I want an opinion on this boost converter 5 volt to 50 volt

[alexaamnda]

Is this circle worth developing, knowing that I am a beginner?

 

[Papabravo]

No, a boost ratio of 10 is impractical to say the least. Small changes in duty cycle result in large changes in the output voltage. The boost ratio of 10 for the voltage means a reduction in the output current by the same factor. So, if you have 5V @ 1 Amp of input power you will have less than 50V @ 100mA of output power because no DC-DC power conversion scheme is 100% efficient. What you want to accomplish is doable with a converter using a transformer, but that might be a level of difficulty you are not ready for.

A more reasonable design for your application might be a 12V to 50V converter. The lower boost ratio will result in a more stable and controllable device.

 

[alexaamnda]

No, a boost ratio of 10 is impractical to say the least. Small changes in duty cycle result in large changes in the output voltage. The boost ratio of 10 for the voltage means a reduction in the output current by the same factor. So, if you have 5V @ 1 Amp of input power you will have less than 50V @ 100mA of output power because no DC-DC power conversion scheme is 100% efficient. What you want to accomplish is doable with a converter using a transformer, but that might be a level of difficulty you are not ready for.

A more reasonable design for your application might be a 12V to 50V converter. The lower boost ratio will result in a more stable and controllable device.

Hello brother, thank you for interacting with me. Can you contact me to help me? I am a beginner.

 

[Papabravo]

Isn't this thread a good way to do that? That way your questions and my answers can be beneficial to a wider audience.

 

[alexaamnda]

Isn't this thread a good way to do that? That way your questions and my answers can be beneficial to a wider audience.

Okay, brother, I want to operate 8 5-volt lamps, each of which has 24 volts and 15 watts. Is this possible or not?

 

[Papabravo]

I don't understand the nature of a 5-volt lamp that has 24 volts on it. When you say "operate 8 5-volt lamps" can you be more specific.

Some numbers:
If you have 8 lamps at 15 watts that is a total of 120 watts. From a 24V output of a DC-DC converter you will need 5 amperes of current because 24 x 5 =120. If we assume that your boost converter is 80% efficient, the required input power will be 120 watts / 80% = 150 watts. Now 150 watts from a +5 volt supply would require 30 Amperes of current. Such a supply would not be within the typical skill level or budget of a backroom hobbyist. At those power levels starting a fire or giving yourself a lethal shock from working on an open frame supply are distinct possibilities.

 

[Tony Stewart]

Your schematic of a 2S16P array with 22.17 V across them mismatches everything you said.
Fine tuning voltage is sensitive/unstable with temperatures.

No this is not a good start.

 

[Karklebapper]

Most of the comments above are correct. I would argue about needing a transformer, but

Volts*Amps output = (Volts*Amps input)*(efficiency = 80% to 95% if not too bad).

I do not know what is supplying your 5V input but above 25 - 30W I would certainly try to start with a higher voltage.

10:1 boost voltage ratio is OK for a discontinuous mode boost converter using a simple inductor, but I would not attempt any of this without help or specific knowledge. Mainly, proper pcb layout is critical and you need an oscilloscope to work on the board once you build it.

For higher output/input boost voltage ratios, I suggest

How2Power.com - Answering Your Questions About Power Design: : Electronics Design Services : Voltage Regulator : Industrial Power Supplies : Semiconductors : Transformer : IGBT Module : Software : Digital Power Supply DC

How2Power brings electrical and electronics engineers extensive and free technical information on Custom Power Supply, Electronics Design Services, Voltage Regulator, Industrial Power Supplies, Semiconductors, Transformer, IGBT Module, Software and Digital Power Supply DC.

www.how2power.com

The above link points to an article about the SEPIC Multiplied Boost Converter

 

[Papabravo]

...

10:1 boost voltage ratio is OK for a discontinuous mode boost converter using a simple inductor, but I would not attempt any of this without help or specific knowledge. Mainly, proper pcb layout is critical and you need an oscilloscope to work on the board once you build it.

...

It is hard to run DCM with a duty cycle of 90%. Also changing the duty cycle by 0.1% in either direction gives boost ratios of approximately 9 and 11. Only a fool would try to make such a boost converter since closing the loop is damn near impossible. In order to do DCM the boost ratio has to be smaller than 10 IMHO. YMMV.

