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[on:]

hi,
please could someone check my power supply,  just wondering if i need a 100nf after the  7805 as well?

my current drain on 5v is 240ma, (im adding the - to my circuit for an op amp) the + - 15 for a single quad op amp

thank you

[Reply #1 on:]

That 7805 by regulating from 15 down to 5 will be dissipating 2.4 watts of heat at 240 ma of load current. That is going to take one very good heatsink to prevent the regulator from shutting down via it's self protection mode. As far as caps, let the datasheet guide you.

Lefty

[Reply #2 on:]

Looks pretty good. Bypass caps right at the v.reg outputs is always a good idea.
One thing to realize is the peak output voltage of the [I assume] 36VAC c.t.
transformer is actually 18*1.414 = 25.5V, so your electrolytics are maybe a
little too low in voltage rating. You could probably get by with a 24VAC c.t.
transformer, and use 7812/7912 v.regs, or even lower voltage for the opAmps.

[Reply #3 on:]

Mostly reiterating some best practices...  Always put 100nF or so at the in and out side of a regulator.  Not a bad idea to put a diode facing from the output to the input to ensure the input is always higher than output.  (It will be reverse-biased when powered on, with input being higher than output.)

Time to look into switching regulators for your 5v load.  My personal rule of thumb:  Any time you go over 100mA, a linear regulator is going to get hot.

[Reply #4 on:]

Mostly reiterating some best practices...  Always put 100nF or so at the in and out side of a regulator.  Not a bad idea to put a diode facing from the output to the input to ensure the input is always higher than output.  (It will be reverse-biased when powered on, with input being higher than output.)

Time to look into switching regulators for your 5v load.  My personal rule of thumb:  Any time you go over 100mA, a linear regulator is going to get hot.

I agree, with the low price for Asian DC/DC converters these days it's really hard to keep using the same old hot running linear regulators.

http://www.ebay.com/itm/251066005460?ssPageName=STRK:MEWAX:IT&_trksid=p3984.m1423.l2649

Lefty

[Reply #5 on:]

ok thanks very much for the input in my power supply lots to think about, just one thing arnt switching reg's noisy looking at the data sheets it mentions oscillators quite a few times ive always thought of switching power supply's as noisy as they are in cased in a metal box?


i started to look at the LM2575T-5.0, think this would solve my 5v problem.

im confused as to how to supply the ic, can you only supply it from the - and + from a bride rectifier, as it states in the data sheet unregulated dc input 7/40v input.

or can you supply it by using a regulated +v be great if i could supply it using using a regulated +v out of my 7815.

thank you

[Reply #6 on:]

You gain little benefit from feeding the LM2575 from the 15V regulated supply since the switching regulator is much more efficient than the LM7815.  Why would you want to have the 7815 linear regulator also be responsible all the current used by the 5V power source? It will most likely need to have a heatsink as a result.  You might be able to get away with not having a heatsink on the 15V regulators at all if you attach the Input side of the LM2575 to the unregulated leg of the + side of the bridge. 

[Reply #7 on:]

hi pw,
yes i did think of doing that shortly after i wrote my last post, thank you

can somebody please explain the data in the highlighted box please,

does that mean voltage has to be 8 to 40v input,  to output those  min max voltages? (and use fixed point grounding)

really dont understand this, 0.2A < ILOAD < 1A, what does the symbol like this mean <

[Reply #8 on:]

Some switching regulator solutions require a minimum "load" for the regulator to perform as expected. As well, it appears that they are clarifying the "drop out" point for a 5V supply is 8V.  They are basically saying that nominal Drop out voltage is 2V... (sort of high for a regulator)

[Reply #9 on:]

0.2a < load < 1.0a means that the current pulled by the load is somewhere between 0.2a and 1.0a.  (0.2a is less than the load which is less than 1.0a.)  So they're saying the output voltage will stay within regulation (between 4.8 and 5.2v) as long as the load is between 200 and 1000mA.  Before settling on this regulator, I would want to know what the absolute minimum supported load is, since a micro alone pulls very little power.  I'm assuming you're only going to be drawing 240mA part of the time...?

