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power supply questions

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csdave

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hi all,
As you might remember, I am designing the power supply part for a small circuit drawing a max of 250mA and have a few more questions.

My initial design was the following:



I computed the 800uF by doing:
Vpeak=9Vac * sqrt(2) = 12.73
the rectifier is quoted as having a max voltage drop per element of 1V (what does per element mean? It's not per diode is it?)
VdropRect=1V (is this correct?
VrippleMax=12.73-7 - 1= 4.73
Then I considered 300mA to be on the safe side...
C= 300mA / (2 * 50Hz * 4.73) = 0.63mF
so I figured 800uF would be fine...

Then however, I realized that a 6V transformer could be enough if I used a larger filter capacitor and was a little bit less conservative so I redid the design by considering a 6Vac 3VA transformer.

here I considered

Vpeak=8.46V
Vdrop=1V
VrippleMax=0.46V
C=300mA / (2 * 50Hz * 0.46) = 6500uF

Would this work?

The power I am asking the transformer to provide is 8.46V*300mA = 2.54W so 3VA should
be enough right?

However, 8.46*400mA=3.4W so I need a smaller fuse for F3, right? (300mA)

A few more questions
1- Was I correct to assume 1V of voltage drop through the rectifier? (cfr http://www.datasheetcatalog.org/datasheet/mcc/W005M.pdf
2- What happens to the current when the filter cap charges and in particular when it charges for the first time? Can't it exceed the 300mA? Do I need a slow-acting fuse? Can this harm the transformer? My equations tells me about the average current in each half cycle of the input sine wave, but aren't there instants when the dissipated power is higher?

3- how big should F1 and F2 be? Back of the envelope tells me 6Vac*300mA = 220Vac*9mA. Should they really be as small as 9mA???
I was thinking of putting the same 400mA or 300mA fuses I have on the other side (they are rated 250Vac), would that be good?

4- if I had designed the circuit to provide 400mA, it would have required a cap of 8.7mF... however 400mA would have been too much for a 3VA transformer, wouldn't it?

Do you have any other suggestions?

Thanks

Davide

---------- Post added at 11:51 ---------- Previous post was at 10:32 ----------

this page Power Supplies seems to give different numbers (apart from the 1.4V on the rectifier) the way they compute the DC voltage is different... I would say mine is more precise, but who's right?
 
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Miguel Gaspar

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Please be more precise, I already read your post but I can't see if it is a transformless non regulated source . Please put a diagram of it.
As you saw in fact there are several ways to do what are you trying. And severals ways to calculate it, some depends of the devices that your are using.
 

csdave

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Hi Miguel,
there should be a diagram. At least I can see it in the post. Otherwise please go to this link: http://img651.imageshack.us/img651/2528/screenshot20110517at952.png

The image refers to the first version with the 9Vac transformer. The changes for the 6Vac transformer are what I put in the text.
Please let me know if it's clearer...

thanks a lot

Davide
 

ark5230

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Q-1.
That is ok
Q-2- What happens to the current when the filter cap charges and in particular when it charges for the first time? Can't it exceed the 300mA?
a. While charging it draws higher current (surge)
b. It can exceed the rating of the bridge used (as is seen in data sheet at the given link)
Q-2 Do I need a slow-acting fuse? Can this harm the transformer?
No need, no harm.
Q-2 My equations tells me about the average current in each half cycle of the input sine wave, but aren't there instants when the dissipated power is higher?
The initial charging currents can be understood but is a different area, so far as power supply goes, ignore this.
Q-3 Fuses are not that critical, why two fuses why not one, what trade off allowed, you can go ahead with what you have.
Q-4 It is matter of permissible ripple, as you are using three terminal regulator this is not major concern.
 

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I see your drawing but I need to log on. If I don't log on I see only text.

* You can use a 6V transformer if your load doesn't draw too much current.

* In operation each diode conducts an average value equal to your load. This consists of a short burst each cycle. The burst may be several times your load. It may produce 0.8-1.0 V across each diode.

