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Turning 5 amps into 12 volts

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superwolf

New Member
This is a tricky one, and will make you think outside the box... ;)

I have a constant current source of 5 amps. In addition to what the 5 amps normally does, I need it to power a 12v fan. The simplest solution would be a resistor, but then you have it consuming 5*12 = 60 watts! :eek: I can handle 20 watts being wasted in addition to the 5 the fan draws, but this is excessive.

So, what would you guys suggest? Since the current through 2 parallel resistors is only dependent on their ratio, I'm considering putting two *small* resistors in parallel, and sort of siphoning off maybe 1/2 an amp with a current mirror. With a 24 ohm resistor, I can get 12 volts while dissipating much less power. Thoughts?
 

Analog

New Member
The 5A current source could be used through a switching element, creating a chopper. This chopper can then feed a transformer that gets you the voltage you want, but you'll also have to rectify it. You probably won't need much filtering, since you are only feeding a fan.
 

superwolf

New Member
In other words, create a switching power supply? mmmmmmmmm........ I'll pass. :eek:

Good idea, tho I'd rather not. KISFSM. (keep it simple for stupid me)

:p
 
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Roff

Well-Known Member
What is the voltage compliance of your current source? Where does the current go when there is no load? Or is this a homework problem?
 
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crutschow

Well-Known Member
Most Helpful Member
What does the 5 amps normally do?
 

superwolf

New Member
Are the 5 Amperes AC or DC ?

DC. (who uses AC?)

What is the voltage compliance of your current source? Where does the current go when there is no load? Or is this a homework problem?

50 volts, but 10 will be used by the load. There will always be a load when it is powered, no worries there. My homework problems were always calculations, nothing practical. That's a topic for another time.
 

crutschow

Well-Known Member
Most Helpful Member
50 volts, but 10 will be used by the load.
Here are some options:

1) Run the fan off the 10V across the load.

2) Use a switching regulator to generate 12V from the 10V across the load.

Note that any current to the fan will reduce the current to the load for options 1 and 2.

3) Design a switching circuit to operate off the current instead of voltage. You could have a MOSFET going to ground in series with the load. The MOSFET drain would be connected to the load and a diode. The other end of the diode would go to a capacitor. The MOSFET would periodically be shut off which would allow the current to flow through the diode into the storage capacitor. The MOSFET would be turned back on when the capacitor voltage reached 12V. This stored capacitor voltage would drive the fan. The size of the capacitor and switching frequency will determine the amount of ripple voltage to the fan. Since the fan is likely quite tolerant of ripple, the cap size should not be critical.

This could be done with a power MOSFET, a comparator, a power diode (preferably schottky), a reference diode, and a capacitor.

This circuit would generate a 12V ripple voltage at the frequency of the MOSFET switch, appearing at the constant current source output, so the effect of that would need to be considered. Putting an inductor in series with the load would reduce that ripple.

This option should have no significant effect on load current.
 

Willbe

New Member
The 5A current source could be used through a switching element, creating a chopper.

Yes, with a ~8% duty cycle. Use a 555 driving a low side or high side transistor.
 

superwolf

New Member
Here are some options:

1) Run the fan off the 10V across the load.

Yes, I think I rather like that option. I can put a small resistor in line to boost the voltage to 12v. Simple. Elegant.

I don't know if the reduction in current will be a problem or not, so there will only be one way to find out. If it is, I'll be back. ;)

BTW, I now realize the current mirror idea wouldn't work. You can only get out the voltage you put in. :(

Thanks for your help guys,
 
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Roff

Well-Known Member
Superwolf, I'm curious about this 5 Amp current source. Is it a real piece of hardware? A linear 5 Amp regulator with 50V compliance, into a short circuit, will be burning up at least 250 watts. With a 2 ohm load, it would dissipate 200W. Is it a switching regulator?
 

crutschow

Well-Known Member
Most Helpful Member

Willbe

New Member
Superwolf, I'm curious about this 5 Amp current source. Is it a real piece of hardware? A linear 5 Amp regulator with 50V compliance, into a short circuit, will be burning up at least 250 watts. With a 2 ohm load, it would dissipate 200W. Is it a switching regulator?

A current source in series with a short dissipates 0 watts.
 

Roff

Well-Known Member
A current source in series with a short dissipates 0 watts.
But a linear current source, in order to have 50 volt compliance, must have at least 50 volts across it when the output is shorted. I was talking about the dissipation of the source, not the load.
I'm talking about hardware here, not a mathematical current source like you find in a spice simulator.
 
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Willbe

New Member
But a linear current source, in order to have 50 volt compliance, must have at least 50 volts across it when the output is shorted.
[open]
I was talking about the dissipation of the source, not the load.
I'm talking about hardware here, not a mathematical current source like you find in a spice simulator.

Now I'm not sure.

A good hardware current source may run off of 9v but incrementally it may behave like a 50v source in series with a 10Ω resistor or a 500v source in series with a 100Ω resistor. The second has higher compliance, and, I suppose, may even run off of less than 9v.

Ideally a current source is a voltage source of infinite voltage in series with a resistor of infinitely high resistance such that V/I = 5A (in this case).

One definition of compliance is "the load current's insensitivity to the load." As the load approaches zero impedance, I guess the apparent compliance of a real world current source, even a bad one, approaches infinity.
:confused:

I could also imagine "negative compliance"; the current actually decreasing with decreasing load impedance, like a power supply with foldback current limiting. With opamps almost anything is possible.
 
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Roff

Well-Known Member
Now I'm not sure.

A good hardware current source may run off of 9v but incrementally it may behave like a 50v source in series with a 10Ω resistor or a 500v source in series with a 100Ω resistor. The second has higher compliance, and, I suppose, may even run off of less than 9v.

Ideally a current source is a voltage source of infinite voltage in series with a resistor of infinitely high resistance such that V/I = 5A (in this case).

One definition of compliance is "the load current's insensitivity to the load." As the load approaches zero impedance, I guess the apparent compliance of a real world current source, even a bad one, approaches infinity.
:confused:

I could also imagine "negative compliance"; the current actually decreasing with decreasing load impedance, like a power supply with foldback current limiting. With opamps almost anything is possible.
The reciprocal of the incremental change in current as a function of voltage is the impedance, and is ideally infinite: dv/di=∞. The compliance is simply how much voltage the current source can tolerate before it begins to limit the voltage. Above this voltage, the current will decrease as the load voltage increases, so the impedance is no longer infinity. As i said before, if a real linear (not switching) current source is to have 50V compliance, it must have more than 50V across it when the output is shorted. This means that a 5A source will dissipate more than 250W under this condition.

I wish superwolf would address the question.
 
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