questions about increasing voltage to power supply

[bonedoc]

I created a flyback converter. The primary side of the transformer and switching mosfet are directly connected to the rail. The rail was always 8 AA batteries, so this never went over 12V. I wanted to experiment with rechargeable options, so had a custom battery made.The rechargeable battery is Ni-Mh and when fully charged is 14.5V.

Well, this made my converter a beast! However, I noticed some jittering in my circuitry. I could smell something getting warm, and it was the 7A PTC I placed on the rail to prevent over current. I tested the current draw on it, and for the most part, it was only at 3-5A. However, there is a fraction of a second where the beast jumps to 20A. So, the PTC was clipping and causing the stutter.

The mosfet was not warm at all, and is designed for this current. So, I thought I would see what happened when I removed the PTC and jumered it with some thick copper. This ended up burning the trace between the transformer and the mosfet. This trace is .120" wide of 1.25 ounce copper, and it is in an external layer. I believe this should handle 8.5A continuous.

So, I need some opinions:

-What is happening in the difference between a 12V rail and a 14.5V rail? The capacitors on the secondary side are charging faster with a 14.5V rail. So, I assume I am drawing more current and burning things up?

-I understand that the traces are rated for about 8.5A continuous. But, there is a chance that a spike is acceptable for a brief second. Where should place the PTC value/. Under 8.5A or at it?

-Should I just say forget it and have another battery made that does not go over 12V?

 

[(*steve*)]

Your ON time is too long. The transformer is saturating, the inductance falls to a low value and the current rises rapidly.

Reduce the ON time, either by increasing the frequency or by limiting your duty cycle.

 

[bonedoc]

-I thought that if this were the issue my mosfet would also be hot? It is not even warm.

-Lets say it is the on time issue. If I have a 14.5V rail and adjust this, I assume the secondary side caps will charge faster. Does this not indeed mean more current is happening? I think I can have a 24V rail with this transformer...but it seems to me that my trace is just too small. You think a .125 wide 1.25 ounce trace can handle the 14.5V if I change the duty/frequency?

 

[(*steve*)]

-I thought that if this were the issue my mosfet would also be hot? It is not even warm.

It may have a resistance much lower than the DC resistance of the transformer.

-Lets say it is the on time issue.

Let's not assume. You should find out.

The easiest way is to measure the voltage across a current sense resistor.

If the voltage moves smoothly up in a ramp, the transformer is not getting saturated. If the curve becomes steeper, this indicates the inductance is falling and the transformer is nearing saturation. If the trace suddenly starts rising much faster, the core is saturated.

Note that the current sense resistor could be that short trace between the mosfet and the transformer.

Removing the PTC resistor would have increased the voltage across the coil for a given current, leading to a faster rise in current and the saturation point being reached much more quickly.

If I have a 14.5V rail and adjust this, I assume the secondary side caps will charge faster. Does this not indeed mean more current is happening? I think I can have a 24V rail with this transformer...but it seems to me that my trace is just too small. You think a .125 wide 1.25 ounce trace can handle the 14.5V if I change the duty/frequency?

If you increase the input voltage you will make the transformer saturate earlier and do even more damage.

The inductor has a limit. The datasheet should give you some values for maximum current and saturation current etc. You must remain below the lower of these (preferably by some margin)

The current rises linearly with time (if the applied voltage and inductance remain constant -- which they are not) Reducing series resistance and/or increasing the input voltage will increase the rate of increase of current and thus reduce the ON time before the transformer saturates.

Depending on the mode of the device, successive ON periods can pump the current higher and higher (this happens if the primary current does not fall to zero during the OFF time)

If you increase the input voltage or reduce the series resistance you will need to reduce the ON time, which may allow you to also increase the frequency. The current in the inductor determines the amount of energy stored, the frequency determines how often you can dump this into the output capacitors. For a given inductor current, increasing the frequency will increase the rate at which energy i transferred and thus the speed at which the capacitors will charge.

