I need to bring 12vdc down to 10v ..

[steveo]

Hey everyone,

I was wondering if someone could give me a simply way to calculate what resistor to use to limit my voltage.

I have a 12v power source & I have a 12v 1amp load i would like to limit the voltage to about 10v...

would i be limiting current with a resistor? will it bring the voltage to 10v?

thanks
-steveo

 

[liveforphysics]

3 1amp diodes in series would drop it by 2.1v, and it wouldn't be current dependent like a resistor solution, which would call for a 2ohm 2w resistor.

Just by chance, if this is for running a string of 3 LEDs, it's going to be a sad fail.

 

[icecube57]

Its funny that you mention leds. I have some cheap bell lights. Ive got them on my dc to dc convertor. They run fine at 12v (2) lights in series but the bulbs break easy. I bought some led replacement bulbs. They come on for a fraction of a second and cut off. I have two lights in series. They should be seing 6v a piece. this should be enough fro the led. They they say these are good up to 6d cells. Ive even tried running them at 3 and 12v. they come on for a fraction of a second.

 

[steveo]

3 1amp diodes in series would drop it by 2.1v, and it wouldn't be current dependent like a resistor solution, which would call for a 2ohm 2w resistor.

Just by chance, if this is for running a string of 3 LEDs, it's going to be a sad fail.

I'm actually trying to reduce voltage/current to a set of 12v .40amp fans to reduce there rpm...

-steveo

 

[liveforphysics]

3 1amp diodes in series would drop it by 2.1v, and it wouldn't be current dependent like a resistor solution, which would call for a 2ohm 2w resistor.

Just by chance, if this is for running a string of 3 LEDs, it's going to be a sad fail.

I'm actually trying to reduce voltage/current to a set of 12v .40amp fans to reduce there rpm...

-steveo

Then 3 1-amp diodes in series will be the optimal method.

 

[full-throttle]

Luke, diodes in parallel won't work (unless they're shottkys) - the one with the lowest Vf will warm-up more than others, decrease its Vf even further and so on.

 

[liveforphysics]

Luke, diodes in parallel won't work (unless they're shottkys) - the one with the lowest Vf will warm-up more than others, decrease its Vf even further and so on.

Series man. Series. Not parallel.

 

[full-throttle]

Righto, must've been a long day..

 

[liveforphysics]

Righto, must've been a long day..

It happens to all of us.

 

[swbluto]

Luke, diodes in parallel won't work (unless they're shottkys)

Schottky's have lower Vf's as they get warmer. They'd also suffer from paralleling due to "current hogging" and thermal runaway, right? Or do I have it wrong(Provide a reference if so)?

 

[steveo]

so what should i use? ... lol .. this is becoming an interesting thread...


2 ohms ... 2w resistor?

thanks
-steveo

 

[liveforphysics]

so what should i use? ... lol .. this is becoming an interesting thread...


2 ohms ... 2w resistor?

thanks
-steveo

No. Resistor is a fail, because your 1amp load at 12v might only be a 0.5amp load at 10v. This is because fan speed power requirements change at the square of RPM. If you were driving a constant current load, than a 2ohm resistor inline would drop it exactly 2v. But you're not.

Use the 3 diodes in series. It's like $0.05 in diodes, and it won't matter if you're pulling 0.01amps or 2amps, it will give roughly the same 2.1v drop you're looking for.

 

[swbluto]

Use the 3 diodes in series. It's like $0.05 in diodes, and it won't matter if you're pulling 0.01amps or 2amps, it will give roughly the same 2.1v drop you're looking for.

Just curious, where are you getting your diodes? I'm looking for diodes to drop 2W each and it seems they typically don't have that high of a power rating without getting past the nickel territory. (2amps*.7volts = 1.4W)

 

[steveo]

so what should i use? ... lol .. this is becoming an interesting thread...


2 ohms ... 2w resistor?

thanks
-steveo

No. Resistor is a fail, because your 1amp load at 12v might only be a 0.5amp load at 10v. This is because fan speed power requirements change at the square of RPM. If you were driving a constant current load, than a 2ohm resistor inline would drop it exactly 2v. But you're not.

Use the 3 diodes in series. It's like $0.05 in diodes, and it won't matter if you're pulling 0.01amps or 2amps, it will give roughly the same 2.1v drop you're looking for.

I don't understand how a diode can reduce voltage though... doesn't a diode push current one way only? .. i didn't thing it can limit voltage .. but voltage drop is i guess what happens when you put 3 diodes in series...

-steveo

 

[liveforphysics]

Use the 3 diodes in series. It's like $0.05 in diodes, and it won't matter if you're pulling 0.01amps or 2amps, it will give roughly the same 2.1v drop you're looking for.

