Buck converter input current vs output current.

D

Daniel Pitts

Guest
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.
 
On Wed, 22 Jan 2014 21:54:42 -0800, Daniel Pitts
<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?
Click to expand...

---
Yes.

If you have a wall-wart that'll put out 12V at 1 amp, that's 12
watts.

If you have a buck converter with a 5V output that can push 2 amps
through a [2.5 ohm] load, that's 10 watts.

Then, since the efficiency of the buck converter is 85%, and

Pout
e = ------,
Pin

rearrange to solve for Pin, and the required power input to the buck
converter will be:

Pout 10W
Pin = ------ = ------ = 11.76 watts
e 0.85

So you're there with about a quarter of a watt to spare.
 
On Thu, 23 Jan 2014 02:15:35 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 22 Jan 2014 21:54:42 -0800, Daniel Pitts
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?

---
Yes.

If you have a wall-wart that'll put out 12V at 1 amp, that's 12
watts.

If you have a buck converter with a 5V output that can push 2 amps
through a [2.5 ohm] load, that's 10 watts.

Then, since the efficiency of the buck converter is 85%, and

Pout
e = ------,
Pin

rearrange to solve for Pin, and the required power input to the buck
converter will be:

Pout 10W
Pin = ------ = ------ = 11.76 watts
e 0.85

So you're there with about a quarter of a watt to spare.
Click to expand...

In other words, not enough margin to guarantee it'll work.

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On 1/23/14 7:02 AM, Jim Thompson wrote:
On Thu, 23 Jan 2014 02:15:35 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 22 Jan 2014 21:54:42 -0800, Daniel Pitts
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?

---
Yes.

If you have a wall-wart that'll put out 12V at 1 amp, that's 12
watts.

If you have a buck converter with a 5V output that can push 2 amps
through a [2.5 ohm] load, that's 10 watts.

Then, since the efficiency of the buck converter is 85%, and

Pout
e = ------,
Pin

rearrange to solve for Pin, and the required power input to the buck
converter will be:

Pout 10W
Pin = ------ = ------ = 11.76 watts
e 0.85

So you're there with about a quarter of a watt to spare.



In other words, not enough margin to guarantee it'll work.
Click to expand...

I agree with that. In my real application, the 2amps is
worst-case-scenario of my MCU shutting-down while leaving my
constant-current chips on full.

I was just more curious in general about the current effects of buck
boost. It seems obvious now that I worked it out, but nothing I read
said it explicitly.
 
On Thu, 23 Jan 2014 08:46:29 -0800, Daniel Pitts
<newsgroup.nospam@virtualinfinity.net> wrote:

On 1/23/14 7:02 AM, Jim Thompson wrote:
On Thu, 23 Jan 2014 02:15:35 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 22 Jan 2014 21:54:42 -0800, Daniel Pitts
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?

---
Yes.

If you have a wall-wart that'll put out 12V at 1 amp, that's 12
watts.

If you have a buck converter with a 5V output that can push 2 amps
through a [2.5 ohm] load, that's 10 watts.

Then, since the efficiency of the buck converter is 85%, and

Pout
e = ------,
Pin

rearrange to solve for Pin, and the required power input to the buck
converter will be:

Pout 10W
Pin = ------ = ------ = 11.76 watts
e 0.85

So you're there with about a quarter of a watt to spare.



In other words, not enough margin to guarantee it'll work.

I agree with that. In my real application, the 2amps is
worst-case-scenario of my MCU shutting-down while leaving my
constant-current chips on full.

I was just more curious in general about the current effects of buck
boost. It seems obvious now that I worked it out, but nothing I read
said it explicitly.
Click to expand...

Maybe dodge that full-on scenario with some extra logic?

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On Wed, 22 Jan 2014 21:54:42 -0800, Daniel Pitts 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?
Click to expand...

2A / (0.85) * (5V/12V) = ?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
On Wednesday, January 22, 2014 9:54:42 PM UTC-8, Daniel Pitts 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?
Click to expand...

Not completely; it's only drawing 1 Amp AVERAGE current from the 10V.

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?
Click to expand...

That's why you need to know avout duty cycle of the converter, and you
need to know the source can handle peak currents of at least the average
current requirement divided by the conduction duty cycle.
And, it matters if your source has foldback current limiting, and if
there is some kind of filter (capacitor or LC, or CLC) on the converter's input.
 
On Thu, 23 Jan 2014 08:02:19 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Thu, 23 Jan 2014 02:15:35 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 22 Jan 2014 21:54:42 -0800, Daniel Pitts
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?

---
Yes.

If you have a wall-wart that'll put out 12V at 1 amp, that's 12
watts.

If you have a buck converter with a 5V output that can push 2 amps
through a [2.5 ohm] load, that's 10 watts.

Then, since the efficiency of the buck converter is 85%, and

Pout
e = ------,
Pin

rearrange to solve for Pin, and the required power input to the buck
converter will be:

Pout 10W
Pin = ------ = ------ = 11.76 watts
e 0.85

So you're there with about a quarter of a watt to spare.



In other words, not enough margin to guarantee it'll work.

...Jim Thompson
Click to expand...

---
No mention was made of surges, so if the wall-wart can put out 12
watts and the buck regulator needs 11.76 watts in to deliver 10
watts to the load, what am I missing?
 
On Thu, 23 Jan 2014 16:24:25 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 23 Jan 2014 08:02:19 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Thu, 23 Jan 2014 02:15:35 -0600, John Fields
jfields@austininstruments.com> wrote:

On Wed, 22 Jan 2014 21:54:42 -0800, Daniel Pitts
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?

