bipolar power trans questions

[9 volts]

Hello, I've got a little dilemma, I've made a biopolar power board with a diode bridge and 7815/7915's etc. The problem is that my trans is a centre tapped rig with 25v ac each side. Is there a good way of reducing the voltage (eg would capacitor / resistor /capacitor ?) to get the voltage down to a more suitable level for the 78..series. After the bridge calculations put the dc voltage up to 35.5 which is right on the threshold of the max input for the 78..series. Or do I have to buy another transformer? Thanks

[9 volts]

I have just read RG's page on bipolar
'Power Supply Basics for Effects'
Quote" Now to the regulators. Three terminal regulators are a breeze. They can with stand up to 40V on their input terminal, they put out the specified output voltage within about 5%, maybe better, and are self limiting for current and thermal problems."
So maybe the voltage of my trans would be ok?

[.Mike]

I've been studying up on regulators and power supplies lately for a multi-effects project I'm working on. I haven't seen anything on specifically what happens if you put too much voltage on the input of the regulator, so I'm not sure if it would handle the extra voltage, if it would shut down, or if it would be destroyed. Maybe someone else knows?

My concern with providing the regulator with so much voltage is that it has to burn off the difference between the input and output voltages as heat. Depending on the current you are pulling, you may need a substantial heatsink to keep your regulators from shutting down.

You're feeding the 7815 with 35.5V and the 7915 with -35.5V? If so, doing a bit of math...

35.5V - 15V = 20.5V to be burned off for each regulator.

Current = Power / Voltage. So, let's see how much current you would need to pull for every watt of heat dissipated...

Current (amps) = 1 (watt) / 20.5 (volts).
Current = 48.78 mA.

For about every 50mA you pull from the regulators, the regulator needs to burn off a full watt as heat. Pull 100mA and you are burning 2 watts. which I believe would put you squarely in the "use a heatsink" range. Try to pull 500mA, half of what the regulator's theoretical max, and you are burning off 10 watts!

So yeah, if you plan on pulling any sort of substantial current from this setup, you might want to do some more research on your options to reduce the voltage prior to the regulators, and make sure you have adequate heatsinking.

I hope that helps (and is right... heh).

Mike

[PRR]

Do you need it to last forever? Get another transformer.

Also: read the datasheets, not stuff you find on the Web. Not even from well-informed folks like R.G. The TI sheet for the 78xx regs says that the 7824 can take 40V, all others are 35V max. It seems likely that they are all the same process/part with one value change. Maybe R.G. has insider confirmation that this is true. However if you put 35.1V into a TI-made 7815, and it dies, TI won't take your complaint. (Not that TI really cares about people who buy less than a truckload; just saying.)

That said: I've done just what you did, 35V-36V on "35V" regulators, and they last many years. I know the datasheet specs has some leeway for idiots like me. I also know how to use a soldering iron and fix it in a few minutes. I would be much less reckless if I was selling a million to non-tech buyers.

Also: I know there is a thing called "40 Volt process". A standard recipe for chips to stand about 40V, different from "5V process and "high voltage process". Most 40V process parts are specced 36V (+/-18V). Or 35V here. There are other parts rated 44V, which are surely "40V" parts carefully made. And ALL parts are "carefully" made today compared to the Early Days when chip-making was young and pretty iffy.

> not sure if it would handle the extra voltage, if it would shut down, or if it would be destroyed.

Many ways to fail. Shut-down is rarely the answer: there is always some voltage which will Break Down a "shut-down" junction. When a junction goes into breakdown, current flow can become Infinite, or very-very-high for long enough to melt the chip.

There is also electromigration. I "had" to run some 28V parts at 35V. They ran for years, and failed. Replaced, ran more years, and failed. High voltage gradient across the small distances on the chip makes atoms move. Can be the aluminum interconnects, can be semiconductor atoms, whatever. When too many atoms are out of place, it stops working.

That's why I say, try it today, but for long-term health, use another transformer.

There is also, as Mike says, the heat from blowing-off excess voltage. The 78xx/79xx are very good about shutting-down when hot, but that still ruins the gig.

