.

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pim.

A potential divider produces a specific voltage *if* the current drawn is negligible compared to the no-load current. Otherwise, it produces a voltage which decreases linearly with the current drawn.

In practical terms, a potential divider is okay if you need a few milliamps or less. Above that, use a linear regulator; when you start measuring in amps rather than milliamps, consider a switching regulator.

If you are lucky and you use standard 3A silicon diodes, it *MAY* get the output voltage in spec for USB. You probably need a small minimum load to keep the voltage down and prevent it from damaging the hub when no devices are connected - a 100 ohm 1/2W resistor across the output after the diodes would be a good idea.

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What diode? Name?

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1N4148, 1N914, etc. - any general-purpose silicon diode.

Have Fun! Rich

Quite expensive a piece. Thanks.

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I won't connect more than 2 USB devices to the usb. SO I am NOT trying to overdraw from the AC adaptor.

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I would probably grab 1N540x series or similar 3A general purpose silicon rectifier diodes. As you only want to connect two devices, unless they both draw the maximum 500mA you could use 1N400x series 1A diodes. If you use too weak a diode it will probably fail short-circuit and blow up the hub and all connected USB devices and maybe your USB port.

If cost is a problem, you could salvage suitable diodes from the mains power input bridge rectifier circuit of just about any scrap equipment or appliance that has a switched mode power supply with its mains input fused at 3A or over - e.g. a scrap PC power supply. If you find the bridge rectifier is a single component, short its positive and negative output terminals and you will have a two diode drop if you put it in series via its input terminals with your 6.2V PSU and load. DON'T FORGET the 100 ohm half watt load resistor I mentioned, and CHECK the maximum output is less than 5.25V BEFORE you connect it to the hub.

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