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Voltage Regulators

J

Joe

Jan 1, 1970
0
Hello,

Has anyone used the AP1501 voltage regulators.
(http://www.diodes.com/datasheets/dsAP1501.pdf)

I have a 3.3v and a 5.0v output version of these v-regs. I was
wondering if I really need the inductor and capacitors for this v-
reg? And also would a 39uH inductor work instead of a 33 uH on the
3.0v-reg? Would the inductor value be different for the 5.0v
regulator? What would the capacitor values be for the the Cin and Cout
shown in the datasheet?

Im sorry for all the questions, but any help in this matter would be
appreciated.

Thanks,
Joe McKibben
 
C

Chuck

Jan 1, 1970
0
Joe said:
Hello,

Has anyone used the AP1501 voltage regulators.
(http://www.diodes.com/datasheets/dsAP1501.pdf)

I have a 3.3v and a 5.0v output version of these v-regs. I was
wondering if I really need the inductor and capacitors for this v-
reg? And also would a 39uH inductor work instead of a 33 uH on the
3.0v-reg? Would the inductor value be different for the 5.0v
regulator? What would the capacitor values be for the the Cin and Cout
shown in the datasheet?

Im sorry for all the questions, but any help in this matter would be
appreciated.

Thanks,
Joe McKibben

Hi Joe,

My initial reaction is that 39 uH may be
near the edge of the chip's frequency
range (~+/-15% of 150kHz).

Cin can probably be 0.1uF. It is there
to filter transients.

Cout is determined by establishing the
ripple you're willing to see in the
output and the maximum load current. The
output from the chip seems to be a
"square wave" between ~0Volts
and Vreg. You choose Cout so that at the
desired load resistance, it discharges
no more than the maximum p-p ripple you
want between the 150kHz pulses.

With luck, someone with direct
experience with the chip will offer some
more detail.

Chuck
 
J

John Popelish

Jan 1, 1970
0
Joe said:
Hello,

Has anyone used the AP1501 voltage regulators.
(http://www.diodes.com/datasheets/dsAP1501.pdf)

I haven't.
I have a 3.3v and a 5.0v output version of these v-regs. I was
wondering if I really need the inductor and capacitors for this v-
reg?

Since these are switching regulators that output pulses, not
steady voltages, the LC output filter is an essential pert
of getting a DC output from them. Since they also draw
current pulses from the upstream supply, if that supply
feeds anything else, you may need to worry quite a bit about
how you filter than side, also, so the operation of this
regulator does not trash the upstream supply for other users.
And also would a 39uH inductor work instead of a 33 uH on the
3.0v-reg?

Almost certainly. The manufacturers are always pushing the
required inductance down to the absolute minimum value,
because storing energy in inductors at low loss is spendy.
Would the inductor value be different for the 5.0v
regulator? What would the capacitor values be for the the Cin and Cout
shown in the datasheet?

All those values on the data sheet tend toward the absolute
minimum that will work. When selecting these capacitors the
equivalent series resistance (ESR) and ripple current
ratings are as important as the capacitance. The input
capacitor sees worse ripple current than the output
capacitor, because it does not have a series inductance
smoothing the pulses into a triangle current wave.

The actual value of capacitance, ESR and ripple current
rating involves the peak current you expect the regulator to
supply and the maximum ripple voltage you can tolerate.

Can you help with any of these specifications?
Im sorry for all the questions, but any help in this matter would be
appreciated.

No apology necessary. This is the right place for such
questions.
 
J

Joe

Jan 1, 1970
0
I haven't.


Since these are switching regulators that output pulses, not
steady voltages, the LC output filter is an essential pert
of getting a DC output from them. Since they also draw
current pulses from the upstream supply, if that supply
feeds anything else, you may need to worry quite a bit about
how you filter than side, also, so the operation of this
regulator does not trash the upstream supply for other users.


Almost certainly. The manufacturers are always pushing the
required inductance down to the absolute minimum value,
because storing energy in inductors at low loss is spendy.


