USB to TP4056 charger - why is the diode there?

P

Peabody

Guest
This is the relevant part of the schematic of an 18650 battery charger and
boost converter (to 5V) module:

https://i.postimg.cc/Tw7X35Fq/USB-to-TP4056.jpg

And here's the module:

https://www.aliexpress.com/item/32870411748.html

Coming from the USB power pin, there's a series Schottky diode and a very
low value resistor.

The resistor is found in the TP4056 datasheet typical application
schematic, and I think I understand why it's there. The TP4056 is a linear
part, and when the battery is mostly discharged, the charger will be
supplying 600mA (in this case), and the voltage drop will be at its
greatest. While the TP4056 has a built-in current reduction if the die
gets too hot, the resistor can take part the voltage drop, and part of the
heat, and allow the charger to continue delivering 600mA.

But I don't understand why the diode is there. Of course it would also
drop the voltage, but if that's its only purpose, the value of the resistor
could have been adjusted to accomplish the same thing - without adding
another part. If it's there as a reverse polarity protection, I would just
say that as a practical matter you aren't going to get a polarity reversal
from a USB port. Of course the 18650 being charged could be inserted
backwards, I don't see how that would affect the USB port.

Can anyone suggest why the diode is there, or if they've seen this in
other charger circuits? I ask because I'm planning to modify the module to
add a load sharing circuit, and the diode would come in handy for that if
it's not really needed in its original position. Thanks very much.
 
Peabody <waybackNO584SPAM44@yahoo.com> wrote:
This is the relevant part of the schematic of an 18650 battery charger and
boost converter (to 5V) module:

https://i.postimg.cc/Tw7X35Fq/USB-to-TP4056.jpg

And here's the module:

https://www.aliexpress.com/item/32870411748.html

Coming from the USB power pin, there's a series Schottky diode and a very
low value resistor.

The resistor is found in the TP4056 datasheet typical application
schematic, and I think I understand why it's there. The TP4056 is a linear
part, and when the battery is mostly discharged, the charger will be
supplying 600mA (in this case), and the voltage drop will be at its
greatest. While the TP4056 has a built-in current reduction if the die
gets too hot, the resistor can take part the voltage drop, and part of the
heat, and allow the charger to continue delivering 600mA.

But I don't understand why the diode is there. Of course it would also
drop the voltage, but if that's its only purpose, the value of the resistor
could have been adjusted to accomplish the same thing - without adding
another part. If it's there as a reverse polarity protection, I would just
say that as a practical matter you aren't going to get a polarity reversal
from a USB port. Of course the 18650 being charged could be inserted
backwards, I don't see how that would affect the USB port.

Can anyone suggest why the diode is there, or if they've seen this in
other charger circuits? I ask because I'm planning to modify the module to
add a load sharing circuit, and the diode would come in handy for that if
it's not really needed in its original position. Thanks very much.
Click to expand...

Microchip's In Circuit Serial Programming (ICSP) implementation:

https://crcomp.net/icsp/TB016.png

uses a Schottky-type diode. The diode isolates the MCLR/Vpp pin from
the application circuit when the chip is programmed.

Thank you,

--
Don Kuenz KB7RPU
There was a young lady named Bright Whose speed was far faster than light;
She set out one day In a relative way And returned on the previous night.
 
On 2/5/20 8:41 am, Peabody wrote:
This is the relevant part of the schematic of an 18650 battery charger and
boost converter (to 5V) module:

https://i.postimg.cc/Tw7X35Fq/USB-to-TP4056.jpg

And here's the module:

https://www.aliexpress.com/item/32870411748.html

Coming from the USB power pin, there's a series Schottky diode and a very
low value resistor.

The resistor is found in the TP4056 datasheet typical application
schematic, and I think I understand why it's there. The TP4056 is a linear
part, and when the battery is mostly discharged, the charger will be
supplying 600mA (in this case), and the voltage drop will be at its
greatest. While the TP4056 has a built-in current reduction if the die
gets too hot, the resistor can take part the voltage drop, and part of the
heat, and allow the charger to continue delivering 600mA.

But I don't understand why the diode is there. Of course it would also
drop the voltage, but if that's its only purpose, the value of the resistor
could have been adjusted to accomplish the same thing - without adding
another part. If it's there as a reverse polarity protection, I would just
say that as a practical matter you aren't going to get a polarity reversal
from a USB port. Of course the 18650 being charged could be inserted
backwards, I don't see how that would affect the USB port.

Can anyone suggest why the diode is there, or if they've seen this in
other charger circuits? I ask because I'm planning to modify the module to
add a load sharing circuit, and the diode would come in handy for that if
it's not really needed in its original position. Thanks very much.
Click to expand...

The diode is there to prevent the charger from powering the USB host and
blowing something up.

CH
 
lørdag den 2. maj 2020 kl. 09.44.36 UTC+2 skrev Clifford Heath:
On 2/5/20 8:41 am, Peabody wrote:
This is the relevant part of the schematic of an 18650 battery charger and
boost converter (to 5V) module:

https://i.postimg.cc/Tw7X35Fq/USB-to-TP4056.jpg

And here's the module:

https://www.aliexpress.com/item/32870411748.html

Coming from the USB power pin, there's a series Schottky diode and a very
low value resistor.

The resistor is found in the TP4056 datasheet typical application
schematic, and I think I understand why it's there. The TP4056 is a linear
part, and when the battery is mostly discharged, the charger will be
supplying 600mA (in this case), and the voltage drop will be at its
greatest. While the TP4056 has a built-in current reduction if the die
gets too hot, the resistor can take part the voltage drop, and part of the
heat, and allow the charger to continue delivering 600mA.

But I don't understand why the diode is there. Of course it would also
drop the voltage, but if that's its only purpose, the value of the resistor
could have been adjusted to accomplish the same thing - without adding
another part. If it's there as a reverse polarity protection, I would just
say that as a practical matter you aren't going to get a polarity reversal
from a USB port. Of course the 18650 being charged could be inserted
backwards, I don't see how that would affect the USB port.

Can anyone suggest why the diode is there, or if they've seen this in
other charger circuits? I ask because I'm planning to modify the module to
add a load sharing circuit, and the diode would come in handy for that if
it's not really needed in its original position. Thanks very much.


The diode is there to prevent the charger from powering the USB host and
blowing something up.
Click to expand...

the datasheet says no diode is required, I think the diode and resistor is
simply there to reduce dissipation in the IC
 
Lasse Langwadt Christensen says...

the datasheet says no diode is required, I think the
diode and resistor is simply there to reduce dissipation
in the IC
Click to expand...

Would it have been possible to eliminate the diode and
instead increase the value and wattage of the resistor to
achieve the same voltage drop? That would have eliminated a
relatively expensive part from the device.

The biggest voltage drop at full charging current is when
the battery is at 3V (below that there will be low-current
trickle charging). And the datasheet says the minimum Vcc
is 4V. So using only a resistor you would have to drop 1V
at 600mA, which would be 1.67R and 600mW. Is a chunky
resistor like that difficult or expensive to source as SMD?
Even so, it seems using two snaller resistors would be less
costly than using the Schottky diode.

Well, I guess I need to figure the voltage drop and
dissipation curves for the two alternatives and see what the
difference is. But I suspect there won't be much
difference.
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top