M
Michael A. Terrell
Guest
Tim Wescott wrote:
At least, and that blows the budget by 50%.
At least, and that blows the budget by 50%.
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Jasen Betts
Guest
On 2014-01-24, John Fields <jfields@austininstruments.com> wrote:
Uh, you seem to be answering some question that wasn't asked.
Won't the 5V 2A wall-wart will have a similar sized core to the 12V 1A
unit just thicker wire on the secondary and fewer turns, and a chunkier
rectifier?
--
For a good time: install ntp
--- news://freenews.netfront.net/ - complaints: news@netfront.net ---
On Fri, 24 Jan 2014 14:41:34 +0100, "petrus bitbyter"
petrus.bitbyter@hotmail.com> wrote:
"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
news:y82Eu.340472$tR7.283464@fx22.iad...
I was wondering about current and buck converters.
For my question, I'd like to assume a 100% efficiency. If I have a 10v
input into a buck converter, and 5v output. If the load draws 2amps, is
the buck converter only drawing 1amp from the 10v? That would make sense
to me, because its the same amount of power.
If that is the case, then what about if the efficiency drops to %80?
5v*2amps = 10watts, 10w on the output, 10/.8=12.5watts from the input,
means 12.5watts/10 = 1.25amps.
The reason I ask, is that I have a wall-wart that will provide me 12v at
1amp, but I need at least a little more amps at 5v. So if I have a
buck-converter to drop the voltage down to 5v, and its 85% efficient,
should I be able to safely draw 2amps?
Thanks,
Daniel.
Theoretically: Yes. But I consider it bad practice. You'd go for a 5V/2A
wall-wart. More reliable, more efficient and cheaper. Or comes your
buckconverter for free?
petrus bitbyter
---
More reliable?
Perhaps.
More efficient?
Apples for apples, the core of the 2 amp unit will be more
voluminous and will require a larger magnetizing current than the 1
amp unit so, for the same power out as the 1 amp unit, will require
greater power in.
Cheaper?
For minimal cost of building either, the larger unit will require a
larger core, larger wire on the primary and secondary, larger
diodes, and a larger cap on the output.
Uh, you seem to be answering some question that wasn't asked.
Won't the 5V 2A wall-wart will have a similar sized core to the 12V 1A
unit just thicker wire on the secondary and fewer turns, and a chunkier
rectifier?
--
For a good time: install ntp
--- news://freenews.netfront.net/ - complaints: news@netfront.net ---
P
petrus bitbyter
Guest
"John Fields" <jfields@austininstruments.com> schreef in bericht
news:3kq5e9h94co5fjgvga91ufd6dmqcf8ac93@4ax.com...
Well, a wall-wart followed by another buck converter seems likely less
reliable then one converter.
Each converter will have its losses. Two likely more then one.
Two converters will likely cost more then one. Unless you don't have to pay
for one of the two.
Apples for apples indeed. You'll have to compare converters of (about) the
same quality.
petrus bitbyter
news:3kq5e9h94co5fjgvga91ufd6dmqcf8ac93@4ax.com...
On Fri, 24 Jan 2014 14:41:34 +0100, "petrus bitbyter"
petrus.bitbyter@hotmail.com> wrote:
"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
news:y82Eu.340472$tR7.283464@fx22.iad...
I was wondering about current and buck converters.
For my question, I'd like to assume a 100% efficiency. If I have a 10v
input into a buck converter, and 5v output. If the load draws 2amps,
is
the buck converter only drawing 1amp from the 10v? That would make sense
to me, because its the same amount of power.
If that is the case, then what about if the efficiency drops to %80?
5v*2amps = 10watts, 10w on the output, 10/.8=12.5watts from the input,
means 12.5watts/10 = 1.25amps.
The reason I ask, is that I have a wall-wart that will provide me 12v at
1amp, but I need at least a little more amps at 5v. So if I have a
buck-converter to drop the voltage down to 5v, and its 85% efficient,
should I be able to safely draw 2amps?