* You can get by with one fuse in the mains supply. A slo-blo type is often better with power supplies. A fast-blo can blow needlessly during startup surge. But you won't find slo-blo as low as your amp load. So maybe a fast-blo is all right. Use several times the amps. You calculated 9mA at the mains. Probably a fuse of 1/10 A would be sufficient.

* You can keep the fuse in the secondary circuit if you wish to protect those components. However a 400 mA fast-blo fuse at F3 is liable to blow the moment you apply power because it will get a few amps surge because the uncharged capacitor is like a short circuit. It's hard to predict how many amps. It depends on transformer windings, diode resistance, etc.
 
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csdave

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thanks all!

I see your drawing but I need to log on. If I don't log on I see only text.
Didn't know that, interesting ;)
* You can use a 6V transformer if your load doesn't draw too much current.
Just how much is too much? Is my reasoning above correct? 300mA fine, 400mA too much for a 3VA 6Vac transformer.

---------- Post added at 23:31 ---------- Previous post was at 23:26 ----------

a. While charging it draws higher current (surge)
b. It can exceed the rating of the bridge used (as is seen in data sheet at the given link)
So does it mean it can blow my bridge? How should I correctly choose a bridge then?
The datasheet mentions a surge overload rate of 50A.

Q-3 Fuses are not that critical, why two fuses why not one, what trade off allowed, you can go ahead with what you have.
I thought of two because I can't tell for sure (given the nonpolarized plug) which is the live and which is the neutral wire.

Q-4 It is matter of permissible ripple, as you are using three terminal regulator this is not major concern.
Mmm, not sure I get this. Won't a large cap limit the ripple at the cost of drawing more current from the transformer than it can safely supply?
 
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chuckey

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It is not good practise to parallel transformer windings as if there is any difference in the number of turns there will be an increased primary current perhaps, leading to overheating. Its a linear law 1% error in number of turns = 1% increase in primary current or 1% less in secondary current. .
When the incoming mains is at a peak, it sends current through two diodes - hence twice the 1 V drop.
" C= 300mA / (2 * 50Hz * 4.73) = 0.63mF ", I like this formula but I don't understand it. C(MF) = amps/ 2 * frequency * voltage ripple. Should there not be a PI in there somewhere? :) where did you get it from ?
Frank
 

csdave

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It is not good practise to parallel transformer windings as if there is any difference in the number of turns there will be an increased primary current perhaps, leading to overheating. Its a linear law 1% error in number of turns = 1% increase in primary current or 1% less in secondary current. .
When the incoming mains is at a peak, it sends current through two diodes - hence twice the 1 V drop.
" C= 300mA / (2 * 50Hz * 4.73) = 0.63mF ", I like this formula but I don't understand it. C(MF) = amps/ 2 * frequency * voltage ripple. Should there not be a PI in there somewhere? :) where did you get it from ?
Frank
wiring windings in parallel was suggested by the guys at the shop and is also on the datasheets of some transformers.

As for the formula, I got it from several places, but this is pretty clear: Full Wave Rectifier and Bridge Rectifier

If I am not mistaken the Pi comes in when you have no cap, once you add a cap the attenuation is less and is Vripple.
 

csdave

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any further insights?

thanks!!

Davide
 

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Just how much is too much? Is my reasoning above correct? 300mA fine, 400mA too much for a 3VA 6Vac transformer.
After AC is rectified, and you put a capacitor across the power leads, your supply can rise to 1.414 x AC nominal minus 2 diode drops. That's with no load.

Once you add load, the supply voltage drops to a level of equilibrium. The greater the load, the more the supply drops.

With your 3W transformer you can expect to get 6 V @ 500mA. Notice that 5% variation in house voltage will give you 5% variation in your supply V.

So does it mean it can blow my bridge? How should I correctly choose a bridge then?
The datasheet mentions a surge overload rate of 50A.
The surge spec is a short burst through a diode. Your diodes won't have 50A.

