Adding to the output of a DC/DC Boost Converter using the same power supply as the input?

Question

I've got a project that requires a variable output between 24-48V. This needs to be continuous-current, not PWM.

Items that I have handy from my 'box of bits', and ready to work with are: an 0.6-36V Variable Boost Converter, and a 12V Regulated Switching Power Supply. Obviously, I'd prefer to use these rather than purchase a specialised supply if possible.

There are two things that I know at this point, therefore:

1. I can use the 12V supply as the input to the Boost Converter, but the output voltage won't be high enough.
2. If I had a 12V battery, I could add this in series with the Boost Converter output, to get a variable output from 12.6-48V.

The diagram below shows 'Point 2'. There is a 12V source in to the Boost Converter, and an additional 12V source in series with the output, leaving me with a variable 12.6-48V. This is great, except for the fact that I only have one 12V source.

Which then brings me to my question, can I use the 12V supply in place of the battery, in series with the output from the Boost Converter? There are now effectively connecting wires going from the output of the boost converter back to the input, which makes me nervous!

The image below shows the same circuit, but with the wires linked back to the original 12V source, rather than to a second one.

I have built the closest circuit I could in Falstad to hopefully illustrate what I'm thinking. They don't have a Boost Converter component, though!

• Could this work as a solution?
• Is there a way to effectively split the 12V source into two separate supplies so I can enact the first diagram?
• Failing the above, do you have a hardware recommendation for a source that's directly variable through the appropriate range?

Answer 1

Why not power the "0.6-36V Variable Boost Converter" with the additional battery and then add the "12V Regulated Switching Power Supply" in series to the output?

This should work, because there are no ground loops in contrast to the other suggestions mentioned so far.

Answer 2

Your 0.6V....36V converter is said to be buck-boost converter in the linked advertisement.

Buck-boost is a well known principle where the same circuit can produce with acceptable efficiency and peak input current as well higher and lower output voltages than the input voltage. The circuit doesn't need a transformer, a single inductor is enough. It has advanced controller IC and the needed switching transistors and diodes for both buck and boost operating modes. The controller selects the needed mode automatically.

Here's an example of the buck-boost idea. In this case it's used for broad input voltage range. https://www.analog.com/media/en/technical-documentation/data-sheets/8390fa.pdf

Boost converter actually adds in series with the input voltage extra DC which is generated as inductive pulses and stored to a capacitor. So, the 12V DC input is already used in series with inductively generated DC. It's no more availabe to be connected in series another time. That would be a short circuit if we assume your linked converter really operates as common boost coverter when it should increase the voltage. No datasheet = the already purchased circuit must be reverse engineered.

But you can make an inverting switch mode converter which generates -12V for you. Between its -12V output and the +36V output there's 48V. Unfortunately we do not know surely is the GND of the 36V converter at the output side same as the GND of the input side. If it's not something unwanted can happen. In addition the resulted 48V wouldn't be against GND, it would be +36V and -12V with GND not connected.

Another possibility is to use your battery idea. Have a flyback converter which has fully isolated output side.

You wrote "No PWM, it must be continuous current" Unfortunately all dc to dc converters eat current pulses from the input supply voltage. They can be smoothed with a LC low-pass filter. Designing it isn't trivial. Actually many converters have a LC input filter, but its smoothing operation is designed only to prevent radio interference. You may need much more effective filtering.

Answer 3

Your question lacks a good understanding of power conversion with the conversion of voltage, current and impedance. So a good answer begs a better question. Always start with good specs like any battery or IC datasheet summary.

When you define a boost regulator the load referred to the source is in theory the same as a passive transformer. with Zin/Zout= N^2 where voltage ratio = N (boost V reduces input Z)

If efficiency was ideal then the input impedance of the converter will be that average. But since it is regulated by pulsed voltages of ramped inductive current the impedance will oscillate about this average. Since your source will now be loaded by this reduced oscillating load impedance , its regulation will be greatly stressed by the impedance ratio of it's source to the transformed load with instabilities due to the real Z(t) ratios of source to load.

Your Falstad simulation is not defined properly in your question or simulation and has several errors.

All you have is a 3x VCVS using +12Vin to make an inverting boost converter to get 12- -36V = 48V. The LED string characteristics were not defined properly and thus the simple LED was taking > 1 kA !.

I won't give you an answer until you defined your question better with good input v. output specs for V,I & Z(t) or ESR or load regulation error.

I replaced it with a 48V 100W bulb which with 4x voltage reduces R by 16.

e.g. that appears as 23 Ohm load steady-state, but the VCVS input transforms that load to the input as 1.44 Ohms=12V^2/100W. Even though VCVS=3x +1 =4 It still has a gain of 4x input voltage thus the input impedance is 4^2 =16x (times) lower than the output load.

Thus for good stability your voltage source impedance needs to be <= 2% of this load and that's just an average and not what you get with a DCDC boost regulator which has to store power in pulses in the choke then switch that power to the output. thus Zin depends on peak/average current ratio.

I suggest you do a lot of reading on boost converters and use a very low impedance source such as a battery or a step down converter from AC.

Answer 4

You will most likely blow up your converter.

As seen above, the output of the buck-boost is connected to the negative battery terminal. Most buck-most converters are using Cuk topology, which has the negative terminals of Input and Output shorted together, and your going to SHORT your battery through the converter.