Thanks,
Daniel.
Theoretically: Yes. But I consider it bad practice. You'd go for a 5V/2A
wall-wart. More reliable, more efficient and cheaper. Or comes your
buckconverter for free?
petrus bitbyter
---
More reliable?
Perhaps.
More efficient?
Apples for apples, the core of the 2 amp unit will be more
voluminous and will require a larger magnetizing current than the 1
amp unit so, for the same power out as the 1 amp unit, will require
greater power in.
Cheaper?
For minimal cost of building either, the larger unit will require a
larger core, larger wire on the primary and secondary, larger
diodes, and a larger cap on the output.
Well, a wall-wart followed by another buck converter seems likely less
reliable then one converter.
Each converter will have its losses. Two likely more then one.
Two converters will likely cost more then one. Unless you don't have to pay
for one of the two.
Apples for apples indeed. You'll have to compare converters of (about) the
same quality.
petrus bitbyter
M
Maynard A. Philbrook Jr.
Guest
In article <y-CdnSW9MN5Y237PnZ2dnUVZ_v6dnZ2d@giganews.com>,
tim@seemywebsite.please says...
The price of bear nuts are about $1.29 a bag.
Deer nuts are just under a buck!
Jamie
tim@seemywebsite.please says...
On Fri, 24 Jan 2014 20:13:36 -0500, Michael A. Terrell wrote:
Tim Wescott wrote:
Or you just need to put an adequate input capacitor on your buck
converter!
That would raise the cost to over a buck. ;-)
A buck-and-a-half converter?
The price of bear nuts are about $1.29 a bag.
Deer nuts are just under a buck!
Jamie
M
Michael A. Terrell
Guest
"Maynard A. Philbrook Jr." wrote:
You are a Buckeye, AKA: A useless nut.
In article <y-CdnSW9MN5Y237PnZ2dnUVZ_v6dnZ2d@giganews.com>,
tim@seemywebsite.please says...
On Fri, 24 Jan 2014 20:13:36 -0500, Michael A. Terrell wrote:
Tim Wescott wrote:
Or you just need to put an adequate input capacitor on your buck
converter!
That would raise the cost to over a buck. ;-)
A buck-and-a-half converter?
The price of bear nuts are about $1.29 a bag.
Deer nuts are just under a buck!
You are a Buckeye, AKA: A useless nut.
J
John Fields
Guest
On Sat, 25 Jan 2014 14:49:49 +0100, "petrus bitbyter"
<petrus.bitbyter@hotmail.com> wrote:
---
Yes, but according to the OP's post, he already has the AC to DC
mains converter and wants to connect a buck converter to its output
to get to 5V, therefore the reliability of the AC to DC converter is
moot.
---
>Each converter will have its losses. Two likely more then one.
---
Depends.
It may be that the losses in the 24 watt wall-wart added to the 10
watt bucks's may be more than the losses in the 12 watt wall-wart
added to the same buck.
---
---
True, and he already has the AC to DC converter on hand.
---
---
But of different size.
<petrus.bitbyter@hotmail.com> wrote:
"John Fields" <jfields@austininstruments.com> schreef in bericht
news:3kq5e9h94co5fjgvga91ufd6dmqcf8ac93@4ax.com...
On Fri, 24 Jan 2014 14:41:34 +0100, "petrus bitbyter"
petrus.bitbyter@hotmail.com> wrote:
"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
news:y82Eu.340472$tR7.283464@fx22.iad...
I was wondering about current and buck converters.
For my question, I'd like to assume a 100% efficiency. If I have a 10v
input into a buck converter, and 5v output. If the load draws 2amps,
is
the buck converter only drawing 1amp from the 10v? That would make sense
to me, because its the same amount of power.
If that is the case, then what about if the efficiency drops to %80?
5v*2amps = 10watts, 10w on the output, 10/.8=12.5watts from the input,
means 12.5watts/10 = 1.25amps.