All those values on the data sheet tend toward the absolute
minimum that will work. When selecting these capacitors the
equivalent series resistance (ESR) and ripple current
ratings are as important as the capacitance. The input
capacitor sees worse ripple current than the output
capacitor, because it does not have a series inductance
smoothing the pulses into a triangle current wave.

The actual value of capacitance, ESR and ripple current
rating involves the peak current you expect the regulator to
supply and the maximum ripple voltage you can tolerate.

Can you help with any of these specifications?


No apology necessary. This is the right place for such
questions.




- Show quoted text -

Theses regulators are a little over kill for my application. On the
3.3v reg. max current might be 2mA (dual axis gyro and tri-axis
accelerometer). And on the 5.0v reg. approx. 150mA (SX-
Microcontroller). Not sure how much ripple voltage I can tolerate, but
I'll try to found out.

Thanks for the help,
Joe McKibben
 
J

John Popelish

Jan 1, 1970
0
Joe said:
Theses regulators are a little over kill for my application. On the
3.3v reg. max current might be 2mA (dual axis gyro and tri-axis
accelerometer). And on the 5.0v reg. approx. 150mA (SX-
Microcontroller). Not sure how much ripple voltage I can tolerate, but
I'll try to found out.

I assume that is a typo and you mean that the load on the
3.3 volt regulator may be as high as 2 amperes.

Does anything else (other than two of these regulators) run
on the up stream source? Are there any RF sensitive devices
anywhere near by?
 
J

Joe

Jan 1, 1970
0
I assume that is a typo and you mean that the load on the
3.3 volt regulator may be as high as 2 amperes.

Does anything else (other than two of these regulators) run
on the up stream source? Are there any RF sensitive devices
anywhere near by?- Hide quoted text -

- Show quoted text -

For the 3.3v it might be close to 10mA. Accelerometer (http://
www.robotshop.ca/PDF/ADXL330_0.pdf), Gyro (http://www.robotshop.ca/PDF/
IDG-300_Datasheet.pdf)
And I dont think there are any RF sensitive devices to really worry
about.

Im not sure what you mean by on the up stream, but I have a 12volt
battery. It supplies the current for two motors and the motor
controllers. The 12 volts also supplies the voltage to the
regulators. I think I might use a seperate battery source for the
regulators to eliminate any voltage spikes from the motors.

Joe McKibben
 
J

John Popelish

Jan 1, 1970
0
Joe said:
(snip)
For the 3.3v it might be close to 10mA. Accelerometer (http://
www.robotshop.ca/PDF/ADXL330_0.pdf), Gyro (http://www.robotshop.ca/PDF/
IDG-300_Datasheet.pdf)
And I dont think there are any RF sensitive devices to really worry
about.

For any load less than about 100 mA, I would go with a
linear regulator, like an LM317. The small efficiency gain
you get with a switcher is not worth the trouble for lower
current loads. The 3.3 volt regulator could get its input
power from the 5 volt output. This will give you
essentially the same efficiency as a switcher producing 3.3
volts from the 12 volt source, with less output noise,
assuming the 3.3 and 5 volt loads share a common negative rail.
Im not sure what you mean by on the up stream, but I have a 12volt
battery.

Yes, from a power flow standpoint, that is up stream of
these regulators.
It supplies the current for two motors and the motor
controllers.

Those will generate some noise and will need their own
storage capacitors to help the battery supply the large
current spikes they need. Something like 1000 uF at 16
volts for each motor.
The 12 volts also supplies the voltage to the
regulators. I think I might use a seperate battery source for the
regulators to eliminate any voltage spikes from the motors.

That shouldn't be necessary, especially for switching
regulators, since they contain their own LC filters.

For the 5 volt 2 amp regulator, I would use a 470 uF 16 volt
Panasonic type FM electrolytic cap on the input and a 330 uF
6.3 volt across the output. Digikey sells these. They have
a low ESR and high ripple current rating. I would probably
also parallel the output cap with a Panasonic V series
(stacked film) 1 uF 50 volt to take care of the high
frequency edges that couple across the inter winding
capacitance of the inductor. In addition, I would add a
ferrite bead on a lead in series with thew 12 volt input,
upstream of the input storage cap, to keep the motor noise
out of the regulator supply.
 
