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They're nice and easy to use. And they're rugged.
LOL
I get into too much trouble with just two transistors! Thank you Harald, that is an excellent post, but above my comprehension at the moment. I ordered some LM2576, about 45 cents US each.
If I am reading this chart correctly, the inductor that is needed for the basic regulation circuit is dictated by the current usage of the circuit and Vinput.
If I have about 28v coming into the circuit and the circuit consumes about, maybe 100mA(unsure?):
Would I need a 1000μH inductor?
What power rating should I select? 1/4 W or higher?
How does one know when they have inductors or resistors if they use the same color code?!
Reading through the datasheet, I noticed that the regulator will operate in discontinuous mode if under 300mA - do I need to design the circuit differently?
If I have about 28v coming into the circuit and the circuit consumes about, maybe 100mA(unsure?):
Would I need a 1000μH inductor?
What power rating should I select? 1/4 W or higher?
How does one know when they have inductors or resistors if they use the same color code?!
Reading through the datasheet, I noticed that the regulator will operate in discontinuous mode if under 300mA - do I need to design the circuit differently?
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I believe that circuit gives you the maximum current vs inductance. Sure, you could use 1000uH (1mH), but you could also get away with 100uH, or anything in between.
No problems with discontinuous mode. It will give a slightly lower frequency ripple, but that is unlikely to be an issue.
edit: the important thing about the inductor is the saturation current. You won't want to use one that looks like a resistor as they're chokes and don't have the low loss characteristics you're after. Ensure that the inductor you choose (toroids are very commonly used) is rated for a current in excess than the current you'll ever need. For 100mA, try to get one rated for 250mA or so -- more is OK, it will just be larger and ore expensive.
No problems with discontinuous mode. It will give a slightly lower frequency ripple, but that is unlikely to be an issue.
edit: the important thing about the inductor is the saturation current. You won't want to use one that looks like a resistor as they're chokes and don't have the low loss characteristics you're after. Ensure that the inductor you choose (toroids are very commonly used) is rated for a current in excess than the current you'll ever need. For 100mA, try to get one rated for 250mA or so -- more is OK, it will just be larger and ore expensive.
davenn
Moderator
also, dropping from 28V to 5V is going to cause a LOT of heat generated by the poor 7805
eg 23V drop x 1A current = 23W of heat !!
in you last example .... 23V drop x 300mA = 7W .... its still gonna get hot
and you are pushing it past its limit ... most datasheets state 25V max input to a 7805
so in reality you don't want to be anywhere near even 25V to give the 7805 a long life expectancy
Dave
eg 23V drop x 1A current = 23W of heat !!
in you last example .... 23V drop x 300mA = 7W .... its still gonna get hot
and you are pushing it past its limit ... most datasheets state 25V max input to a 7805
so in reality you don't want to be anywhere near even 25V to give the 7805 a long life expectancy
Dave
eep... LOL I will update the schematic, its supposed to read LM2576 - switch mode supply.also, dropping from 28V to 5V is going to cause a LOT of heat generated by the poor 7805
eg 23V drop x 1A current = 23W of heat !!
in you last example .... 23V drop x 300mA = 7W .... its still gonna get hot
and you are pushing it past its limit ... most datasheets state 25V max input to a 7805
so in reality you don't want to be anywhere near even 25V to give the 7805 a long life expectancy
Dave
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I thought chokes and inductors were synonymous? Shame too, I don't seem to find an assortment of toroidal style or coil style inductors on e-bay. I did find the resistor type. Will this type not work at all?
Thank you
LOL well there's always a good reason for these things
interestingly ... all I saw was the 7805 label .... didn't even notice the extra 2 legs on the device
D
No worries, I do worse... have you seen some of my schematics?
davenn
Moderator
OK ... next point
your diag. says 29.2V on secondary, so 29.2VAC rms x 1.414 = 41.3 Vpk (peak) - say ~2V drop through the diodes
you are going to have 39-40V DC going into the regulator. Not the 28V you have stated
Ohhh and I would up C1 to at least 1000uF. 100uF just aint enough
Dave
your diag. says 29.2V on secondary, so 29.2VAC rms x 1.414 = 41.3 Vpk (peak) - say ~2V drop through the diodes
you are going to have 39-40V DC going into the regulator. Not the 28V you have stated
Ohhh and I would up C1 to at least 1000uF. 100uF just aint enough
Dave
OK ... next point
your diag. says 29.2V on secondary, so 29.2VAC rms x 1.414 = 41.3 Vpk (peak) - say ~2V drop through the diodes
you are going to have 39-40V DC going into the regulator. Not the 28V you have stated
Ohhh and I would up C1 to at least 1000uF. 100uF just aint enough
Dave
I need to clarify this, I just couldn't find an appropriate component in eagle. The input is really 29.2VAC as measured on my DMM. It's coming out of a 24V wall adapter.