The reason I ask, is that I have a wall-wart that will provide me 12v at
1amp, but I need at least a little more amps at 5v. So if I have a
buck-converter to drop the voltage down to 5v, and its 85% efficient,
should I be able to safely draw 2amps?
Thanks,
Daniel.
Theoretically: Yes. But I consider it bad practice. You'd go for a 5V/2A
wall-wart. More reliable, more efficient and cheaper. Or comes your
buckconverter for free?
petrus bitbyter
---
More reliable?
Perhaps.
More efficient?
Apples for apples, the core of the 2 amp unit will be more
voluminous and will require a larger magnetizing current than the 1
amp unit so, for the same power out as the 1 amp unit, will require
greater power in.
Cheaper?
For minimal cost of building either, the larger unit will require a
larger core, larger wire on the primary and secondary, larger
diodes, and a larger cap on the output.
Well, a wall-wart followed by another buck converter seems likely less
reliable then one converter.
---
Yes, but according to the OP's post, he already has the AC to DC
mains converter and wants to connect a buck converter to its output
to get to 5V, therefore the reliability of the AC to DC converter is
moot.
---
>Each converter will have its losses. Two likely more then one.
---
Depends.
It may be that the losses in the 24 watt wall-wart added to the 10
watt bucks's may be more than the losses in the 12 watt wall-wart
added to the same buck.
---
Two converters will likely cost more then one. Unless you don't have to pay
for one of the two.
---
True, and he already has the AC to DC converter on hand.
---
Apples for apples indeed. You'll have to compare converters of (about) the
same quality.
---
But of different size.
J
John Larkin
Guest
On Wed, 22 Jan 2014 21:54:42 -0800, Daniel Pitts
<newsgroup.nospam@virtualinfinity.net> wrote:
Sounds risky.
A buck switcher has a negative input impedance; the lower the input voltage, the
higher the input current. So if the 12 volt wart current-limits at 1 amp, and
its output is at some instant, say, 8 volts, the input current will be about 1.5
amps. If it current limits there, it can't pull its output up to 12.
It's a little tricky, driving a switcher from a switcher, or from any
current-limited supply without a lot of margin. We make sure our gear has
soft-start, so external warts don't bog down at startup.
Some switching warts current limit and some shut down and retry later when
overloaded. Either can fail to pull up a load that they could handle
steady-state.
And don't believe the nameplate currents, especially if the warts are made in
C****.
--
John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com
Precision electronic instrumentation
<newsgroup.nospam@virtualinfinity.net> wrote:
I was wondering about current and buck converters.
For my question, I'd like to assume a 100% efficiency. If I have a 10v
input into a buck converter, and 5v output. If the load draws 2amps,
is the buck converter only drawing 1amp from the 10v? That would make
sense to me, because its the same amount of power.
If that is the case, then what about if the efficiency drops to %80?
5v*2amps = 10watts, 10w on the output, 10/.8=12.5watts from the input,
means 12.5watts/10 = 1.25amps.
The reason I ask, is that I have a wall-wart that will provide me 12v at
1amp, but I need at least a little more amps at 5v. So if I have a
buck-converter to drop the voltage down to 5v, and its 85% efficient,
should I be able to safely draw 2amps?
Thanks,
Daniel.
Sounds risky.
A buck switcher has a negative input impedance; the lower the input voltage, the
higher the input current. So if the 12 volt wart current-limits at 1 amp, and
its output is at some instant, say, 8 volts, the input current will be about 1.5
amps. If it current limits there, it can't pull its output up to 12.
It's a little tricky, driving a switcher from a switcher, or from any
current-limited supply without a lot of margin. We make sure our gear has
soft-start, so external warts don't bog down at startup.
Some switching warts current limit and some shut down and retry later when
overloaded. Either can fail to pull up a load that they could handle
steady-state.
And don't believe the nameplate currents, especially if the warts are made in
C****.
--
John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com
Precision electronic instrumentation