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Author Topic: AC voltage divider  (Read 1041 times)
KD4FNB
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Posts: 17




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« on: November 22, 2007, 08:37:22 AM »

Hello,

I'm trying to come up with a simple voltage divider to convert from 12 VAC to 9 VAC.  The load will be a maximum of 2 amps at 9 VAC and probably much less continuous.

Now I know how to calculate the resistance values necessary to get the values I want but what I don't understand is what power rating I will need for the resistors.  It seems that the higher resistance values I use the less power handling capability they will require?  V / (R1 + R2) or something like that?

But there must be a limit somewhere, right?  Assuming some super high resistance I could use 1/4 watt resistors  to power a 200 amp load which doesn't make any sense.  I don't think I'm understanding how this works.  In the end I'm really just trying to figure out if I need some large power resistors or if I can use higher resistance 1/2 watt pieces for this particular project.
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KD4FNB
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Posts: 17




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« Reply #1 on: November 22, 2007, 08:48:12 AM »

Oh wow, after posting this I had a revelation in that the load's resistance is what affects the values I need to use for the divider.  If I use too high of resistance resistors then the loads voltage will drop significantly.

Hmmm, so in theory maybe I just need one high power handling resistor.  I guess the main problem is I don't know exactly what the load's resistance is or if I can even figure that out (it's basically a black-box complicated electronic circuit).
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N2AXZ
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Posts: 90




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« Reply #2 on: November 22, 2007, 09:08:21 AM »

A resistive divider is a very inefficient way to change AC voltages ---- a transformer is a much better device to use.  How much power are you attempting to deliver to your load?

David, N2AXZ
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N2AXZ
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Posts: 90




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« Reply #3 on: November 22, 2007, 09:16:15 AM »

I read your initial post a bit more closely ---- if you are trying to deliver 2 amps at 9 volts, this is 18 watts of power, which is considerable.  The problem with using a resistive divider is that you have absolutely no regulation --- in other words, the voltage across your load will change as the current through it changes.  The impedance of your load at maximum current draw is (9/2) =  4.5 ohms, but at lower currents the load impedance will be higher.  This load impedance appears in parallel with the shunt resistor in your voltage divider, so the division ratio of the divider (r2 / (r1 + r2)) will change with the change in load current.

David, N2AXZ
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KD4FNB
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Posts: 17




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« Reply #4 on: November 22, 2007, 09:16:51 AM »

Yes, it's inefficient but all I have is a 12 VAC transformer and so far it has been impossible to find a cheap 9 VAC transformer that will support 2000 mA.

The load is 9 VAC at 2A.  I do not know the resistance of the load.
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KD4FNB
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Posts: 17




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« Reply #5 on: November 22, 2007, 09:21:23 AM »

Thanks.  Having exactly 9 VAC isn't critical.  It just needs to be within +/- 10% or so.  Even without regulation if R1 and R2 are relatively low resistance then the output voltage will not vary much will it?
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N2AXZ
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Posts: 90




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« Reply #6 on: November 22, 2007, 09:41:35 AM »

If you use a 1 ohm resistor as the series arm of the divider, and a 9 ohm resistor as the shunt arm, at a load current of 2A you will have 9 volts across your load.  The power dissipation in the divider will be 9 watts in the series arm and 9 watts in the shunt arm.

Under no load conditions (where your load draws no current), the voltage at the output of your divider will be 10.8 volts.

Hope this helps!

David, N2AXZ
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N2AXZ
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Posts: 90




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« Reply #7 on: November 22, 2007, 09:46:29 AM »

An important caveat to the solution I gave you ---- at your maximum load current your 12 volt transformer will be supplying 3A.

David, N2AXZ
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KD4FNB
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Posts: 17




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« Reply #8 on: November 23, 2007, 07:47:11 PM »

Thanks!  Are you sure about the 9 watts in each resistor even though they are different values?

I would think the wattage in the arm is going to be the full load or 18 watts.
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W5RKL
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Posts: 888




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« Reply #9 on: December 01, 2007, 09:08:23 AM »


A voltage divider network with 2 resistors in series
to ground, R1 and R2, a source voltage of 12 volts
and a current of 2 amps.

In order to arrive at the target voltage of 9 volts,
R1 must drop "3 volts".

Therefore, R1 = 12 - 9 / 2 = 1.5 ohms. There is 2 amps
of current flowing through R1, therefore, R1's wattage
is P = 2 x 3 = 6 watts.

To determine the resistance of R2, we must know the
voltage and current flowing through it which is
9 volts at 2 amps.

Therefore, R2 = 9 / 2 = 4.5 ohms. Power consumed
through R2 is P = 2 x 9 = 18 watts.

The difference in wattage is based on the voltage drop
across each resistor, 3 volts on R1 and 9 volts on R2
and 2 amps of current through each resistor. There
is 2 amps of current flow at the junction of R1 and R2,
therefore the same current flows through R1 as it
does through R2.

73's
Mike
W5RKL
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W5RKL
Member

Posts: 888




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« Reply #10 on: December 01, 2007, 09:22:03 AM »

In low current circuits of a few milliamps, a
voltage divider works rather well. However, in
high current demand circuits, voltage dividers
are a huge waste of power. Higher current demands
put on the divider network can cause reduction in
the target voltage and possibly damage to components.
Therefore, I agree, it would be much wiser to use
a transformer that is rated for the target voltage
and current rating.

73's
Mike
W5RKL
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KD4FNB
Member

Posts: 17




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« Reply #11 on: December 01, 2007, 07:03:34 PM »

Thanks for the help.  I managed to get it working long enough for my tests.  Turns out the continuous load was about 1.4 amps at 9 volts.

In the midst of all this I also managed to find a 9 volt transformer rated at 2 amps so I should be good to go.
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HA6SST
Member

Posts: 110




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« Reply #12 on: December 04, 2007, 01:07:30 PM »

Grab an old mains transformer and strip it down. Then you wind a new transformer with 240 turns on the primary and 180 turns on the secondary. Use slightly heavier wire for the secondary.

HA6SST
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K4DPK
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Posts: 1077


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« Reply #13 on: January 04, 2008, 04:12:49 PM »

Is the load going to vary during operation or not?

If the load is constant, then a simple series resistor of the value (3 / steadystate current amperes) is all that is needed and regulation isn't necessary.

The wattage of the resistor should be (steadystate current squared X Ohms x 2) for safety margin.

73
Phil C. Sr.
k4dpk
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