But diodes will get a surge because the capacitor draws a lot of current on power-up. How much surge current is hard to predict. It depends on transformer resistance, windings, etc. It's only for a fraction of a second.

A good rule of thumb is to size components for 2x expected use. Therefore a 1A bridge rectifier should be sufficient.

All any of us can say is, if the bridge blows someday, replace it with a higher rating.

I had to do that with my power supply. Transformer is 12V @ 3 A. I powered an automobile headlight for several minutes. Blew the 4A diode bridge. I replaced it with a 6A bridge. Used heat sink grease (as I should have with the 4A bridge). No problems in the years since.

I thought of two because I can't tell for sure (given the nonpolarized plug) which is the live and which is the neutral wire.
No harm using two fuses in the primary. It's all up to your perception of safe practice.

Mmm, not sure I get this. Won't a large cap limit the ripple at the cost of drawing more current from the transformer than it can safely supply?
In normal operation your load is the only thing that determines how much current comes from the transformer.

You'll see the voltage settle to an average running value (which you read on a meter).

The size of the cap determines only the AC component of ripple, above and below your average running voltage. (Which you can read on your meter as AC if you have a setting to show AC with no DC component.)
 

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Thanks for your reply!

After AC is rectified, and you put a capacitor across the power leads, your supply can rise to 1.414 x AC nominal minus 2 diode drops. That's with no load.

Once you add load, the supply voltage drops to a level of equilibrium. The greater the load, the more the supply drops.

With your 3W transformer you can expect to get 6 V @ 500mA. Notice that 5% variation in house voltage will give you 5% variation in your supply V.
Do you mean that if the 7805 is trying to draw 400mA then the voltage will drop even if I put a huge filter cap and the transformer will not overheat? To my understanding, voltage will indeed drop, but the transformer will likely overheat because I am trying to get too much out of it, won't it?

Anyway, my reasoning is: if it can provide 6V at 500mA, then if I am trying to draw 7V, then the maximum current will be 420mA. if I am trying to draw 8V, so that I get 7 after the diode drop, then max current will be 375mA. Is this correct?

The surge spec is a short burst through a diode. Your diodes won't have 50A.
The way I get the source spec is that the diodes will withstand a surge of 50A, that is a very short burst of current of up to 50A, if the burst is long they'll fry, and if the burst needs more than 50A, they'll also fry. Isn't it this way?

But diodes will get a surge because the capacitor draws a lot of current on power-up. How much surge current is hard to predict. It depends on transformer resistance, windings, etc. It's only for a fraction of a second.

A good rule of thumb is to size components for 2x expected use. Therefore a 1A bridge rectifier should be sufficient.

All any of us can say is, if the bridge blows someday, replace it with a higher rating.

I had to do that with my power supply. Transformer is 12V @ 3 A. I powered an automobile headlight for several minutes. Blew the 4A diode bridge. I replaced it with a 6A bridge. Used heat sink grease (as I should have with the 4A bridge). No problems in the years since.
so would you go with the 9V 5VA or the 6V 3VA transformer in the above example if I need 300mA?
 

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In the old days there were charts showing the parameters for power supply design e.g.
Dayal M. (1964): Power Rectification with Silicon Diodes; Mullard Technical Communication 7,
230-262, and 9, 46-47, 1966.
Unvala B.A. (1968): Power Supply Circuits Using Silicon Rectifiers; Texas Instruments
Application Report D4.
but these may be difficult to find online. However, since circuit simulation is now so accessible, get hold of a free simulator and run transient simulations to give you all the information you need. Try for example the Linear Technology version available online at Linear Technology - Home Page.
 

csdave

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I indeed learned old-day electronics in school LOL ;). In any case, I've started playing with ltspice. It's pretty straightforward, but I am not sure how I should correctly model the 3VA transformer.