Think of the inductor as a bucket and the mosfet as a valve. If you leave the valve open too long the bucket will overflow (this is saturation and this is what is most likely happening). If you increase the pressure in the pipe, or remove something that limits the rate of flow, then the bucket will fill faster and will overflow sooner.

 

[bonedoc]

Thanks for the long explanation! You sure are smart. When I originally did calculations I based everything off a 12v rail and 300v output. Many parts can handle higher ratings....only unfortunate part is that the snubber was pretty specific in values and I'm afraid if I change the switching voltage and frequencies that my snubber values will be off and I already ordered in the hundreds.

When using a flyback, you sure get spoiled when you try highers input voltages and see what they do. It is easy to forget how current limits are different for wires and magnetics. My first assumption was just not enough copper.

I will do some testing when I get home. Changing the duty is simple enough. I fear that my current rating in the inductor may soon be surpassed, but the companies I have seen allow a 24v rail, so maybe not.

I may want to have them also make another battery. 10 cells making 14.5v is 1.45v per cell. Maybe I can have an 8 cell made and things will be comparable as to when I had a 8 AA battery pack.

 

[bonedoc]

One question. If I have done calculations of frequency/duty where there is a 12v supply and 5A passing through the inductor.... I cannot charge any faster, correct? If I understand you, raising the voltage will need shorter on times, an the goal is not to pull more current than the inductor can handle.

In other words, despite the voltage, duty and frequency....if all are set according to the max current the inductor can handle, won't the output be the same?

If I am pulling max current at 12v, I can't charge any faster if I pull max current at 24 volts, otherwise I would be explodig the bucket. Right?

 

[nepow]

Steve has the right answer... otherwise you could try placing diodes with an appropriate current handling capacity in series between your batteries and supply. The forward voltage drops may just overcome your problem!! not the best idea... but may be worth a try.

 

[bonedoc]

Am I correct in saying that it will charge the same speed weather I am at 12v or 24v if I am at the optimal frequency and duty for that voltage, as I would be pulling the max current in either situation?

 

[(*steve*)]

A higher input voltage will result in a higher di/dt in the inductor since V = L*di/dt.

L remains the same, and time passes at the same rate , so the rate of change of current is proportional to voltage.

As the current increases beyond the point at which the core saturates, the inductance (L) drops dramatically and thus the rate of change of current increases *much* faster.

The best way to operate the circuit is to monitor the inductor current and to leave the mosfet on only as long as it takes to reach a certain inductor current. That way your circuit will be largely independent of supply voltage.

 

[bonedoc]

I wanted to post this calculator I used for the calcs:

http://www.daycounter.com/Calculators/Flyback-SMPS/Flyback-SMPS-Calculator.phtml

For whatever reason, it keeps giving me 600kHz frequency, and the duty cycle units make no sense. So what I did is just used the ON time calculation. For 12v, the may on time is 4.17us. With a 50% duty, that is an 8.34us period =1/.00000834 = 119,904 = 120kHz.

To keep it away from continuous mode or saturation, I did this frequency at a 40% duty at 12V.

A 40% duty at this frequency gave an on time of 3.34us, which is below what the online calculator recommended. That was 3.45us.

I attached pics of my results. Interestingly, the 14,5V version is pulling less current, according to the calculator, but I understand how you say that current will raise faster under higher voltages. Maybe I need to check my code. There is a chance my waveform is backwards and may be active low.

 

[(*steve*)]

You are right, some of those figures make no sense.

However your interpretation appears sound.

 

[bonedoc]

Ok, here is what I have done. I had them make a new battery and it is just below the 12V supply I have used. In the past, I used 8 AA batteries. It puts out right at 12V when new. I am now using 8 rechargeable NIMH batteries that are putting out 11.63V when fully charged.

Guess what...this causes massive overheating (not as bad as the 14.5V did) !

