Current Amplifier Circuit with low input current

Hello I am attempting to building a circuit using a 30mAh lipo battery or perhaps a pair of batteries connected in parallel (60mAh). I'm still unsure of the initial power requirements. I need the circuit to amplify the output current to 150mAh.

Can any of the Adafruit products provide the type of current amplification requirements I am looking for? Or perhaps someone could recommend what I think needs to be a MOSFET circuit to accommodate a battery with such a low mAh.

Thank you! All help is greatly appreciated.

-Jesse

unknown5591

Posts: 75
Joined: Mon Jun 13, 2016 8:19 pm

Re: Current Amplifier Circuit with low input current

You can only increase current if you decrease voltage (physics!). So 3.7V at 60mAh can be 3.7*0.06*0.85/0.15=1.25V at 150mA at best (using 85% efficient conversion).
If you need more currect at the same voltage, you need to get bigger cells. If size is a requirement, don't forget to look at the round cells like 14500 ones, they can be quite powerfull for the size.

blnkjns

Posts: 630
Joined: Fri Oct 02, 2020 3:33 am

Re: Current Amplifier Circuit with low input current

I need the circuit to amplify the output current to 150mAh.

mAh is not a measure of current. It is a measure of charge capacity. More specifically, mAh is mA (milliamps) times h (hours). 150mAh means 150 milliamps times one hour.

To understand the current capabilities of a cell, you need to look at the datasheet. For example, this LIR2450 coin cell:

It is rated at 110mAh - at a 0.2C discharge rate. That means it is designed for a discharge rate of 0.2 * 110 = 22mA. At that rate, your run-time would be 5 hours.

It has a maximum discharge current of 2C - so it can deliver as much as 220mA. But the effective capacity at that rate will be lower, so your run time would likely be well under 1/2 hour.

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Re: Current Amplifier Circuit with low input current

Hi adafruit_support_bill and blnkjns, Thank you!

Its been a little while since I've been on the circuit playground and I'm certainly no expert.

Size restraint is a major problem in the design. The batteries that I've currently selected are AirPods replacements found on eBay https://www.ebay.com/itm/384061835350?hash=item596bde8856:g:h2IAAOSwiTNf~bzM . I am attempting to make a sleek coil lighter. This is for a product I recently unveiled that currently features a Pull Strike Butane Lighter. If you are interested you can see more at https://www.iozzio.life/parts/7wpi3nt6d3m4wo342rc50mssgizwf6 to see video of its function check out my Instagram https://www.instagram.com/p/CNXPZfMByk7/

I believe that batteries are Panasonic Pin Cells I contacted them regarding a sample and I've attached a screenshot of there general features
CG-320B CG-420A CG-425A General Features
Screenshot (59).png (71.18 KiB) Viewed 147 times

https://industrial.panasonic.com/ww/products/batteries/secondary-batteries/pin-li-ion

From what I can understand from the CG-425A discharge graphs it can output a maximum of 150mA(5.0C) ? As you can probably tell I'm still unsure about the power requirements of the device. I biased the requirements on a coil lighter I purchased on amazon which had a 150mAh Lipo battery. https://www.amazon.com/Rechargeable-Windproof-Cigarette-Fingerprint-Indicator/dp/B07MGXDTFQ/ref=sr_1_3?dchild=1&keywords=coil+lighter&qid=1620311477&s=hpc&sr=1-3
CG-425A discharge graphs
Screenshot (62).png (328.35 KiB) Viewed 147 times

In my latest test I connected 3 batteries in parallel and connected them to a 0.9-5V To 5V DC-DC Step-Up Power Module Voltage Boost Converter Board - US Ship https://www.ebay.com/itm/0-9-5V-To-5V-DC-DC-Step-Up-Power-Module-Voltage-Boost-Converter-Board-US-Ship/324318535523?_trkparms=aid%3D1110006%26algo%3DHOMESPLICE.SIM%26ao%3D1%26asc%3D20201210111314%26meid%3Db5747256f9f14b51a99508043871fc30%26pid%3D101195%26rk%3D1%26rkt%3D12%26mehot%3Dsb%26sd%3D323970386799%26itm%3D324318535523%26pmt%3D1%26noa%3D0%26pg%3D2047675%26algv%3DSimplAMLv9PairwiseUnbiasedWeb%26brand%3DUnbranded&_trksid=p2047675.c101195.m1851

