What's the function of the following diode in the diagram?

Thread Starter

JosXD

Joined Mar 16, 2022
16
diode.png
The diode is in the first arrow, it is reverse polarity protection?, if yes, shouldn't be placed in the second arrow?
 

Papabravo

Joined Feb 24, 2006
19,578
It is how a boost converter works. The diode blocks the voltage on COUT from returning to the Lithium Cell so it continues to build up when the switch is open, and the inductor charges the capacitor. When the switch is closed, the capacitor can only discharge into the load. This process repeats each switching cycle.

You didn't say which part you were using in your example. The steady state behavior may be different from the transient behavior.
 
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Thread Starter

JosXD

Joined Mar 16, 2022
16
It is how a boost converter works. The diode blocks the voltage on COUT from returning to the Lithium Cell so it continues to build up when the switch is open, and the inductor charges the capacitor. When the switch is closed, the capacitor can only discharge into the load. This process repeats each switching cycle.

You didn't say which part you were using in your example. The steady state behavior may be different from the transient behavior.
Hi Papabravo, is the SD6201 Boost Converter, datasheet: https://datasheet.lcsc.com/lcsc/1804250821_SHOUDING-SD6201-AF_C171633.pdf

In the Datasheet, you can see that the first diagram says typical aplication circuit and in that diagram there is not diode and also there is not path between Vout and Inductor SW.

So my question, is the diode in the first arrow for reverse polarity protection or what is it for?
 

crutschow

Joined Mar 14, 2008
31,123
In the Datasheet, you can see that the first diagram says typical aplication circuit and in that diagram there is not diode
I believe that data sheet figure is in error and the diode should be there, since it is shown in two other circuit examples.
 

michael8

Joined Jan 11, 2015
333
Interesting. I notice that there isn't any block diagram of what's inside the IC.

The datasheet mentions both an NMOS switch and a PMOS switch.

page 7: VOUT (Pin 5): Output Voltage Sense Input and Drain of the Internal Synchronous Rectifier MOSFET.

Also:
VIN (Pin 6): Battery Input Voltage. The device gets its start-up bias from VIN. Once VOUT exceeds VIN,
bias comes from VOUT. Thus, once started, operation is completely independent from VIN.

A guess: Perhaps the diode is to get voltage to the output (which runs the chip!) for startup under certain
input voltage conditions?
 

Papabravo

Joined Feb 24, 2006
19,578
If the part is indeed a synchronous boost regulator, then operation without the diode is possible. In such a regulator it is not uncommon to add a diode in parallel with the synchronous switch and its associated body diode. Nothing wrong with that at all.
 

k1ng 1337

Joined Sep 11, 2020
701
If the part is indeed a synchronous boost regulator, then operation without the diode is possible. In such a regulator it is not uncommon to add a diode in parallel with the synchronous switch and its associated body diode. Nothing wrong with that
Why add a diode if the FETs do the work?
 

Thread Starter

JosXD

Joined Mar 16, 2022
16
So guys, then this SD6201 has that diode bypassed like when you add reverse current protection to a LDO?
ldo-diode.png

also because is a synchronous boost converter and uses like an internal p mos? here is the SX1308 which doesn't have a pin for Vout, the Vout goes directly from the inductor:
sx1308-diode.png
SX1308 datasheet: https://datasheetspdf.com/pdf-file/921054/Suosemi/SX1308/1
 
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boostbuck

Joined Oct 5, 2017
260
It appears from the second paragraph in the datasheet that the SD6201 is able to run at low output voltages without an external shottky bypass diode.
 

crutschow

Joined Mar 14, 2008
31,123
So the diode is apparently needed only for an output greater than 4.3V.
It's not clear to me how the diode helps raise the output voltage.
 

Papabravo

Joined Feb 24, 2006
19,578
So the diode is apparently needed only for an output greater than 4.3V.
It's not clear to me how the diode helps raise the output voltage.
It is possible that with the external diode you have more range on the possible duty cycle values. Since the output voltage is inversely proportional to (1-D), a larger duty cycle results in a larger boost ratio. Without some kind of an internal block diagram or a simulation model it will be hard to tell for certain.
 
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