 

[alexaamnda]

I don't understand the nature of a 5-volt lamp that has 24 volts on it. When you say "operate 8 5-volt lamps" can you be more specific.

Some numbers:
If you have 8 lamps at 15 watts that is a total of 120 watts. From a 24V output of a DC-DC converter you will need 5 amperes of current because 24 x 5 =120. If we assume that your boost converter is 80% efficient, the required input power will be 120 watts / 80% = 150 watts. Now 150 watts from a +5 volt supply would require 30 Amperes of current. Such a supply would not be within the typical skill level or budget of a backroom hobbyist. At those power levels starting a fire or giving yourself a lethal shock from working on an open frame supply are distinct possibilities.

Yes, I will develop this circuit and republish it, but thank you for your response and thank you, my brother, for your beautiful response.

 

[alexaamnda]

Your schematic of a 2S16P array with 22.17 V across them mismatches everything you said.
Fine tuning voltage is sensitive/unstable with temperatures.

No this is not a good start.

I am a beginner, my brother, and I am trying to develop this circle so that it has a high percentage of success. Thank you for your response, my brother.

 

[MrAl]

Is this circle worth developing, knowing that I am a beginner?

View attachment 146636

Hi,

You can make a boost converter with a higher ratio like 1:10 but you have to pay attention to detail and what has been done in the past. There are boost converters out there that do 1:10 but they are more carefully designed, and above all, HAVE FEEDBACK.

That's one major thing that your design is missing, feedback. You have to have some mechanism that can measure the output voltage and can provide a signal to the control part of the circuit that allows it to adjust the output voltage so that it stays constant. However, with that said, LED drivers are best when they can provide a constant current not a constant voltage. You may get by with a constant voltage though as long as you do not see the current change too much as the LEDs heat up.

So, the first thing wrong with the circuit is there is no feedback which is almost mandatory for this kind of circuit. If you do not use feedback, you have to put up with changes in input voltage and output impedance. It's not that that will never work, but you'd have to test it very carefully to make sure nothing changed too much.

The second thing is the MOSFET driver. A transistor and 220 Ohm resistor is not going to cut it. The MOSFET will most likely not turn off fast enough. A lot of the power in a MOSFET can end up being due to the switching times, the rise and fall times, when there is both current through the MOSFET and voltage across it. That means you should have fast switching times. This also plays in with the "total resistance" issue with boost converters which I'll get to in a minute. So the second thing you have to do is use a better MOSFET driver. There are all kinds on the web you can do a search. The simplest would be a two transistor setup which turns the MOSFET on with one transistor and turns it off with another transistor.

The third thing is the "total resistance" issue that plagues all boost converters. This is the combined resistance of all the power dissipating devices that appear in the circuit. This limits the input-to-output ratio and duty cycle.
For higher input-to-output ratios you need a low total internal resistance.
If you look at the attachment you'll see a blue plot and a red plot. The voltage gain ratio is shown on the left and the duty cycle across the bottom. You have to increase the duty cycle to get higher input-to-output ratio but if you go too far (see where the blue plot falls when it gets close to a duty cycle of 1) the output actually falls, and it falls fast. That means you no longer get a boost in voltage. At the top peak of the blue plot you can see the voltage ratio is close to 13, meaning it would be possible to get a ratio of 10. That's with a fairly low total internal resistance. The red plot on the other hand is with higher total internal resistance, and we can see the peak of that only gets up to about 9 before it falls fast. That means with that higher resistance a gain of 10 would not ever be possible no matter what the duty cycle is.
This is one of the main problems with boost circuits, they can only boost so much, and that is mainly governed by the total internal resistance of the circuit that dissipates power. This would be due to almost every power handling part in the system including the transistor, the inductor, and the filter capacitor. The output load also has some influence over this because it causes more current to flow through all the elements. This means attention to these components is paramount to getting a boost converter to work right. Of course the control circuit also has to be right. It cannot be allowed to go 'off the deep end' where the control circuit wants to keep increasing the duty cycle and does not know where to stop.

So there you have three main issues that have to be addressed.
Feedback, MOSFET driver, and Total internal resistance.

Most designers these days choose a boost converter controller chip to take care of most of the control.