[9 volts]

Thanks for the info, I did test it and it worked fine, but yeah was thinking about the long term effect. Why not add a resistor across the rail and one down to ground (voltage drop) before it hits the 7815. Eg 100k would maybe drop it by about half? (maybe place it between two large caps before it hits the 78.15 etc), Much like the way amps reduce voltage for different parts eg pre...Long term I'll look out for a trans  but for now.....

[R.G.]

I did test it and it worked fine, but yeah was thinking about the long term effect. Why not add a resistor across the rail and one down to ground (voltage drop) before it hits the 7815.

Not a bad idea. Power wasting resistors are a time-honored (if not particularly power-efficient) way to keep power generation in active regulators down.

Eg 100k would maybe drop it by about half? (maybe place it between two large caps before it hits the 78.15 etc), Much like the way amps reduce voltage for different parts eg pre

But this won't work except for truly microscopic currents.

Remember that Ohm's law thing? V=I*R? If you have a voltage source V and you put a resistor divider R1 and R2 across it, then the output voltage with no load is Vout = Vin * R2/(R1+R2). No problem so far. But to any load, this acts like a voltage source of Vout with a resistor of R1 in parallel with R2 in series with it. So the voltage sags through this equivalent resistance when you pull any current.

If you use 100K/100K to drop 38.5V, you get 19.25V in series with 50K. If you pull 50ma through this, you get a loss of V = I*R of 0.05A*50K =2500V. Ooops... that's bigger than 38.5V, which means this is impossible. In fact if you short the middle of the divider, you can only get I = V/R = 19.5/50K = 390uA.

What you need to do is lower the resistors to get the voltage drop from the dropping resistors down. Doing this takes some math and some electronics.

If you want to provide no less than 17V to your 15V regulators (because the datasheet says they need at least 2V higher input than they're providing output) then you need to find some combination of resistors that divide 38.5V down and also keep the regulator from overheating; also, you will find that the current through the resistors to ground are a problem. They eat a current which loads down your bulk power supply but provide no output to your load at all.

You're free to pick any two resistors. They need to lower the voltage to the regulator to keep it cooler, but also not waste much more of your transformer's power than you can afford. The pertinent design equations are in this post.

[9 volts]

Thanks RG, this has given my some insight.
Using this webpage
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html
I determined that if R2 (parallel to load) is 1m, and the input voltage is 35.5, assuming the load (this is the bit thats new to me) is 50k, a 50k resistor running in series with a value of 50k would bring the unloaded voltage to 33.8k and the loaded voltage down to 17.3 volts. But thats assuming the load is 50k. I have seven pcbs connected in series. I'm wondering if I could use a a trim as R1 with a value of 50k as a means of working out what the load actually is? I could be right off the track with this as the 1m may consume to much current.

[PRR]

> assuming the load (this is the bit thats new to me) is 50k

You have not said (seem to not-know) what your load is.

Mike pulled this example out of his hat: "For about every 50mA you pull from the regulators".

That probably isn't the "right" answer. But it probably isn't very far wrong.

If a load (your "seven pcbs") takes 15V and 50mA, it acts a lot like a resistor of 15V/50mA= 300 ohms.

Note:

300 is a lot less than the 100K and 50K numbers you have been penciling. Whatever conclusion you find with 100K is probably way-wrong for loads nearer 300 ohms.

If we assume just a 35V starting point, your 100K series dropper, and a 300 ohm load (forget the regulator for a moment), the divider ratio may be figured from R.G.'s post. It's about 0.2V, which is obviously not a satisfactory result.

Bah, forget high-value resistors. Use a 1 ohm resistor and a 2 ohm resistor to divide 35V to about 23V. A 300 ohm load hanging on the 2 ohm leg makes hardly no difference at all! BUT the total 3 ohms hanging on a 35V source pulls over 35V/3= 10 Amperes. You probably have a 0.5A or 1A transformer. The 10X overload will make it very unhappy. Although unless you use gigantic 1r and 2r resistors, the resistors will smoke violently.

[9 volts]

Ha, the joys! When I was doing calculations I noticed that the lower the resistors the more drain on the transformer. Thats why I up'd them. The solution (easiest) seems to buy a better suited trans, the trouble is that the only one I seem to be able to get locally is a centre tapped
24 v 150 mA (australia) and I'm not sure it would give out enough current. (basic modular synth). That said it's a situation like this that brings on some learning/better understanding. Thanks again everyone.