Genius! Please tell me the theory which apply on this

I came across a similar supply in a hi-tech refrigerator. The series resistor burnt when it was used with a modified sine wave inverter ( because i = C dv/dt) and I used the information in AN954 Transformerless rc psu to calculate the value. These supplies usually have a 1 watt or higher zener diode connected directly to the output of the bridge rectifier to limit the peak voltage.

Timescope

MrAl said:

4. You can not specify just a time for charging an Li-ion battery at a known current unless you know the capacity of the
cell. The capacity of the cell goes down with age so it may be hard to judge this parameter.
The best i can do here is tell you that in order to avoid problems the voltage has to be monitored and charging has to stop
when the cell reaches 4.200 volts, but a better safer voltage is 4.150 volts which takes into account the inaccuracy of some
meters. The very high efficiency Li-ion cells can go up to 4.35, but that's a different kind of cell really and you better
be sure you have that type before you allow the voltage to go as high as that.
Now theoretically if a cell has capacity of 1AHr and it is totally discharged, then we can charge it with 100ma for 10 hours
and it still wont be charged to full capacity so we're safe. But if we misjudge some measurement or something else happens
that we didnt think of it could mean disaster. So avoid any charging of Li-ion cells unless there is some way to monitor
the voltage.
I've actually done this in the past, but i was willing to sit there and watch the volt meter waiting and waiting for the
voltage to get close to 4.150v, but that's because i was doing a test and i didnt mind sitting there all that time. That's
what it takes though, unless we use a monitoring circuit of some type. These days the circuit of choice is usually the
microcontroller chip, one type or another.


So just to put this into perspective for you...
You can get away with overcharging several types of batteries, but not the Li-ion type. That also goes for discharging
because you can not discharge the Li-ion too low either without risk of damage in the form of electroplating of the
electrodes which makes the two terminals of the battery physically closer together and thereby increases the risk of
internal short (an external protection circuit does nothing here to stop this either). An internal short causes fire and
possibly explosion.

Click to expand...

Lithium Ion batteries are dangerous for more reasons than being overcharged. They also 'blow up' when you over drain them. A 'dead' lithium ion cell is about 2V. Draining them beyond this liquifies the metal and when you recharge they tend to blow up. This is one of the main problems they were having with the early liion batteries. The good thing about Li-ion cells (unlike other battery types), is that they hold their voltage for about 90% of their charge while being drained, and then the voltage falls off rather sharply.

Li-ion cells are either 4.1V, or 4.2V fully charged, depending on their configuration. They are charged at a constant current until they hit about 90% charged, then they are charged at constant voltage (4.1 or 4.2V) until they are fully charged, with the current going down until it hits about 5% and they are considered full. We built a charger/monitor that charged and monitored a stack of 4 cells based on the Linear Technologies series of Li-ion charging ICs. We were awarded a patent for the way we monitored not the stack, but each cell independently. And yes, we had a fire in our lab within the first few days of testing (the manufacturer asked, "what, was that your first explosion?"). The project engineer wanted to charge them overnight, but I had convinced him to wait until "bright and early" the next morning. This was back in Jan 2000, when LiIon batteries were still unheard of (mostly). And how we found out about the never less than 2.0V drain condition.

yes i grew up with a lot of "universal" stuff that could be run off of 120V AC or DC. there were TV's, guitar amps, stereos, AM/FM radios. these were called "hot chassis" devices, and yes, were dangerous if plugged in backwards. these devices had polarized plugs to facilitate connecting them the right polarity in both AC outlets and DC outlets. one problem with them was that not everybody would connect AC outlets properly, or people would connect their radio/whatever using a nonpolarized extension cord, and end up with the chassis at 120Vac instead of near ground. these devices used tubes, and the plate voltage was rectified directly off the line (or when connected to DC, the rectifier diode would just pass the DC straight through) and all of the filaments were connected in series across the line (this is why you had weird filament voltages like 50, 33, 27, 17, etc..., the series string always totaled about 120V)

I have a couple of old bush valve radio's, they have a mains 240v power connector on the back, this can be plugged in either way around, one way making the chassis and all the knob spindles on the front live.
Not one of their better ideas, not the kind of thing you'd get away with these days.

MrAl said:

Hi again,


Before i start, i have to ask the question that's on everyone's mind here i think...
That is, why dont you want to use a wall wart type transformer?.

Click to expand...

Thanks again for your explanations !!!

Your Question : Why i need to struggle with a CAP .... yes i agree, there are so many limitations. so i can ask question and get clarified some theoretical matters by showing them to experts like you guys (Electronic is not my main field ...but one of my hobby) if it's a perfect diagram ...no way to ask questions.

i think if i need to go ahead with this, the best and safe solution is to use a circuit removed from a common mobile phone charger (5V . 200mA). the current and the voltage can be limited with resisters and Zener diodes. what do you think of it ?


i have another question of transformers :

A I have a 230V/12V, 3A step-down transformer.