Why would 2 different batteries at the same voltage be that different? Do you suppose that the NIMH batteries are putting out more current than the standard AA, and this is allowing more current in the system that was not attainable before....allowig me to not notice my problem before?

I thought I would show you a video a scope placed at the gate, so you could see my duty cycles. This is 12V AA supply in the video. The device charges extremely fast, so to prevent overshooting of the target voltage, I have 3 different duties so that slower charging happens if there is a lower target voltage. I show all 3 duties in the video.

Interesting...my code is a 40% duty, but I created a 160kHz duty on accident....so my periods are even shorter than I thought. If my signal is accidentally inverted at this frequency and I have a 60% duty, this is 3.75us. At the 14.5V supply voltage, this is not enough, as stated in the calcs I would need at <3.34us. This makes sense.....


But, on the 12V version it is is way less than the max on time of 4.17us. However, this only leaves .0000025us to fall. So, the question is: why would 2 different battery types at the same voltage cause different behavior? One overheats and the other does not. More current in one, and not enough falling time?

 

[bonedoc]

And here is a link in which I inverted the signal....guess what? The NiMH works fine since I have done this. Also...when I charge at high voltages, it stays way cooler than low voltages. Thats because the lower voltages are going into saturation now with the inverted signal.

Thanks! I think you got me on the right track. Im nervous to do it, but I am going to try to adjust this to fit the 14.5 supply and see what happens ;-)

new duty cycle

 

[bonedoc]

Ok, so here is the deal. It looks like the 12V NiMH is working fine with adjustment of the frequency and duty. But, I cant seem to do anything to make the 14.5V work. With the resolution, I cant hit the 6.9us (144,928Hz) period exactly. The next lowest programmable is 142,800Hz and that was measured with a frequency analyzer. The duty is 40%. This is still getting insanely hot on the 14.5V.

I believe these are good numbers. I dont know what else to do. I actually tried a 1MHz frequency a a 50% and it STILL gets hot. I dont see how the bucket could be full....unless I am not emptying it long enough. Any advice from here?

 

[(*steve*)]

At those high frequencies you are probably not allowing enough time for the mosfet to turn on (and possibly off).

This means it is operating in its linear region (i.e. between on and off) for longer and longer (as a proportion of time) and you are most likely seeing the effects of switching losses.

 

[bonedoc]

good point! I actually just tried a super long period with a very low duty and it actually worked without getting hot! But...was a low charge....just need to find the limits ;-)

 

[(*steve*)]

As I said, keep the frequency the same (or close to the same -- 100kHz is probably fine) and alter the on time to suit the input voltage. Better to do this automatically, but if you want to do it manually...

 

[bonedoc]

AWESOME!! Its working great! Thanks for the help! The 8 cell AA charges to 300V in 10s. The 8 cell NiMH is 7 seconds. The 10 cell is 2.5-3 seconds!

I had to just test each duty bit by bit. Even though it is 10 bit resolution, there is a 1-2 bit resolution that puts it over the boarder. I am doing 100kHz right now, and I think that is more than enough time for switching losses, and I also think there is more than enough time for the current to drop in the transformer, so I may keep the current ON time and decrease the period. Thanks again!

 

[bonedoc]

2 related questions:

-is there a risk in setting up something at a 14.5V as I just did and then later using 12V with the same parameters? I assume so, but assume it would just be slow.

-my current switch has a GS voltage of +-15V. Is there a danger in driving it with 14.5V? I could always fins a 20V mosfet, but like what I have.

 

[(*steve*)]

It all depends on what is driving the mosfet. If it is a device running from 5V, then no.

However if it runs directly from the supply voltage then you are getting close.

Sometimes you'll see a zener diode (say 12V) placed between gate and source after the gate resistor. You need to be careful and ensure that the higher current which flows if something tries to pull the gate higher than the zener voltage does not cause additional problems.

edit, oh, and yes -- set it up for 14.5V and run it from 10, it should just be slower.