The coil did begin to smoke but did not become redhot. The charge also didn't last very long and connections to the batteries were only made via finger pressure. I believe 3 batteries in parallel is the max length the device can accommodate two would be ideal for size requirements.

Thanks for providing a space for circuit work through. I hope you like my project/product!

-Jesse

unknown5591

Posts: 75
Joined: Mon Jun 13, 2016 8:19 pm

Re: Current Amplifier Circuit with low input current

If you have multiple cells, you may be better off connecting them in series. That will give you a higher voltage without the inefficiency of a boost converter. What are the voltage and current requirements for your load?

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Re: Current Amplifier Circuit with low input current

Bill,

Interesting thought, I'm currently unsure of the load requirements. I'm having a hard time finding datasheets for the coils. However I did connect the 150mAh 3.7V Lipo battery directly to the coil and it turned it redhot.

That is why I tried to put the batteries in parallel. Knowing that 3.7V is enough voltage to heat the coil I figured increasing the amps
would get the coil hot enough.

Up for any and all suggestions on the matter.

Upon further research I found a Nichrome wire calculator. https://www.easycalculation.com/engineering/electrical/nichrome-wire-calculator.php

Theoretical Nichrome Wire
Screenshot (63).png (91.42 KiB) Viewed 140 times

Potential Power Requirements
Screenshot (64).png (37.38 KiB) Viewed 140 times

unknown5591

Posts: 75
Joined: Mon Jun 13, 2016 8:19 pm

Re: Current Amplifier Circuit with low input current

Knowing that 3.7V is enough voltage to heat the coil I figured increasing the amps
would get the coil hot enough.

With a resistive load such as a heating coil, the current is determined by Ohm's Law: I = V/R. Use a multimeter to measure the resistance of the heating coil. Then use the multimeter to measure the battery voltage when connected to the coil (most likely, it will drop below 3.7v).

Upon further research I found a Nichrome wire calculator.

For that to be useful, you would need to know the exact length and gauge of the wire on your coils.

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Re: Current Amplifier Circuit with low input current

Previously when connecting the coil to the 150mAh battery the voltage across the coil was 3V.

The resistance across the coil alone reads 1.8 Ohm. Given the formula you provided I = V/R ; 3V/1.8 Ohm = 1.66A ~ 1660mA

Does that sound correct? Seems very high.

unknown5591

Posts: 75
Joined: Mon Jun 13, 2016 8:19 pm

Re: Current Amplifier Circuit with low input current

Not familiar with the cells you are using. But I've never seen a 150mAh cell that could deliver anywhere close to 11C of current with only a 20% voltage dip. I'd double check those measurements.

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Joined: Sat Feb 07, 2009 10:11 am

Re: Current Amplifier Circuit with low input current

Bill thank you for your patience! Bare with me. The battery has been handled quite a bit and the sn numbers are faded. It is a 3.7V 150mAh.

Fully charged the multimeter read 4.35V when the coil was connected directly to the battery the V dropped and stabilized around 3.55V.

I will also note that the Ohm setting options on my multimeter are 200 ohm, 2000 ohm, 20 kohm, 200, kohm, 2000 kohm. The 1.8 ohm reading I received was while the dial was turned to 200 ohm. Is that the appropriate setting or is my decimal in the wrong place.

I = V/R ( 3.55V/1.8ohm = 1.97A)

unknown5591

Posts: 75
Joined: Mon Jun 13, 2016 8:19 pm

Re: Current Amplifier Circuit with low input current

It is a 3.7V 150mAh.