Just curious, where are you getting your diodes? I'm looking for diodes to drop 2W each and it seems they typically don't have that high of a power rating without getting past the nickel territory. (2amps*.7volts = 1.4W)

Holy retard morning.

1amp diodes. Like the kind you get in a 20pack at radio shack for $0.99. I think they call them 4001N, or N4001 if my memory serves me.
0.7v drop across each diode. 1 amp. 700mw per diode... Things in series share the heat between them, if each diode released 2.1w of heat, you would be an instant billionare for beating the law of conservation of mass and energy, and giving the world unlimited clean energy.

 

[steveo]

Use the 3 diodes in series. It's like $0.05 in diodes, and it won't matter if you're pulling 0.01amps or 2amps, it will give roughly the same 2.1v drop you're looking for.

Just curious, where are you getting your diodes? I'm looking for diodes to drop 2W each and it seems they typically don't have that high of a power rating without getting past the nickel territory. (2amps*.7volts = 1.4W)

Holy retard morning.

1amp diodes. Like the kind you get in a 20pack at radio shack for $0.99. I think they call them 4001N, or N4001 if my memory serves me.
0.7v drop across each diode. 1 amp. 700mw per diode... Things in series share the heat between them, if each diode released 2.1w of heat, you would be an instant billionare for beating the law of conservation of mass and energy, and giving the world unlimited clean energy.

the 3 diodes worked great!!

thanks
-steveo

 

[liveforphysics]

Great Steve Now you've got a new trick up your sleeve for working with low power electronics.

 

[Nimbuzz]

Is this a similar but higher voltage situation? I have a controller that is designed to not start if the voltage is over 47.5V however the 48 V pack off the charger is 51V. Can I limit this voltage to not go above 47.5 in a similar manner that you just did the 12V to 10 V with diodes?

2 -- PS; A question in this thread that seemed to go unanswered << I don't understand how a diode can reduce voltage though... doesn't a diode push current one way only? .. i didn't thing it can limit voltage .. but voltage drop is i guess what happens when you put 3 diodes in series... steveo -- >>

Thanks for info

 

[Gregory]

Thread jack, I have the same question with a Dc-DC converter that says absolute max voltage is 80V (that outputs 3-5V, at a 6 to 9 amps)
My pack is 82V off the charger.
I would like to use it to power a LED flashlight mounted on the handlebar.

Looking around it looks like the diode 0.7V drop is only good at low to modest amp ratings. Anyone know if I can use the same trick for 5 to 10 amps?
And 10A06, 10A10 which are 600V or 1000V and 10Amps and still seem to have a 1V drop?

Thanks!
Greg

 

[BiGH]

2 -- PS; A question in this thread that seemed to go unanswered << I don't understand how a diode can reduce voltage though... doesn't a diode push current one way only? .. i didn't thing it can limit voltage .. but voltage drop is i guess what happens when you put 3 diodes in series... steveo -- >>

Thanks for info

I have a limited understanding of this, but here's my guess: Each component in a circuit uses some electricity, each more than others (internal resistance). By using the laws of conservation of energy some of the energy passed through the diode gets turned into heat etc. This is what causes the voltage drop.

http://en.wikipedia.org/wiki/Voltage_drop

i'm probably wrong, but feel free to correct me. I'm starting to get very interested in electronics, especially as I'll be building my own controller for the car soon .

 

[swbluto]

2 -- PS; A question in this thread that seemed to go unanswered << I don't understand how a diode can reduce voltage though... doesn't a diode push current one way only? .. i didn't thing it can limit voltage .. but voltage drop is i guess what happens when you put 3 diodes in series... steveo -- >>

Thanks for info

I have a limited understanding of this, but here's my guess: Each component in a circuit uses some electricity, each more than others (internal resistance). By using the laws of conservation of energy some of the energy passed through the diode gets turned into heat etc. This is what causes the voltage drop.

http://en.wikipedia.org/wiki/Voltage_drop

i'm probably wrong, but feel free to correct me. I'm starting to get very interested in electronics, especially as I'll be building my own controller for the car soon .

Or is it the voltage drop and current that causes the heat? :wink:

My understanding of the diode is that it loses voltage due to a PN junction with a "band-gap". For electrons to cross this band gap requires at least however many volts it takes to jump it. Once they jump it, they lose that voltage on average, and that voltage and current determines the power loss which is basically heat. Now, it's true that diodes have a resistance, but the voltage drop is usually determined by the bandgap voltage decrease, so most of the heat comes from this and not its "resistance". Diodes are funny creatures in that they aren't ohmic: voltage drop doesn't depend on current and some fairly static resistance, it depends on current and whatever voltage they happen to drop at that given current(And the temperature of the diode).

