ElectronMan1
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Posts posted by ElectronMan1
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Thanks, that puts things in perspective - I checked out the datasheet of Energiser AAA and even if I only want 100mA at 5V (0.5W) and I draw 400mA-500mA at an average of 1.25V to get there, my mAh capacity of the AAA would only be around 450mA in total as per datasheet capacity discharge curve. That's a huge drop, as it could be as high as 1100mAh at 25mA, and 900mAh at 100mA.
I think it could be fun to build a joule thief and use it as a little night light though
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Thanks, wow I'm going to check out the MAX8815A datasheet. If I can get 500mA from a AAA battery @ 5V that'd be awesome
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Hi,
I have some AA and AAA batteries lying around and looking for some ways to use them. Like when you buy a 4 pack of batteries, and the product only needs 3. What to do with the left over one?? I try not to mix batteries so I'm looking for some ideas on how to use them.
Some ideas like a single AAA flashlight, maybe a clock? Any other ideas?
I bought a 0.9V-5V DC-DC boost converter but it sucked, could not deliver even 50-100mA without large drop in output voltage
Also is it really a big deal if I mix two brands of Alkaline batteries, with similar levels of charge??
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Thank you audioguru.
Did you derive the saturation current ratio of 1/20 from the "Conditions - Base-Emitter Saturation Voltage" - collector current 10mA and base current 0.5mA from the datasheet?
I think I understand your calculations, I have redrawn to confirm that I understand you correctly.
I would like to explore the Darlington transistor - or two BC548 arranged into a Darlington pair.
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Hi,
This will sound like a homework question. All I can say is that it isn't, and I'm not looking for any answers so I can pass any tests (those days are behind me, I hope!) For those interested I'm an IT professional dabbling in electronics.
I am designing this Dark Activated Light for fun and I'm having a problem understanding how to calculate the value of the resistor that goes into the Base of the Transistor.
The item is R2 (as below).
Some calculations below
I just made up a figure of 300uA, seemed small enough not to waste battery but large enough to turn on the transistor.
This is the part I am struggling with:
How do I calculate the value of R2, to give the right voltage drop of 0.7V when the LDR has a higher resistance.
I tried blocking off the 4k (say when dark) and focus on the VR1 + R2 path to give 0.7V -> 6V - 0.7V = 5.3V -> 5.3V / 200uA = 26500 ohms. I have 16000 ohms above, so I have about 10000 ohms which I plugged in to make up 26000-ish ohms.
However wouldn't this always switch on the transistor as the voltage drop across the parallel path will be equal?
I calculated the current through this leg of the circuit, determined the ESR to be 2857 ohms for the 4k || 10k part, which was 318uA. The LDR leg had 227uA and the Base of the transistor 91uA. I checked the BC548 data sheet but couldn't find if it's enough current to switch on the transistor.. just the voltages. https://www.fairchildsemi.com/datasheets/BC/BC547.pdf
But am I on the right track, I just starve the Base of current so it can't turn on? Technically it has the voltage but not enough current.. ? Then when the LDR gives resistance it gets the full 318uA which is enough to switch it on?
Also is this how electrical engineers do these calculations? Is there is a better way? I'm really just flying solo with these calculations, and you can probably tell
Batteriser - Evaluating their claim to get 80% more energy from batteries
in Electronics chit chat
Posted · Edited by ElectronMan1
Hi there!
I think this new product called Batteriser is a bunch of marketing hype and won't stand up to any real tests.
I would not recommend anyone purchase this product and instead wait for the reviews to come out and to be very skeptical
It's a stainless steel sleeve that goes over a battery and has a micro DC booster to step up the voltage to 1.5V to "get 80% more out of battery" - they used to say 800% longer
Their website http://batteriser.com/
I try to keep an open mind and interested in any thoughts people may have.
Dave Jones over at the EEVblog did a debunking https://www.youtube.com/watch?v=4iEshd6izgk
My thoughts:
One thing I don't know and would like to know is, would a product hit its "cut off voltage" and turn off if it dipped under that voltage of say 1.1V only for a moment? Say with a camera when it takes a flash and charges the capacity, there could be a voltage sag which takes it under 1.1V for a moment, but then would shoot back up. Would this Batteriser then be useful to stop this? With a new battery that voltage may only drop momentarily to 1.3V, but over time it drops to 1.1V for that very short period of time..
Interested in your thoughts!