- #1
Dixo
- 18
- 1
Hi all
I have this Maxitonix 500-1 lab thing which I bought years ago and never got around to doing anything with it until this year.
Now it has a breadboard and along the top, six different supply rails labelled V1 - V6. Each ones goes up in 1.5V increments from 1.5V to 9V.
I'm just curious how this is done? In the battery compartment, there are 6 AAs and while they are paired (eg 3 pairs connected) that wouldn't explain how it could use a voltage divider with an unknown current consumption.
So, any ideas please? I'm just wondering how in the real practical world, you would achieve this when you have no idea what currents might be used in a circuit (eg, how do power supplies achieve this?).
Initially I figured that it must be a potential divider but I had a hunch it couldn't be so I put that to the test. I drew up a simple 3 resistors in series with a 12V supply and wanting 6V and 3V. With 500mA I calculated I'd need 12Ω, 6Ω and 6Ω. Of course the moment you change that current to more or less the voltage figures change so that doesn't work!
So how do they do it?
Many thanks!
Dixo
I have this Maxitonix 500-1 lab thing which I bought years ago and never got around to doing anything with it until this year.
Now it has a breadboard and along the top, six different supply rails labelled V1 - V6. Each ones goes up in 1.5V increments from 1.5V to 9V.
I'm just curious how this is done? In the battery compartment, there are 6 AAs and while they are paired (eg 3 pairs connected) that wouldn't explain how it could use a voltage divider with an unknown current consumption.
So, any ideas please? I'm just wondering how in the real practical world, you would achieve this when you have no idea what currents might be used in a circuit (eg, how do power supplies achieve this?).
Initially I figured that it must be a potential divider but I had a hunch it couldn't be so I put that to the test. I drew up a simple 3 resistors in series with a 12V supply and wanting 6V and 3V. With 500mA I calculated I'd need 12Ω, 6Ω and 6Ω. Of course the moment you change that current to more or less the voltage figures change so that doesn't work!
So how do they do it?
Many thanks!
Dixo
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