Constant-Current-Load

Electronics and electrotechnical stuff
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Orngrimm
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Constant-Current-Load

Post by Orngrimm »

I just wanted to drop a short post on how to create a super simple constant current-load.
If you google it, you often find a schematic which works but is a bit too complicated if you think about it:
A closely defined voltage feeds a OpAmp and this opamp feeds a FET or BJT to load a resistor with current. The current thru the resistor generates a voltage which is taken back to the OpAmp. This ensures the voltage over the resistor is = the voltage supplied to the other imput of the OpAmp.
Something like Image
Granted, this approach is nice if you want a variable constant current-load as you can just adjust the voltage supplied to the OpAmp and Bob's your uncle.

But i find myself often in the situation to just have a given current and i have to hold that for until the supply is exhausted.

Now, if we look at the basic function of the schematic i explained above:
- Take supply-voltage and drop it down
- create a constant voltage at the output
- to create a constant current over a constant reistor
this is awfully close to what a LDO (Low Dropout Regulator) does! Basically the complete regulation and feedback and FET is within the LDO already and ready to use without any adjustments needed

So basically my goto-Solution is:
- Pick a super efficient LDO with like 25-100nA quiescent current to minimise errors
- The LDO should have a fixed output-voltage, again, to make life easyer and more precise
- Feed the LDO your source to be measured (Like a small LiPo or one of your scavenged 18650)
- Add a resistor to the output which generated the desired current
- Formula for current: Uldo / Rload


Example:
This is especially handy as very often in your simple projects, you have a LDO anyway to take down the voltage (lets say of 2S 18650 = 7-8V) to 5V or 3.3V ans therefore exactly the usecase we also measure.
LDO = TPS7A0320PDQNR with 2V Output, 25nA Iq and 200mA Iout
R = 100 Ohm for 20mA accoding to Ohm's law I = U / R = 2V / 100 Ohm = 0.02A = 20mA
Now, as soon as you connect the battery, the LDO drops down the voltage of your single cell to 2V and loads it with 100 Ohms = 20mA. A normal load-level for an ESP32 and thelike.
Now you simply time how long the battery can sustain the 20mA till it is dead and needs to be recharged and you have the mAh of the specific battery.

Just as a heads up as i figured, a lot of people dont know this handy trick to make a constant current-load...
Builder of stuff, creator of things, inventor of many and master of none.
Tinkerer by heart, archer by choice and electronics engineer by trade.
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Orngrimm
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Re: Constant-Current-Load

Post by Orngrimm »

https://www.instructables.com/DIY-BATTE ... ITY-TESTER is such an example where the schematic could be simplified by this quite substantially...
Builder of stuff, creator of things, inventor of many and master of none.
Tinkerer by heart, archer by choice and electronics engineer by trade.
Downunder35m
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Re: Constant-Current-Load

Post by Downunder35m »

Interesting...
I wonder if the chinaman is doing exactly that in their cheap load testers ?
For under 20 bucks and with the low part count that must doing at least something very similar.
Exploring the works of the old inventors, mixng them up with a modern touch.
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