3.7v to 5v dc to dc boost converter

[sumit garg]

Hlo
i want to make a 3.3v to 5.5 v dc to dc boost convertor
before buying the components i made one circuit on proteus
but its not working
components:
22uH inductor
ss16 shottky diode
12uF capacitors
ME2108A33M3g ic
If there is any alternative circuit for converting 3.3v-5v please tell
thanks

 

[KlausST]

Hi,

3.3V or 3.7V? Or is there a valid input voltage range?

If you want to get 5.5V output voltage then choose the according converter IC. Your´s is designed for 3.3V output.

But still there´s something wrong with the simulation. Even if the IC is not switching (likely because of wrong choice) there needs to be about 3.3V at the output due to the schottky diode.
0V as shown in your screenshot is impossible.

Klaus

 

[Akanimo]

Hi,

Can you connect a load resistor and try again?

 

[danadakk]

The datasheet does not call out a min load current spec, but the test conditions for startup
and hold are @ 1 mA so calc your resistor load to do 1 mA min at Voutmin..

Regards, Dana.

 

[KlausST]

Hi,

the output node shows 0V. So if I connect a resistor of any value (except zero) I´d expect zero current, too.
And zero current shouldn´t affect the circuit behaviour.

But I´m not a friend of those circuit simulators at all. I don´t say they are useless.
In my eyes they can´t replace a real circuit (regarding testing). If a circuit works in a simulation it does not meant that it will work in reality, too .... and vice versa (as we can see here).

I mean: what can one gain from simulation like above? (following the datasheet recommendations)
If it works: it´s just as expected.
If it does not work: Most probalby the simulation is wrong.
So one rather tests the simulation tool than the real circuit.


The biggest benefit I see: They prevent form smoke and fire, when one does something wrong.

Klaus

 

[danadakk]

Sims are very useful in determining how well your model of a device is
in describing its performance. Especially for non linear circuits where
analytical non-linear math ability limited in engineering community training.

Like the standard equations we use for biasing, they come from a model
that is a simplification of device behavior. Works well enough that jillions
of circuits have been designed using those inaccurate bias equations.

Its great for showing AC performance as well. Especially in complex networks,
of many poles and zeroes. And if it deviates from measured then modal was
insufficient/inaccurate.

The real problem in industry is models, as supplied by vendors purposely, in many
instances, not complete to hide proprietary design techniques and methods in their
IC.

And there is the propensity of users to not read the simulator setup properly.
like time step, initial conditions, BLAH BLAH BLAH.

So I blame the models (vendors) and users for most sim limitations, not the
simulator.

Of course this does not discuss measurement error and imperfect test systems used to
confirm bench results, a whole other topic of errors and limited capability.

Regards, Dana.

 

[sumit garg]

Hi,

3.3V or 3.7V? Or is there a valid input voltage range?

If you want to get 5.5V output voltage then choose the according converter IC. Your´s is designed for 3.3V output.

But still there´s something wrong with the simulation. Even if the IC is not switching (likely because of wrong choice) there needs to be about 3.3V at the output due to the schottky diode.
0V as shown in your screenshot is impossible.

Klaus

sorry for typing worng actually its 3.7v bcz while making real circuit i will use 3.7v lipo battery.

 

[stenzer]

Hi,

you are using a step-up converter IC to generate a lower output voltage, at least in your simulation (ME2108a33 -> Vout = 3.3 V). So you should use the proper model (5 V output voltage).

Has your inductor a resistance, or is it an ideal one? Can you measure the current into the LX node (pin 1)? You could also try to perform a time-domaine based simulation, you might see how the IC is struggling.

If there is no 5 V version, try to run your simulation with an input voltage of e.g. 2 V (Vin < Vout) instead of 3.7 V, to test if the simulation/IC-model works at all.

BR