 

[alexaamnda]

Hi,

You can make a boost converter with a higher ratio like 1:10 but you have to pay attention to detail and what has been done in the past. There are boost converters out there that do 1:10 but they are more carefully designed, and above all, HAVE FEEDBACK.

That's one major thing that your design is missing, feedback. You have to have some mechanism that can measure the output voltage and can provide a signal to the control part of the circuit that allows it to adjust the output voltage so that it stays constant. However, with that said, LED drivers are best when they can provide a constant current not a constant voltage. You may get by with a constant voltage though as long as you do not see the current change too much as the LEDs heat up.

So, the first thing wrong with the circuit is there is no feedback which is almost mandatory for this kind of circuit. If you do not use feedback, you have to put up with changes in input voltage and output impedance. It's not that that will never work, but you'd have to test it very carefully to make sure nothing changed too much.

The second thing is the MOSFET driver. A transistor and 220 Ohm resistor is not going to cut it. The MOSFET will most likely not turn off fast enough. A lot of the power in a MOSFET can end up being due to the switching times, the rise and fall times, when there is both current through the MOSFET and voltage across it. That means you should have fast switching times. This also plays in with the "total resistance" issue with boost converters which I'll get to in a minute. So the second thing you have to do is use a better MOSFET driver. There are all kinds on the web you can do a search. The simplest would be a two transistor setup which turns the MOSFET on with one transistor and turns it off with another transistor.

The third thing is the "total resistance" issue that plagues all boost converters. This is the combined resistance of all the power dissipating devices that appear in the circuit. This limits the input-to-output ratio and duty cycle.
For higher input-to-output ratios you need a low total internal resistance.
If you look at the attachment you'll see a blue plot and a red plot. The voltage gain ratio is shown on the left and the duty cycle across the bottom. You have to increase the duty cycle to get higher input-to-output ratio but if you go too far (see where the blue plot falls when it gets close to a duty cycle of 1) the output actually falls, and it falls fast. That means you no longer get a boost in voltage. At the top peak of the blue plot you can see the voltage ratio is close to 13, meaning it would be possible to get a ratio of 10. That's with a fairly low total internal resistance. The red plot on the other hand is with higher total internal resistance, and we can see the peak of that only gets up to about 9 before it falls fast. That means with that higher resistance a gain of 10 would not ever be possible no matter what the duty cycle is.
This is one of the main problems with boost circuits, they can only boost so much, and that is mainly governed by the total internal resistance of the circuit that dissipates power. This would be due to almost every power handling part in the system including the transistor, the inductor, and the filter capacitor. The output load also has some influence over this because it causes more current to flow through all the elements. This means attention to these components is paramount to getting a boost converter to work right. Of course the control circuit also has to be right. It cannot be allowed to go 'off the deep end' where the control circuit wants to keep increasing the duty cycle and does not know where to stop.

So there you have three main issues that have to be addressed.
Feedback, MOSFET driver, and Total internal resistance.

Most designers these days choose a boost converter controller chip to take care of most of the control.

Thank you very much for this advice. It helped me a lot. I will use it and download a new plan.

 

[Karklebapper]

It is hard to run DCM with a duty cycle of 90%

In DCM you can run the duty cycle at a lower number. That is the main reason why I chose DCM. At 100 KHz you could have 5 uSec ON time, 500 nSec discharge time, and 4.5 uSec dead time (approximately, for example.)

 

[Papabravo]

In DCM you can run the duty cycle at a lower number. That is the main reason why I chose DCM. At 100 KHz you could have 5 uSec ON time, 500 nSec discharge time, and 4.5 uSec dead time (approximately, for example.)

Then how does the duty cycle control the boost ratio?

 

[Karklebapper]

The loop dynamics are different from Continuous Conduction Mode, but each "ON" time charges the inductor up to an energy level determined by 0.5*L*(I^2). Switching Frequency*energy=Power. The output voltage will depend on both the peak current (determining energy) and the load. But...keep in mind that even in CCM if you reduce the load the output voltage will climb. If you are operating in CCM and a load reduction causes an output voltage increase you will eventually enter DCM, unless the ON time is increased.

I have found this (I have not examined it but a TI paper should be reliable.)

https://www.ti.com/download/trng/docs/seminar/Topic_3_Lynch.pdf