What will happen at primary side of transformer if I attach a 12V/5A device at secondary side? Will primary side attributes (such as voltage or current) change?

you have a couple of options with that transformer. with a bridge rectifier and filter cap, you will get about 15Vdc @3A. you can then use a DC-DC converter to get 5V@ 6 to 8A. if your load doesn't really draw 5A, but 3A or less, you can use a linear regulator to get 5V.

ranatungawk said:

A I have a 230V/12V, 3A step-down transformer.

What will happen at primary side of transformer if I attach a 12V/5A device at secondary side? Will primary side attributes (such as voltage or current) change?

Click to expand...

DO NOT overload the transformer like that. It might catch on fire, which will melt its insulation then make even more heat!

When you increase the secondary current then the primary current will also increase.

audioguru said:

DO NOT overload the transformer like that. It might catch on fire, which will melt its insulation then make even more heat!

When you increase the secondary current then the primary current will also increase.

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Thanks you audioguru!! you have touch the correct point !!

"When you increase the secondary current then the primary current will also increase" --- > is it so ??? when considering a step down transformer, will primary side attribute (such as voltage/current) be changed according to the changes that have been done at the secondary side?

actually i have no deep understanding of the theoretical matters on this area, that's why i need to get clarified these things ...

as far as i know, according to primary theories which apply on transformers :A varying current in the primary winding creates a varying magnetic flux in the transformer's core and thus a varying magnetic flux through the secondary winding. --- > this is OK , but can it act vice versa????

What uncle jed was trying to say is that if you use a switching converter then the current from the trans will increase according to the drop in voltage minus the efficciency of the converter.
In other words if you had 12v at 3a, and you had a converter that stepped this down to 6v, then you'd have a little under 6 amps.

If you use a linear regulator the current out would be the same, in this case 3a.

The power into a transformer is almost the same amount as the power out of a transformer.

With no load the power out of a transformer is nothing then the power into it is almost nothing. Add a load to the output and it draws power. Then the power into the primary of the transformer also increases because it cannot be nothing.

Power is the voltage and current multiplied.

ranatungawk said:

Thanks again for your explanations !!!

Your Question : Why i need to struggle with a CAP .... yes i agree, there are so many limitations. so i can ask question and get clarified some theoretical matters by showing them to experts like you guys (Electronic is not my main field ...but one of my hobby) if it's a perfect diagram ...no way to ask questions.

i think if i need to go ahead with this, the best and safe solution is to use a circuit removed from a common mobile phone charger (5V . 200mA). the current and the voltage can be limited with resisters and Zener diodes. what do you think of it ?


i have another question of transformers :

A I have a 230V/12V, 3A step-down transformer.

What will happen at primary side of transformer if I attach a 12V/5A device at secondary side? Will primary side attributes (such as voltage or current) change?

Click to expand...

Hi,


The basic theory behind the transformer isnt very difficult at all. It starts by understanding what the "turns ratio" is. That tells you a lot.

The turns ratio for a transformer like this is simply the output voltage divided by the input voltage, which is 12/120 which if you do the math comes out to just 0.1 (which is the fraction 1/10). That simply means that when you input a voltage like 120v, you get one-tenth out, which is 12v. So if you input 100v you'd get 10v out. The limitation here is that you cant go over the rating of the transformer, which in this case the primary is 120 and the secondary is 12.

So far we've talked about voltage, but the turns ratio also affects the current. There's a limit here too though that we have to respect. If the output current rating is 3 amps, then we can not go over that rating without risking the possibility of burning up the transformer. So this rating is used mostly for the output, but if you want to calculate the input current then you multiply the output current by the turns ratio which is 0.1 (or the fraction 1/10). Multiplying 3 amps by 0.1 gives us 0.3 amps input current, so that's the input current. We do end up with a little more than that because the transformer excitation takes a little current too, but that's usually a good enough approximation for the simpler applications. What is more is that since the input current we calculated to 0.3 amps then that's the approximate rating of the primary windings.

The idea here is to not exceed any of ratings either the current or the voltage of either primary or secondary. If you want 2 amps out that's not a problem because that is lower than 3 amps (the rating) but if you want 4 amps out that's a problem because that exceeds the current rating of the secondary winding.

So you see it's not too hard to figure out what works and what doesnt. You just have to calculate the turns ratio and then take it from there.

Just to recap the two basic 'formulas' for the transformer:
1. The output voltage is equal to the input voltage times the turns ratio.
2. The input current is equal to the output current times the turns ratio.

If you need more examples that's fine.