# Boost battery voltage without regulating it

I'd like to boost the voltage of a lithium battery while having the converted voltage reflect the battery's state of charge (such that the output voltage equals the input voltage plus a set amount).

From what I understand, available boost/buck converters also regulate the voltage. Is there any way around this?

Would it be possible electrically to somehow combine/offset the converted output voltage with the unconverted voltage from the battery?

I should add that I have fairly good experience with the practical aspects of electronics (knowledge of components, circuit boards, soldering etc.), but very lacking knowledge when it comes to electrical engineering and circuit design. So I need all the help I can get here :)

• Sure, it's possible to make a converter that outputs twice the input voltage for example. The converter chips are very flexible; they use feedback to get the desired output. With different feedback you can make them do constant voltage, constant current, or in your case, double the input voltage. You would need to make your own converter (starting from a control chip), not just get one designed to output a certain voltage, which has the feedback built-in. Jun 2, 2022 at 17:24
• Have you looked at charge pumping? That tends to be a multiple of the input voltage. Jun 2, 2022 at 17:34
• Synchronous switcher oriented as a boost and forced to run at 50% duty cycle without feedback. Jun 2, 2022 at 18:40
• @PerryWebb I haven’t come across charge pumps before (though I realize now that they are a common thing), but in theory that seems to be the solution I’m looking for. However, are they an option also for higher currents? I’ll be using the battery to power a laptop and need it to deliver 2 A at ≈6.5 V (from a ≈3.7 V lithium polymer battery).
– Misa
Jun 2, 2022 at 22:13

output voltage equals the input voltage plus a set amount

A way to manipulate the feedback node of DC-DC converters is to inject a current into it.

Here "X1" represents your boost converter. R1 and R2 set the output voltage. R5 converts the input voltage to a current, which is mirrored, and pulled from the feedback node. Higher input voltage will sink more current from the feedback node, causing output voltage to rise.

You can set R1, R2 and R5 to get the effect you want.

You can also use an opamp, but the idea here is that, to cause the output voltage to increase, a current has to be pulled from the feedback node. With a negative supply, this could be achieved with an inverting opamp and a resistor, but without a negative supply, a current mirror is a convenient alternative.

Mohamed Gharib demonstrating the circuit.

• Thank you for the detailed explanation. A lot of new terminology and concepts here, and a bit over my head, but after some quick research I think I follow along the basic idea. A couple of questions: 1) Is it a simple matter to identify the feedback node on any commercially available boost module? (I assume this is the one labelled ”sense” in your diagram) 2) How much drain would this extra circuity add to the battery, and will it be draining the battery even when it’s not in use? 3) Can R1 and R2 be omitted if I get a boost module with a variable output?
– Misa
Jun 2, 2022 at 22:15
• More specifically, I’m looking to convert the circa 4.0–3.5 V range of a 10,000 mAh lithium polymer battery to circa 6.8–5.65 V (upper range can vary between 6.8 and 7.6 V).
– Misa
Jun 2, 2022 at 22:17
• 1) yes it's usually labeled "feedback pin". 2) you need a power switch anyway because the boost will draw some current even without a load at the output. 3) this circuit depends on the values of R1-R2 so it won't work with variable output. What's the ultimate goal? Jun 3, 2022 at 6:14
• I can’t seem to find any markings other than the usual VIN, VOUT etc. when looking at a few modules online. Could you point to an example? Or are you talking about documentation for boards or individual ICs?
– Misa
Jun 3, 2022 at 9:14
• I've added it to the answer. Thanks! Jul 14, 2022 at 23:14

From what I understand, available boost/buck converters also regulate the voltage. Is there any way around this?

Yes, this simply can be done but maybe in a way that might not be acceptable to you i.e. instead of the output equalling the input battery voltage plus an offset, a simple solution is to make the output proportional to the input voltage.

This can be easily done because normal regulator circuits tend to use a fixed voltage reference thus achieving output voltage regulation so, instead, use a proportion of the battery voltage to act as the reference then, as the battery voltage lowers, the output (boosted) also lowers but in proportion to the battery voltage.

Of course, this can be modified to what you appear to want by adding a fixed voltage reference in series with the battery before making a proportion that acts as the reference. The booster circuit would still use raw battery for its power source.

• Thank you for your response. Am I understanding you correctly if I interpret the solution you suggest here to be the same as the one bobflux detailed in his response?
– Misa
Jun 2, 2022 at 22:20
• I haven't read it sorry. Jun 2, 2022 at 22:33