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.
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
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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.
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.
For the 3.3v it might be close to 10mA. Accelerometer
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Gyro
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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.
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.
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 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.
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.
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
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.
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
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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.
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.
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