J

Joe

Jan 1, 1970
0
For any load less than about 100 mA, I would go with a
linearregulator, like an LM317. The small efficiency gain
you get with a switcher is not worth the trouble for lower
current loads. The 3.3 voltregulatorcould get its input
power from the 5 volt output. This will give you
essentially the same efficiency as a switcher producing 3.3
volts from the 12 volt source, with less output noise,
assuming the 3.3 and 5 volt loads share a common negative rail.


Yes, from a power flow standpoint, that is up stream of
these regulators.


Those will generate some noise and will need their own
storage capacitors to help the battery supply the large
current spikes they need. Something like 1000 uF at 16
volts for each motor.


That shouldn't be necessary, especially for switching
regulators, since they contain their own LC filters.

For the 5 volt 2 ampregulator, I would use a 470 uF 16 volt
Panasonic type FM electrolytic cap on the input and a 330 uF
6.3 volt across the output. Digikey sells these. They have
a low ESR and high ripple current rating. I would probably
also parallel the output cap with a Panasonic V series
(stacked film) 1 uF 50 volt to take care of the high
frequency edges that couple across the inter winding
capacitance of the inductor. In addition, I would add a
ferrite bead on a lead in series with thew 12 volt input,
upstream of the input storage cap, to keep the motor noise
out of theregulatorsupply.

Cool, thanks for all the help. Ill try to get a linear regulator for
the 3.3volt. and the other components I need.

Joe McKibben
 
J

John Popelish

Jan 1, 1970
0
Joe said:
Cool, thanks for all the help. Ill try to get a linear regulator for
the 3.3volt. and the other components I need.

The LM317 adjustable linear regulator requires a couple
resistors to program its output voltage. An input and
output capacitor is also recommended. See the data sheet.

http://www.fairchildsemi.com/ds/LM/LM317.pdf
 
J

John Popelish

Jan 1, 1970
0
Joe said:
Why wouldn't I just get a fixed reg. instead of an adjustable?

5 volt linear regulators are common as dirt, but 3.3 types
are a lot harder to find. I would avoid low drop out types,
since they are harder to stabilize against oscillations.
The 317 is very common and is available in a low power
version that comes in a TO-92 transistor package.

But a type that does not require setting resistors (and is
not LDO) would be:
LM78L33ACZ
http://www.st.com/stonline/products/literature/ds/2145/l78l05ab.pdf
 
J

Joe

Jan 1, 1970
0
5 volt linear regulators are common as dirt, but 3.3 types
are a lot harder to find. I would avoid low drop out types,
since they are harder to stabilize against oscillations.
The 317 is very common and is available in a low power
version that comes in a TO-92 transistor package.

But a type that does not require setting resistors (and is
not LDO) would be:
LM78L33ACZhttp://www.st.com/stonline/products/literature/ds/2145/l78l05ab.pdf

Do you know of a good place to find the LM78L33ACZ v-reg? I cant find
it anywhere.

Joe McKibben
 
J

John Popelish

Jan 1, 1970
0
Joe said:
Do you know of a good place to find the LM78L33ACZ v-reg? I cant find
it anywhere.

No. Hence the LM317 suggestion.
 
J

Joe

Jan 1, 1970
0
No. Hence the LM317 suggestion.

Oh, ok thanks. I found some surface mount versions of the LM78L33, but
that doesn't help me me to much.

How do you choose the resistors for the adj regulator? I went through
the formula under the typical application drawing using 3.3volts, but
I am not sure what do with the value. I need to do some brushing up on
my DC circuits understanding, well probably more than brushing up.

Thanks again,
Joe McKibben
 
J

John Popelish

Jan 1, 1970
0
Joe said:
How do you choose the resistors for the adj regulator? I went through
the formula under the typical application drawing using 3.3volts, but
I am not sure what do with the value. I need to do some brushing up on
my DC circuits understanding, well probably more than brushing up.

