Circuit Simulation of LTC3703

[electronicsLearner77]

I am trying to simulate the circuit which is a smaller part of the bigger circuit of the buck converter IC LTC3703 data sheet https://www.analog.com/en/products/ltc3703.html, the smaller circuit i have attached, which is on page 17, i created the LT spice model for the same, my expectation is the output shall be 12V but i get some mV. Please advise what is my mistake.

 

[Harald Kapp]

Have you tried to use the example that is supplied with LTSpice?

 

[electronicsLearner77]

Thank you for the reply I am not able to find this example model. I am able to see only LTC3703 part. Could you please share me the path or the model itself?

 

[crutschow]

I am not able to find this example model.

You right-click on the model icon and it will show you the path to the test fixture example (below):

 

[electronicsLearner77]

I have simulated the circuit, and it gives as expected 12V output,

Few questions i want to ask is, as referred in the data, can i use the output voltage to provide power to Vcc and DRVcc? I am trying it but not successful.

I get something like this


2. There is also requirement if the input power is provided an LED shall glow, how do i calculate what resistance value to use? If you could tell, i would do the same for the output as well. Please advise.

 

[crutschow]

What is Vin?

 

[electronicsLearner77]

Vin is 24V to 100V. Vout is 12V and 10Amps.

 

[Harald Kapp]

can i use the output voltage to provide power to Vcc and DRVcc? I am trying it but not successful.

Small wonder or no wonder at all: Where should the output voltage come from at startup? When powered on, V_out is 0 V, therefore the chip has no power and cannot start up to supply V_out.
You would need at least a special startup circuit. But why make the circuit more complex than necessary?

 

[crutschow]

Here's the simulation with 24V input, 10A load, and Vcc generated from the output 12V.

The simplified circuits you posted are just to indicate the basic circuit configuration, not to build an operating circuit.

The series resistor values for an LED are:
(Vs-Vd) / Id, where Vs is the supply voltage, Vd is the LED forward drop voltage, and Id is the desired LED current (usually a few mA, depending upon how much brightness you want).

 

[crutschow]

ou would need at least a special startup circuit. But why make the circuit more complex than necessary?

Because the Vcc and DRVcc bias voltages are limited to 15V max., so with a 24V-100V input, some means is needed to reduce the voltage for those inputs.
This requires either a separate supply or the Zener-Transistor kickstart circuit shown, to provide Vcc until the output can supply the Vcc voltage.

 

[Harald Kapp]

O.K., I see.

 

[electronicsLearner77]

Here's the simulation with 24V input, 10A load, and Vcc generated from the output 12V.

The simplified circuits you posted are just to indicate the basic circuit configuration, not to build an operating circuit.

The series resistor values for an LED are:
(Vs-Vd) / Id, where Vs is the supply voltage, Vd is the LED forward drop voltage, and Id is the desired LED current (usually a few mA, depending upon how much brightness you want).

Thank you very much. i will simulate and update the results, i have few other questions may be once i am ok with the results i will ask.

 

[crutschow]

Note that for a 100V input, you will need a higher voltage rated MOSFET (at least 120V).

Also Q1 will need to be mounted on a heat-sink.

 

[electronicsLearner77]

I have updated with higher MOSFET rating IC of 120V Vds. I also added for indication of the LED at the input and output side, but the simulation does not give 12V output when i add these components. The circuit works without input and output LED circuit.



One other question is since it is operating at high voltage of 100V, what other protections shall i implement? Short circuit protection will it be taken card in the LT3703, or i need to add additional circuit.

 

[crutschow]

Not surprising the circuit doesn't work.
The LEDs go from the input and output voltages to ground, not in series.
And why did you put one in the Vcc bias line?
You are trying to disobey Ohm's law.
How do you expect amps of load current to go through the 1kΩ resistor and LED?

Only the upper MOSFET needs the higher voltage rating.

Note that 100V is the Absolute Maximum Rating for the LTC3703.
For good reliability it should be operated at no more than about 75% of that or 75V.

The internal short short circuit protection should be sufficient.

 

[electronicsLearner77]

Not surprising the circuit doesn't work.
The LEDs go from the input and output voltages to ground, not in series.

The idea here is one LED will be ON when input voltage is powered on. One other LED will be ON when the output voltage has reached 12V.

And why did you put one in the Vcc bias line?

I thought lot about it, where to place the LED, the main idea was the Vin voltage could vary from 24V to 100V and i did not want to effect the LED brightness/ current due to this voltage variation. I wanted to have fixed voltage coming to LED hence i have put in the Vcc bias line.

You are trying to disobey Ohm's law.

I am not sure where it disobeyed.

How do you expect amps of load current to go through the 1kΩ resistor and LED?

Yes this doubt i had, i was not sure where to place LED.

 

[crutschow]

I wanted to have fixed voltage coming to LED hence i have put in the Vcc bias line.

But that bias is powered from the output voltage, so it's no different than one on the output.\

I am not sure where it disobeyed.

According to Ohm's law, to put 10 amps through a 1k resistor requires 10,000 volts.

i was not sure where to place LED.

Do you not understand what I mean by saying you should put them from the voltage being monitored to ground?

 

[electronicsLearner77]

Sorry for delay, i was bit confused, finally i got the output matching with 12V

I attached the schematic. Could you please confirm if it is correct?

 

[crutschow]

Could you please confirm if it is correct?

It is not.
You still are having problems with LEDs.
You placed an LED (D5) across R6, part of the voltage divider that provides feedback that determines the output voltage.
Since the FB reference voltage is only 800mV, the LED will never light.
If it did, it would cause an incorrect output voltage.
What made you think you could put it there?

As I've stated several times (are you not understanding me?) you need to add the LED with a series resistor at the output to ground.
Randomly adding parts without understanding the circuit seldom leads to a good results.

 

[electronicsLearner77]

Sorry ok I understand now. I will update and post it.