My understanding is probably a little simplified, though. It doesn't explain why the voltage drop is lower at really low currents.

 

[liveforphysics]

2 -- PS; A question in this thread that seemed to go unanswered << I don't understand how a diode can reduce voltage though... doesn't a diode push current one way only? .. i didn't thing it can limit voltage .. but voltage drop is i guess what happens when you put 3 diodes in series... steveo -- >>

Thanks for info

I have a limited understanding of this, but here's my guess: Each component in a circuit uses some electricity, each more than others (internal resistance). By using the laws of conservation of energy some of the energy passed through the diode gets turned into heat etc. This is what causes the voltage drop.

http://en.wikipedia.org/wiki/Voltage_drop

i'm probably wrong, but feel free to correct me. I'm starting to get very interested in electronics, especially as I'll be building my own controller for the car soon .

It's not the resistance at all, and anything resistance related would vary linearly with the amount of current. Once you're passing the needed current level to reach the Vf of the diode (0.7v in this case), it holds that 0.7v drop roughly fixed solid unless you drastically over current the device.

A non-technical way to think about it is like this.

A resistance loss is kinda like an income tax. You lose an amount of your money (voltage) based on the amount of current you're passing. You pass 1amp through something with 1ohm of resistance, you loose 1volt. You pass 10amp, you loose 10volts. You pass 100amps you loose 100v. etc etc

A diode Vf drop is more like a toll booth on a road. If you're driving a $400 geo metro through, you pay your $0.70 toll. If you're driving your 1.5million dollar Bugatti Veyron through the toll booth, you still just pay your $0.7 toll. As long as you don't exceed the current limitations of the diode, you always just pay the 0.7v drop, and it doesn't matter if you're passing 1amp or 100amps (as long as the diode is rated to handle 100amps of course).

 

[BiGH]

2 -- PS; A question in this thread that seemed to go unanswered << I don't understand how a diode can reduce voltage though... doesn't a diode push current one way only? .. i didn't thing it can limit voltage .. but voltage drop is i guess what happens when you put 3 diodes in series... steveo -- >>

Thanks for info

I have a limited understanding of this, but here's my guess: Each component in a circuit uses some electricity, each more than others (internal resistance). By using the laws of conservation of energy some of the energy passed through the diode gets turned into heat etc. This is what causes the voltage drop.

http://en.wikipedia.org/wiki/Voltage_drop

i'm probably wrong, but feel free to correct me. I'm starting to get very interested in electronics, especially as I'll be building my own controller for the car soon .

It's not the resistance at all, and anything resistance related would vary linearly with the amount of current. Once you're passing the needed current level to reach the Vf of the diode (0.7v in this case), it holds that 0.7v drop roughly fixed solid unless you drastically over current the device.

A non-technical way to think about it is like this.

A resistance loss is kinda like an income tax. You lose an amount of your money (voltage) based on the amount of current you're passing. You pass 1amp through something with 1ohm of resistance, you loose 1volt. You pass 10amp, you loose 10volts. You pass 100amps you loose 100v. etc etc

A diode Vf drop is more like a toll booth on a road. If you're driving a $400 geo metro through, you pay your $0.70 toll. If you're driving your 1.5million dollar Bugatti Veyron through the toll booth, you still just pay your $0.7 toll. As long as you don't exceed the current limitations of the diode, you always just pay the 0.7v drop, and it doesn't matter if you're passing 1amp or 100amps (as long as the diode is rated to handle 100amps of course).

liveforphysics, swbluto - thanks for the excellent explainations that analogy is perfect

 

[full-throttle]

Nice analogy Luke, however Bugatti pays a higher 'toll' and the 'toll' decreases with temperature


swbluto, I couldn't find any trustworthy reference on why Schottky diodes share the current better than plain rectifiers, perhaps its a myth.
The only time I've seen it being used is with 6CWQ04FNPbF, but its more likely to do with single substrate than Schottky.

BTW In GaAs diodes (LEDs) the energy required to jump the band-gap is emitted as photons and not just heat.

 

[syntax_x]

Stevo, Im assuming that you didnt want to use the 7v power source in your pc for fear of feeding back to much power to the psu ( the difference between the 5 and 12v lines)

Im making the same set up but with switches to control speeds


Luke i had a few quick questions regarding diodes.

1. do they have a polarity?
2. is there a single diode that will do the job of the 2ish volt step down, or the 3 diodes are the way to go?
3. Do they heat up much?

Im choosing this method over the Pot and MOSFET method because i want to keep heat in the case to a bare minimum, and the chance of fires even lower, Which is why i asked if the diodes produce much heat.