Yes, switching PSUs are noisy.  They work by chopping the incoming voltage into pulses, and then filtering it through a lowpass filter to approximate a lower steady voltage.  The lowpass filter is made up of the inductor (choke), a cap, and the load resistance -- which is why it needs a certain amount of load to work properly.  If this lowpass filter is designed and built properly, the output can be just as quiet - or better - than a linear regulator.  (Some datasheets specify a second 2-pole filter, with another inductor + cap, to reduce output ripple further.)

You see metal cases around PSUs for various reasons -- one of which is to reduce radiated noise.  Another is to reject noise from outside sources.  Yet another is to provide a safety ground in case of a fault.  (The case will conduct an unintended short at the chassis to ground, which will blow a fuse rather than electrocuting the user.)  Finally, they tend to withstand (and shed!) heat better than, say, wood or rubber.

[Reply #10 on:]

One thing that's not indicated on the diagram in reply #5 is that considerations about
the actual layout, component positioning, trace lengths and widths, ground planes, etc,
are extremely critical in getting such ckts to work properly.

Research this before proceeding.

[Reply #11 on:]

One thing that's not indicated on the diagram in reply #5 is that considerations about
the actual layout, component positioning, trace lengths and widths, ground planes, etc,
are extremely critical in getting such ckts to work properly.

Research this before proceeding.

hi dan
i did have a go at laying this out kept all tracks to a min length shortest possible route, and gnd is fixed point as possible, there is a bit in the data sheet about layout will have a really good look, think i will double up on the tracks top/bottom to be sure, its only layed out on a single layer at the moment
thanks

[Reply #12 on:]

0.2a < load < 1.0a means that the current pulled by the load is somewhere between 0.2a and 1.0a.  (0.2a is less than the load which is less than 1.0a.)  So they're saying the output voltage will stay within regulation (between 4.8 and 5.2v) as long as the load is between 200 and 1000mA.  Before settling on this regulator, I would want to know what the absolute minimum supported load is, since a micro alone pulls very little power.  I'm assuming you're only going to be drawing 240mA part of the time...?

Yes, switching PSUs are noisy.  They work by chopping the incoming voltage into pulses, and then filtering it through a lowpass filter to approximate a lower steady voltage.  The lowpass filter is made up of the inductor (choke), a cap, and the load resistance -- which is why it needs a certain amount of load to work properly.  If this lowpass filter is designed and built properly, the output can be just as quiet - or better - than a linear regulator.  (Some datasheets specify a second 2-pole filter, with another inductor + cap, to reduce output ripple further.)

You see metal cases around PSUs for various reasons -- one of which is to reduce radiated noise.  Another is to reject noise from outside sources.  Yet another is to provide a safety ground in case of a fault.  (The case will conduct an unintended short at the chassis to ground, which will blow a fuse rather than electrocuting the user.)  Finally, they tend to withstand (and shed!) heat better than, say, wood or rubber.

ok thanks i understand now,

well the current does fluctuate a bit on for 20 seconds 90ma then circuit then switches on the 2 relays then 240ma, ive reciently added a bit more to the circuit so its probably running at over 300ma all the time, but there could be a situation where its only the 90ma.

so its possible the circuit is idle @ 90ma, thought maby of adding a dummy load to keep the current over the 200ma maby when idle say a 45 ohm 1w resistor or near, do you think this is a good idea?

using this LM2575T really does sort my problem with my power supply, cant think of another way round it

[Reply #13 on:]

i did have a go at laying this out kept all tracks to a min length shortest possible route, and gnd is fixed point as possible, there is a bit in the data sheet about layout will have a really good look, think i will double up on the tracks top/bottom to be sure, its only layed out on a single layer at the moment

I would also round up a few more datasheets on buck and boost converters and look at
their layout suggestions too. If not done perfectly well, the thing will generate more EMI
than you know what to do with.

[Reply #14 on:]

You may not need to burn that much power in a dummy load.  Does the data sheet specify a minimum load to maintain tolerance?  It could be that it just requires 200mA to be wtihin +/- 0.2v, but you don't necessarily need it to be THAT tight.

oric_dan -- Any suggestions on how you would know?  I built my first switcher recently.  I have no idea how to measure for EMF, or any other metric to determine if my layout is any good.  The suggestions I've seen in some TI and Maxim datasheets are all rather sparse and tend to rely on the designer to know a thing or two already.  "Ensure your layout uses best practices" and stuff like that.  I certainly don't have any windy twisty traces, but if they need to be aligned to the phase of the moon, that went right over my head.