Voltage divider

[raphaelriv]

hello, i am trying to use a voltage divider to power up a radio for testing purposes, i actually thought this was to be more simple than this but my input is a 12V battery, Load Resistance 90K, I need 3amps at my output with just 8V. I tried a R1 = 10K and a R2 = 27K according to this formula things should work but they don't:

Vout = Vin*(R2//RL)/(R1+(R2//RL))

Vout = R2 * Vin / R1+R2+(R1*R2/RL)

but as soon as i hook the load the radio cuts out, not enough current nor voltage but if the calculations are right then i should and do have 8.1volts at my output with that load resistance.

 

[poopeater]

That's not the correct way to supply power to a device. Use an LM317 regulator.

 

[Nigel Goodwin]

No, you can't power something from a voltage divider, it would be VERY inefficient and require massive wirewound resistors that will get VERY hot.

Your calculations are also completely wrong - for a start, where did you imagine the 90K load resistance from? - assuming your load is constant (and it won't be) 3A at 8V is only 2.66 ohms.

As usual, give more details, we've no idea what you're trying to do, or why - or even if it might work?.

 

[raphaelriv]

that adjustable regulator also cut out when i set the output to 8volts, still what is the difference? since the outcome is the same

 

[dknguyen]

The outcome is not the same since the regulator will output a relatively constant voltage regardless of the load impedence and input voltage (both of which are NOT constant and changes during circuit operation- batteries lower in voltage as they drain, AC power supply lines fluctuate a lot, and transistors, capacitors, and other things have different impedences depending on the signals that are running through them). If your input voltage dips by a certain amount, so too will the output voltage on the divider since it is just "fractioning" the input voltage. A regulator's output will stay at a fairly constant voltage given that the input voltage is within it's range of operation. The regulator does it much more efficiently and independently of temperature than the voltage divider.

For every bit of current the load draws in a voltage divider, a bunch of current is also shorted to ground- this doesn't happen in a regulator.

 

[Nigel Goodwin]

that adjustable regulator also cut out when i set the output to 8volts, still what is the difference? since the outcome is the same

Do the maths! - it's simple enough. A potential divider should pass at least 5 times the load current, so if the load is 3A maximum, then the current down the top leg of the potential divider is 15A, and 12A down it's bottom leg. That's a massive waste of power, and you will need very low value, very high wattage, very expensive, resistors. It also won't give very good regulation - again, simple ohms law will work it out for you.

You still haven't told us what you're trying to feed?, so we might all be completely wasting our time anyway?.

 

[dknguyen]

What you are looking for is to use a zener diode with a resistor rather than a voltage divider.

Diodes basically allow current to only flow one way through them, but they can only hold back current against so much voltage, so when this voltage is exceeded the diode enters breakdown and allows current to flow backwards through it. WHen this happens the voltage across the diode is pretty much constant with changes in current (unlike a voltage divider driving a load). Zener diodes are specifically designed to work in breakdown and they have specific breakdown voltages that you use in parallel with your load to clamp the voltage down and regulate it. (If the voltage is lower than the breakdown voltage, it will not clamp it since this will not push the diode into breakdown and the diode is basically an open circuit.).

Google and read up on it. What you basically do is put a zener diode directioned to be in reverse bias with the source and to be in parallel with the load. And then you put a resistor so it is in series with the POWER SUPPLY (and the parallel combintion formed by the zener and load). The excess voltage from the zener diode clamping the voltage will appear across this resistor and be dissipated. The resistor should be sized so that if the load is removed, you won't fry the zener diode.

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/zenereg.html

 

[Nigel Goodwin]

The zener suggestion isn't much better!, just like the voltage divider you should run it at 5 times the load current, so still have 180W of heat to get rid of (assuming you can find a zener of high enough wattage?).

The answer, as already suggested, is a SERIES regulator, and not a parallel one - but it's design really depends on what it's for?, which is still apparently a secret?.

BTW, the difference is 36W maximum dissipation, as opposed to 180W all the time!.

 

[Hero999]

that adjustable regulator also cut out when i set the output to 8volts, still what is the difference? since the outcome is the same

Post your schematic.

I assume you've either cocked up or one of the componants isn't working.

 

[raphaelriv]

re:

i have this radio:

http://www.motorola.com/governmenta...388i/id_2227i/id_1832i&accessorypath=id_2359i

the point of this exercise is to power it up to just transmit and receive with a power supply set at 12V (simulating a battery) but to regulate the voltage to 8V with a current max of 3Amps. I used the LM317 regulator
(my set up was the following: **broken link removed**)
( of course setting the resistors values to give me 8V)
and it didn't seem to do its job so i went online and i looked for the most easiest way to regulate voltage but is obvious that i can't go in this direction b/c of the above mentioned, i appreciate all the input! this is a pain on my ass but thankgod to this forum

 

[Dr.EM]

Well, a LM317 can't supply 3A, so that's why it won't work if that was the reg you were using. My reccomendation, several 3A+ diodes in series with the 12v supply, add them till you have 8v at the output. Fine for testing.

EDIT: you posted as I did, sounds like you actually need a regulated supply? Use the LM338 regulator if you do.

 

[Nigel Goodwin]

Yes, the LM338 should be fine, as would a discrete regulator using something like a 2N3055 - but in either case you will need a fairly substantial heatsink to dissipate the 12W if you draw 3A. As it's only on transmit you get the high current, and transmit duration is usually much less than receive, you may be able to use a smaller heatsink, but keep an eye on it's temperature.

If you build it in a metal box, the casing may be heatsink enough?.