Need some advices about USB powering.

Diode USB +5V ------- ->|------------------ | -------------- PIC Chip | | Diodes | ----- Capacitor LM7805 +5V ------- ->|-->|--------------- ----- | GND A simple diode switch will work. The voltage drop when USB is unplugged will be about 0.3V for schottky diodes.

Yes, they do. However, there are many low-power variants that have ground currents in the 10s of microamperes.

Mark Borgerson

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Thank you!! So that would be 2 diodes on the output of the LM7805? I think the drop woulb be too big, somethink like 0.6V for both diodes. Thats too close to the limit (4.2V).

Any recomendation for low power 7805? Thank you!

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The 2 diodes on the LM7805 supply are to guarantee it doesn't turn on at all if the USB voltage is a bit lower than 5V.

If you pick the right Schottky diodes, you should have only about

I use the MIC5205 from Micrel. There are LOTS of others.

Mark Borgerson

You may want to look at the USB specifications on

Download the USB 3.0 specification and look at Chapter 11. Interoperability and Power Delivery/

The nominal voltage is 4.45 - 5.25 volts but could be as low as 4.0volts in some cases and still meet USB spec's

Regards,

Walter

-- Walter Banks Byte Craft Limited

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A linear regulator draws a few milliamps when no output current is consumed. The schematic below shows how I did it in a commercial product:

The diode in the ground line of the regulator compensates for the loss of the other diode in the output line. On the other hand, as someone else already pointed out, the USB bus voltage can be as low as 4V and I have never experienced any trouble with this product, even though the design was made for 5V...

Meindert

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A very simple question, at you design, what would make the USB have higher voltage over the regulator as you are using a diode on the regulato ground to compensatethe loss of the output diode? And if there is n external supply the USB wont be able to supply power too right? Becous there will be no voltage at the FET Gate.

Second, if i make a boost converter for USB voltage to something like 13V will i have any problem to charge a Lead-Acid Battery?

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You won't be able to charge a very large battery at anything more than a trickle. USB ports are generally limited to 500mA or less. by the time you boost that to 13.5V, you will be lucky to get

Mark Borgerson

The battery iself will be a very small one just for backup for the USB. But will that work? Is there any problem to charge a 12V lead-acid battery with a switched converter? I will be using a LT1618.

Thank you!!

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Not sure what you mean with this? The diode on the regulator output prevents current from the USB to flow into the regulator output. The FET acts as a switch to block the USB current when the regulator is supplying current.

I used a p-channel FET, which conducts when the gate is negative with respect to the source. So when no external power is supplied, the source is at Vusb level and the gate is tied to ground through the 22k resistor. Hence the gate is negative to the source and the FET conducts. Applying external voltage to the regulator puts the gate at the same level as the source and the FET is pinched off.

No, but there will be only a very low current to charge that battery.

Meindert

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Thank you for you explanation!! Now i got the idea. Clever one actually. And the low current is not a problem. Thank yo for the idea!!

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In theory, it should work. In practice, I would be wary of connecting a switching converter to the USB +5V without good filtering. The switching noise might interfere with other devices on the same USB hardware.

You might be better off with a set of 4 NIMH cells and a linear charging circuit. The actual design will depend a lot on how much current will be drawn and for how long when the USB is not connected.

Mark Borgerson

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I am afraid of that too. As i am for of being an expert on switched converters but I chose this specific part number becouse of the current limit and becouse the typical application is a "USB to 12V Boost Converter" hehe. The datasheet is here if you to take a look:

I am following the design on the datasheet. Any further sugestion? The battery itself is not part of the project. It is a requirement. The company is planing to change the battery on future designs. But for now thats all i got. As the battery will be a backup for the circuit, it is expected that the USB will be powering the system 99.99999% of the time, and the battery will remain charged, so there will be almost no current flowing to the boost converter. For switching between both i will be using a TPS2110A. So when the USB voltage gets bellow 4.5V the Battery will start to power the system. For rgulation, i will be using a linear one. The eficiency of the battery powerng the system will be about 30% but thats ok, as the current will be something close to 50-70mA and the battery will be supplying the circuit almost never. Any further sugestion?

Thank you!

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