DC to DC boost converter malfunctioning

[electricalguy]

Hi, I would like some assistance with a DC to DC boost converter that isn't powering the load I require. The boost converter is one I purchased online. It is a DROK numerical control regulator, 900 watt boost converter.

Boost converter specs
Input voltage DC 8-60 volts
Input current DC 0-15 amps
Output voltage DC 10-120 volts
Output current DC 0-15 amps
Conversion efficiency 85%
Working frequency 150khz
The inductor value is 53uHenry
The input capacitor is 470 uFarad
The output capacitor is 0.1uFarad

There are 3 smoothing caps on the output which are 330uFarad each
The mosfet part number is:
1638AJ FDH 055N15AE
There was a part number on the diode but it is really small and hard to read, I will put down the number that I think it is but could be wrong:
5TP520J30CT
It has a forward voltage drop of 0.247 volts

The issue I am having is the boosting circuit isn't working for the load I have and with my limited expertise in electrical circuits I am hoping one of the experts can let me know how to adjust the components to make it work. So I took some measurements across the various components during the testing. My load has a resistance of 2 ohms and what I am trying to achieve is a voltage of 8 to 9 volts and an amperage of 0.4 to 0.7 amps if that is at all possible with the value of the load resistance.

The inductor voltage is 0.618 volts AC
The frequency across the components is 16.51khz
The voltage across the 470uF cap is 20.89 volts DC, the voltage across the output caps is 20.97 volts DC, the voltage across the mosfet is 20.80 volts DC, the voltage across the diode is 0.235 volts DC. The power supply I am using is a sky toppower STP3010 300 watt DC variable power unit. I was putting 21.3 volts at roughly 9 amps into the circuit. Across the load there was 19.7 volts DC at 7.2 amps DC.

I know the voltage is higher than what I am looking for but was trying to get the system to work by putting a high amount of power in the circuit. If anyone can give me some suggestions it would be very much appreciated.

 

[trurle]

I am trying to achieve is a voltage of 8 to 9 volts

The voltage across the 470uF cap is 20.89 volts DC,

The boost converted do not work in this way. You cannot get output less than input. Either use 8V input or use another, buck-type converter.

 

[Tom.G]

My load has a resistance of 2 ohms and what I am trying to achieve is a voltage of 8 to 9 volts and an amperage of 0.4 to 0.7 amps if that is at all possible with the value of the load resistance.

Seems to be a disconnect between what you are after and what is physically possible. There is a relation between Volts, Amps, and Ohms. It is called Ohms Law. Simply stated it is:

Volts= Current x Resistance

This is often shown as E= I x R.

Your values are Current= 0.4 to 0.7 and Resistance= 2
Plugging these values into Ohms Law gives: Volts= 0.4 x 2 or 0.8Volts (0.7 x 2 or 1.4Volts). This shows that to put 0.4Amps thru 2 Ohms you need only 0.8V. A higher voltage will result in a higher current.

To turn it around a bit, if you apply 8V to 2 Ohms you will have 8/2 or 4Amps flowing.

Or maybe you really want a resistor that will allow 0.4A to flow when 8Volts are applied. This would need a resistor of Volts/Amps, or 8/0.4 Ohms, which is 20 Ohms.

Do some reading about Ohms Law and decide what is possible that will satisfy your requirements.

Cheers,
Tom

 

[jim hardy]

Candidate for DIY thread ?

 

[electricalguy]

Thank you for the help, I had a misconception about the functionality of boost converters. I was thinking they increased the overall output impedance of the circuit and load, and since my load does have a reactive component to it I thought it would effect it. But I was able to find an alternative. Thanks again.

 

[davenn]

Thank you for the help, I had a misconception about the functionality of boost converters. I was thinking they increased the overall output impedance of the circuit and load, and since my load does have a reactive component to it I thought it would effect it.

no, they BOOST the voltage

But I was able to find an alternative. Thanks again.

so don't leave us hanging ... what have you found ?Dave

 

[Kelly23]

no, they BOOST the voltage

so don't leave us hanging ... what have you found ?Dave

What you find? Can you please share its alternative?

 

[Kelly23]

The boost converted do not work in this way. You cannot get output less than input. Either use 8V input or use another, buck dc dc converter.

I'm confused that how a boost converter can't work in this way whereas buck dc type converters can work. Please elaborate it a bit more.

 

[berkeman]

What you find? Can you please share its alternative?

The OP has not been to PF since 2020, so it's unlikely that they will reply to you or Dave.

I'm confused that how a boost converter can't work in this way whereas buck dc type converters can work. Please elaborate it a bit more.

Work in what way? It looks like the OP was wanting to use the converter to change the impedance of his voltage source to better match his load. That is not the primary goal of a DC-DC converter.

Can you say exactly what your current question is? The better you ask your question, the better we will be able to answer it. Thanks.

 

[DaveE]

I'm confused that how a boost converter can't work in this way whereas buck dc type converters can work. Please elaborate it a bit more.

This is a nice introduction. But, it may be more than you want. I would only look at the Buck, Boost, and Flyback topologies to start with.
https://ww1.microchip.com/downloads/en/appnotes/01114a.pdf