There are generally two constraints. The series pair of
resistors must consume at least the regulator minimum output
current, if the regulator must work all the way to zero
external load. And the divider voltage must be the desired
output voltage minus the nominal difference voltage between
the output and the reference pin.

A secondary constraint is that the range of current passed
through the reference pin does not distort the divider
voltage too much.

For the LM317L
http://www.onsemi.com/pub/Collateral/LM317L-D.PDF

the maximum value of the minimum output current allowed if
you expect the regulator to regulate is 10 mA, so if you
will not always have at least that much external load, then
the divider must consume it. Since the nominal difference
between the reference pin voltage and the output voltage is
1.25 volts, there will be that much voltage across the top
resistor of the reference divider, so it must be no more
than 1.25/.01=125 ohms. If you use the typical value of
minimum load current or 3 mA, instead of the worst case
(usually okay for a single system that will not be
manufactured in high volume) that resistor rises to
1.25/.003=417 ohms. Usually, any value between those will
work. I notice that the example on the sheet uses 240 ohms.

Then the bottom divider resistor must drop the rest of the
desired output voltage, or 3.3-1.25=2.05 volts, in this
case, while it also passes the reference pin (adjustment
pin) current of about 50 uA. Since the first resistor is
delivering 3 to 10 mA, the additional current from the
reference pin may be within the tolerance effect of the
resistors.

Ignoring the reference pin current and assuming you use 240
ohms as the upper resistor
(delivering about 1.25/240=5.2 mA), the lower one would be
2.05/.0052=394 ohms.

All this process is described in detail starting on page 8.
 
J

Joe

Jan 1, 1970
0
There are generally two constraints. The series pair of
resistors must consume at least the regulator minimum output
current, if the regulator must work all the way to zero
external load. And the divider voltage must be the desired
output voltage minus the nominal difference voltage between
the output and the reference pin.

A secondary constraint is that the range of current passed
through the reference pin does not distort the divider
voltage too much.

For the LM317Lhttp://www.onsemi.com/pub/Collateral/LM317L-D.PDF

the maximum value of the minimum output current allowed if
you expect the regulator to regulate is 10 mA, so if you
will not always have at least that much external load, then
the divider must consume it. Since the nominal difference
between the reference pin voltage and the output voltage is
1.25 volts, there will be that much voltage across the top
resistor of the reference divider, so it must be no more
than 1.25/.01=125 ohms. If you use the typical value of
minimum load current or 3 mA, instead of the worst case
(usually okay for a single system that will not be
manufactured in high volume) that resistor rises to
1.25/.003=417 ohms. Usually, any value between those will
work. I notice that the example on the sheet uses 240 ohms.

Then the bottom divider resistor must drop the rest of the
desired output voltage, or 3.3-1.25=2.05 volts, in this
case, while it also passes the reference pin (adjustment
pin) current of about 50 uA. Since the first resistor is
delivering 3 to 10 mA, the additional current from the
reference pin may be within the tolerance effect of the
resistors.

Ignoring the reference pin current and assuming you use 240
ohms as the upper resistor
(delivering about 1.25/240=5.2 mA), the lower one would be
2.05/.0052=394 ohms.

All this process is described in detail starting on page 8.

Ok, thanks again. That data sheet is different than the first one you
posted for the LM317. It is more helpful than the first.
Now I just have to get the LM317, capacitors, inductors, and resistors
and I can finally get this thing hooked up.

Thanks for being patient with me,
Joe McKibben
 
J

John Popelish

Jan 1, 1970
0
Joe said:
Ok, thanks again. That data sheet is different than the first one you
posted for the LM317. It is more helpful than the first.
Now I just have to get the LM317, capacitors, inductors, and resistors
and I can finally get this thing hooked up.

Before you send for parts or at least before you solder
anything together, perhaps you should send me a schematic of
the system, so I can correct any misunderstandings. We have
covered a lot of ground, since your first post.
 
J

Joe

Jan 1, 1970
0
Before you send for parts or at least before you solder
anything together, perhaps you should send me a schematic of
the system, so I can correct any misunderstandings. We have
covered a lot of ground, since your first post.

OK, I'll get one drawn up.

Joe McKibben
 
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