Not all 3.7 150 mAh cells are created equal. Details of the cell design and construction will vary depending on the intended use. Cells designed for very high discharge rates tend to have electrodes with increased surface area and are constructed to minimize internal resistance.

Fully charged the multimeter read 4.35V when the coil was connected directly to the battery the V dropped and stabilized around 3.55V.

That is about an 18% voltage drop. Looking at the discharge curve charts posted earlier, that is substantially more of a drop than the 5C line on the chart. So a >10C discharge rate does not seem out of the question for the cell you are using.

I will also note that the Ohm setting options on my multimeter are 200 ohm, 2000 ohm, 20 kohm, 200, kohm, 2000 kohm. The 1.8 ohm reading I received was while the dial was turned to 200 ohm. Is that the appropriate setting or is my decimal in the wrong place.

That would be the right setting for your meter. Although you are extreme low end of the measurement range. So unless your meter has been recently calibrated, some error is possible. And even a small change in the reading could make a significant difference in the calculated current.

But one thing is certain, adding batteries in parallel won't increase the current through your coil by any more than about 18-20%.

What you need to figure out is how much current is required to get your coil hot enough. Then use Ohm's Law to figure out what voltage you need to sustain to make that happen.

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Joined: Sat Feb 07, 2009 10:11 am

Re: Current Amplifier Circuit with low input current

Not all 3.7 150 mAh cells are created equal. Details of the cell design and construction will vary depending on the intended use. Cells designed for very high discharge rates tend to have electrodes with increased surface area and are constructed to minimize internal resistance.
I never knew!

That is about an 18% voltage drop. Looking at the discharge curve charts posted earlier, that is substantially more of a drop than the 5C line on the chart. So a >10C discharge rate does not seem out of the question for the cell you are using.

I have difficulty understanding the C ratings of batteries.

That would be the right setting for your meter. Although you are extreme low end of the measurement range. So unless your meter has been recently calibrated, some error is possible. And even a small change in the reading could make a significant difference in the calculated current.

It is a relatively new meter only a few months old but yeah has limited settings.

What you need to figure out is how much current is required to get your coil hot enough. Then use Ohm's Law to figure out what voltage you need to sustain to make that happen.

I have a 30V / 10A variable power supply I can connect the coil!

unknown5591

Posts: 75
Joined: Mon Jun 13, 2016 8:19 pm

Re: Current Amplifier Circuit with low input current

I have difficulty understanding the C ratings of batteries.

A '1C' discharge rate is the rate that would theoretically discharge the battery completely in 1 hour. For a 150mAh cell, 1C would be 150mA. For a 2000mAh cell, 1C would be 2000mA.

In theory, an 'ideal' cell discharging at a 2C rate should give you 30 minutes of run time. At 4C, you would get 15 minutes. But real-world batteries are usually optimized for the discharge rates required by the intended application, so performance may suffer when operated outside the intended range.

Cells designed for devices like cell phones and cameras typically are rated for discharge rates in the 0.5C to 1C range. Cells designed for flashlights or 'power-banks' might be rated for 2C or 3C. Cells designed for extreme applications like RC race cars and planes are often rated for 30C-50C discharge rates.

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Re: Current Amplifier Circuit with low input current

Hi Bill,

I managed to get the coil connected to the power supply today.

The coil was red hot at 3.5V 1.85A.

Less hot but still red 3.0V 1.58A.

Barely red 2.8V 1.37A

So I guess the question now is how can I produce 1850mA from these 30mA batteries.

The CG-425A has a maximum continues discharge current of 60mA(2C)

unknown5591

Posts: 75
Joined: Mon Jun 13, 2016 8:19 pm

Re: Current Amplifier Circuit with low input current

So I guess the question now is how can I produce 1850mA from these 30mA batteries.

You are confusing mA and mAh again. They are 30mAh cells and are capable of putting out considerably more than 30mA - for a short time at least.

The CG-425A has a maximum continues discharge current of 60mA(2C)

How long do you need to maintain the 1850mA?