In the old days there were charts showing the parameters for power supply design e.g.
Dayal M. (1964): Power Rectification with Silicon Diodes; Mullard Technical Communication 7,
230-262, and 9, 46-47, 1966.
Unvala B.A. (1968): Power Supply Circuits Using Silicon Rectifiers; Texas Instruments
Application Report D4.
but these may be difficult to find online. However, since circuit simulation is now so accessible, get hold of a free simulator and run transient simulations to give you all the information you need. Try for example the Linear Technology version available online at Linear Technology - Home Page.
 

BradtheRad

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Do you mean that if the 7805 is trying to draw 400mA then the voltage will drop even if I put a huge filter cap and the transformer will not overheat?
Correct.

Anyway, my reasoning is: if it can provide 6V at 500mA, then if I am trying to draw 7V, then the maximum current will be 420mA. if I am trying to draw 8V, so that I get 7 after the diode drop, then max current will be 375mA. Is this correct?
Pretty much. Note that the diode drops will allow no more than 7.2V at the capacitor.
My simulator shows the last case will not be as high as 375 mA. Perhaps not as much as 300.

The way I get the source spec is that the diodes will withstand a surge of 50A, that is a very short burst of current of up to 50A, if the burst is long they'll fry, and if the burst needs more than 50A, they'll also fry. Isn't it this way?
Correct.

so would you go with the 9V 5VA or the 6V 3VA transformer in the above example if I need 300mA?
If it were me I'd get the 6V. I've run a simulation showing the 6V will be adequate.

So much the better if it turns out to be 6.3. That's been a traditional transformer value.

The diodes are only dropping .84 V. (Not 1V.)

The 7805 will have about 1.8V higher input than the output. Datasheet graphs show that as the minimum margin to yield regulated 5V.

This is your chief obstacle. The 7800 series require a couple volts higher input than the spec output in order to maintain regulation.

Be advised a 6V transformer will give you borderline feasible operation. If your load draws more than 300mA, you'll get less than 5V from the 7805.

If your house voltage is below norm, you'll get less than 5V from the 7805.

On the other hand TTL devices can operate down to 4.75 V.

And you can always get slightly higher output V by installing a pass transistor in parallel with the regulator.
-----------
 
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csdave

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Hi all,
I tested my circuit on ltspice. I modelled the 9V transformer secondary with a 12.73 peak sine wave generator at 50Hz. I then put a 35 ohm resistance that gives me around 300mA.

To my surprise, the RMS current through the generator is 585mA... almost twice as much as what I expected.
Can anyone explain to me why?

Here is a screenshot



the red line is the current through the resistor
the blue line is the voltage across the generator
the green line is the current through the generator

thanks!!!

Davide
 

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I magnified the image and was able to see that the zero V axis isn't aligned with zero A axis.

As a result the green A plot is centered at +1V. It should return to zero V between current bursts.

Could that be where extra current is accidentally showing up?
 

csdave

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well it does not surprise me that the current is not zero between bursts as the resistor is drawing some from the diode bridge and some from the capacitor that discharges between bursts.

but I guess the issue is that the capacitive component of the circuit is too important
 

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Well, I did a bit of testing and a few calculations and it all makes sense. The capacitor is associated with quite a lot of reactive power in the circuit, so all my calculations in the opening post were wrong. the 6V 3VA transformer will never be enough.
If I put a 6V supply in the simulator I get an RMS current through it of 800mA with a 1800uF cap, which is not enough to get the little ripple I need.
With the 9V supply, the 470uF cap is enough, but my current is 584mA: so this would mean I need a 5.3VA transformer.
What I have is a 9V 6VA transformer. Is it enough or is it too close to the limit?
 

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I really don't see how the current can be that high. The average current flow can be no more than your load pulls.

The diodes can conduct only at points in the cycle where transformer voltage exceeds the charge on the capacitor (factoring in 2 diode drops).

The result is that diode conduction occurs at no other times than those brief bursts of current in your time graph.

In between those bursts the current flow has to be zero (unless you have a blown diode).
 
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csdave

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current between bursts is zero. The current scale is on the right hand side of the plot.

---------- Post added at 08:16 ---------- Previous post was at 08:08 ----------

average current going into the DC side is 302mA, but rms is 584mA.
 

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