Looking at the entire circuit, if you're just using the power supply to power a couple of LEDs, a switchmode power supply is going way over the top.
You could probably run the whole lot from the rectified and filtered AC, possibly using some transistors capable of 60V.
Unless you actually need a source of 5V for a particular reason...?
You could probably run the whole lot from the rectified and filtered AC, possibly using some transistors capable of 60V.
Unless you actually need a source of 5V for a particular reason...?
Those transformer based wall adaptors typically drop in voltage as they are loaded by a current. Measure the output voltage of the adaptor at no load and at max. load (max. current drawn). The voltages you measure give you the input range for which the regulator needs to be designed.
As you go from >20V to 5V this should be no issue hre - just wanted to notice.
Another issue: If you measured 29.2V AC you probably used the AC range of your meter. This means 29.2V are RMS. The peak value of a 29.2V RMA voltage is 29.2V*sqrt(2)= 41V! This is the peak voltage you will see across C1!
This means you would need the HV version of the LM2576 as the standard version is rated only up to 37V input voltage. Build a mockup consisting only of transformer, rectifier and Capacitor and measure the real voltage across C1.
As you go from >20V to 5V this should be no issue hre - just wanted to notice.
Another issue: If you measured 29.2V AC you probably used the AC range of your meter. This means 29.2V are RMS. The peak value of a 29.2V RMA voltage is 29.2V*sqrt(2)= 41V! This is the peak voltage you will see across C1!
This means you would need the HV version of the LM2576 as the standard version is rated only up to 37V input voltage. Build a mockup consisting only of transformer, rectifier and Capacitor and measure the real voltage across C1.
I had 7805's on hand so that seemed the easiest way to produce the 5v. Why 5v? - I figured with two LED's I might need the Vdrop - but now realize I would only need about 3v since they would never be in series.
I had given thought to a capacitor fed p.s. - but remembered that the wall adapter might give varying voltage (under load I have seen several volts dropped with two different adapters). Unless the voltage is consistent cap-fed supplies are not suitable since they have no v-reg (LED's would burn out). I could be wrong, but that is what I thought from my limited experience.
Can you elaborate on what you are thinking of? Rectified and filtered I get (bridge rectifier and caps) - that would still have us at a peak of ~39VDC. What would the transistors do?
You may be best off if you use another wall wart that has a 5V output.
Alternatively you can reduce the voltage after the rectifier by putting an impedance in series with the rectifier. The impedance will create a voltage drop such that the output voltage of the rectifier is lower. If you put the impedance before the rectifier, it will see an AC current. You can then use a capacitor for impedance. This has the advantage that it dissipates much less energy as heat than a resistor. Here's a simplified example:
Green is the input voltage (+-30V from the transformer), blue is the output voltage across C2. By varying the value of C1 you can adjust the output voltage. Note that R2 (crossed red) is for simulation purposes only and should not be part of a real circuit.
Green is the input voltage (+-30V from the transformer), blue is the output voltage across C2. By varying the value of C1 you can adjust the output voltage. Note that R2 (crossed red) is for simulation purposes only and should not be part of a real circuit.
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The VDC looks pretty stable there! I didn't think it would be because of the source...Alternatively you can reduce the voltage after the rectifier by putting an impedance in series with the rectifier. The impedance will create a voltage drop such that the output voltage of the rectifier is lower. If you put the impedance before the rectifier, it will see an AC current. You can then use a capacitor for impedance. This has the advantage that it dissipates much less energy as heat than a resistor. Here's a simplified example:
Green is the input voltage (+-30V from the transformer), blue is the output voltage across C2. By varying the value of C1 you can adjust the output voltage. Note that R2 (crossed red) is for simulation purposes only and should not be part of a real circuit.
Am I correct in assuming then that with a X2 or Y2 rated cap for c1, I can build a suitable supply for the two transistors and leds without a regulator?
1) you don't need an X and/or Y rated capacitor for C1. C1 is on the low voltage side of the transformer so (almost) any capacitor is suitable.
2) you sjould use a regulator anyway to acommodate for variations in input voltage, load, temperature etc. By lowering the input voltage to the regulator via C1, you reduce the power loss in the regulator.
Looks pretty much like my schematic with diffferent values. Of course you have to change the values to adpat the circuit to your needs (in term of voltage and current).
Note that C1 sees AC. You cannot use a polarized capacitor (elko) for this one.
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