---
Yes.

If you have a wall-wart that'll put out 12V at 1 amp, that's 12
watts.

If you have a buck converter with a 5V output that can push 2 amps
through a [2.5 ohm] load, that's 10 watts.

Then, since the efficiency of the buck converter is 85%, and

Pout
e = ------,
Pin

rearrange to solve for Pin, and the required power input to the buck
converter will be:

Pout 10W
Pin = ------ = ------ = 11.76 watts
e 0.85

So you're there with about a quarter of a watt to spare.



In other words, not enough margin to guarantee it'll work.

...Jim Thompson

---
No mention was made of surges, so if the wall-wart can put out 12
watts and the buck regulator needs 11.76 watts in to deliver 10
watts to the load, what am I missing?
Click to expand...

Nothing. I just don't like such tight margins. And, what if its an
RS-quality wall wart ?:)

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On 1/23/14 5:40 PM, whit3rd wrote:
On Wednesday, January 22, 2014 9:54:42 PM UTC-8, Daniel Pitts 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?

Not completely; it's only drawing 1 Amp AVERAGE current from the 10V.
Click to expand...
Good point. I'll keep that in mind if I need to build something that
does this.
 
"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.
Click to expand...

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
 
On 1/24/14 5:41 AM, petrus bitbyter 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?
Click to expand...

Mostly I was just curious. Some day (in the far future) I may end up
designing high-powered projects (a la Jim Thompson's design work in
action), so I was just making sure I understand the theory ;-)
 
On Thu, 23 Jan 2014 17:40:34 -0800, whit3rd wrote:

On Wednesday, January 22, 2014 9:54:42 PM UTC-8, Daniel Pitts 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?

Not completely; it's only drawing 1 Amp AVERAGE current from the 10V.

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?

That's why you need to know avout duty cycle of the converter, and you
need to know the source can handle peak currents of at least the average
current requirement divided by the conduction duty cycle.
And, it matters if your source has foldback current limiting, and if
there is some kind of filter (capacitor or LC, or CLC) on the
converter's input.
Click to expand...

Or you just need to put an adequate input capacitor on your buck
converter!

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
news:GtxEu.60255$J12.52429@fx10.iad...
On 1/24/14 5:41 AM, petrus bitbyter wrote:
"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
news:y82Eu.340472$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?

Mostly I was just curious. Some day (in the far future) I may end up
designing high-powered projects (a la Jim Thompson's design work in
action), so I was just making sure I understand the theory ;-)
Click to expand...

Well, if you want to catch up with Jim you have a long, looong way to go.
But don't give up :)

The most efficient converters are still the ordinary transformers. As long
as you have AC off course. The same theory applies to both transformers and
(other) converters: The amount of energy that goes in, will have to come out
some way. Hopefully as much as possible in the requested format, but there
are always losses that come out as heat.

petrus bitbyter
 
On 1/24/14 11:05 AM, petrus bitbyter wrote:
"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
Mostly I was just curious. Some day (in the far future) I may end up
designing high-powered projects (a la Jim Thompson's design work in
action), so I was just making sure I understand the theory ;-)

Well, if you want to catch up with Jim you have a long, looong way to go.
I would say so. He's been doing this stuff since before my mom was born ;-)
Click to expand...

> But don't give up :)
Never do. I've got a deep software background, and hope to be able to
apply that perspective to EE. Of course this is all just a hobby for me
now, so I may never get to be a great EE.
 
On Fri, 24 Jan 2014 11:28:04 -0800, Daniel Pitts
<newsgroup.nospam@virtualinfinity.net> wrote:

On 1/24/14 11:05 AM, petrus bitbyter wrote:
"Daniel Pitts" <newsgroup.nospam@virtualinfinity.net> schreef in bericht
Mostly I was just curious. Some day (in the far future) I may end up
designing high-powered projects (a la Jim Thompson's design work in
action), so I was just making sure I understand the theory ;-)

Well, if you want to catch up with Jim you have a long, looong way to go.
I would say so. He's been doing this stuff since before my mom was born ;-)
Click to expand...

That could well be... I've "officially" been a graduate engineer for
51 years, plus tinkered around in my father's radio and TV repair shop
for at least 10 years before that.

But don't give up :)
Never do. I've got a deep software background, and hope to be able to
apply that perspective to EE. Of course this is all just a hobby for me
now, so I may never get to be a great EE.
Click to expand...

I'm the other way 'round... now learning software tricks so that I can
devise nice behavioral models for complex analog systems.

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
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
Click to expand...

---
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.
 
On Thu, 23 Jan 2014 15:28:42 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Thu, 23 Jan 2014 16:24:25 -0600, John Fields
jfields@austininstruments.com> wrote:

On Thu, 23 Jan 2014 08:02:19 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

In other words, not enough margin to guarantee it'll work.

...Jim Thompson

---
No mention was made of surges, so if the wall-wart can put out 12
watts and the buck regulator needs 11.76 watts in to deliver 10
watts to the load, what am I missing?

Nothing. I just don't like such tight margins. And, what if its an
RS-quality wall wart ?:)

...Jim Thompson
Click to expand...
---
Buy two.

John Fields
 
Tim Wescott wrote:
Or you just need to put an adequate input capacitor on your buck
converter!
Click to expand...

That would raise the cost to over a buck. ;-)


--
Anyone wanting to run for any political office in the US should have to
have a DD214, and a honorable discharge.
 
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. ;-)
Click to expand...

A buck-and-a-half converter?

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
 
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