Diodes - 24 VAC to 9 VDC

With something like a soldering iron it's always good to be galvanically isolated, since you might be soldering on something connected to ground.

I couldn't quite make out what you're talking about with regards to the power supply, but, if you know the voltage that the display requires, then, yes, you can use an LM317, or any other device to create it.

If it was me, I'd look through my box of old wallwarts to find one that supplies 12VAC or so, and build a supply from that.

To do what you want to do, you need to identify the output drives to the segments. Buffer each of these outputs (you neglected to inform us of how many displays you are needing to copy, was it a single, double or triple???) these with a resistor and a small transistor x 3, one for each of the new displays.

Assuming a 3 x 7 segment display, you need to get a buffered signal from each segment.

Plus you might need to know if the display is multiplexed or not....(this is a method of "scanning" the display quickly to reduce the current and number of lines needing to be driven. Each display is only on for a very short time, but refreshed so quickly, the eye is fooled into thinking that it is continuous) , then you could reduce the number of transistors and the complexity of your project.....

To give you real proper help we need to see the schematic....or at least a scanned image of the display , including the chip/transistor numbers if possible, but no promises mind....

The least of your problems at this time is what power supply to use.......!!!!!

The ICL7106 is driving an LCD display so the current demand of the circuit is very low probably 10 ma would be in the ball park.You can measure this yourself.

Your bridge diodes of 500ma and 100v general purpose would be OK. The 8.4 v zener regulating from 9Vac 250 MW rating would do. If you use an LM317 you would need a 12vac supply a 240 or 120v to 12v transformer is probably more common than 24v to 9v transformer and the smallest one you can find will do the job 6VA rating good quality for high voltage isolation [ as the sensor is in the heater ] is plenty.

That would be 10 mA per simultaneously lit LED. Andy's comment about knowing how the circuit functions is most critical now is right on the mark. (Yes I gave Andy a GA. ) For a three digit seven segment display, depending on how the circuit functions, the peak current could be just 10 mA or as high as 210 mA.

You are mistaken the LC7106 is an LCD display driver and overall consumption would not exceed 10ma. my comments are correct.

You are half right, the ICL7106CPL can drive LCDs directly, but needs interfacing to drive LEDs directly, whereas the ICL7107CPL will drive LEDs directly......

But either can drive LEDs......its part of the description. See:-http://www.datasheetcatalog.net/de/datasheets_pdf/I/C/L/7/ICL7106CPL.shtml

The OP did not really describe things well I feel, so its an uphill job helping him....

Have a great day in spite of me!!

Thank you all for reply.

I have read them all but only now I can give you more details.

The 24vac is the supply for the temperature control IC and heater of my soldering iron. In parallel by using a transformer (labeled as 24/9v) and a rectifing bridge is the temperature display (3 full digit LCD) driven directly by ICL7106CPL. The current in primary is 25mA (full load), in secundary I have about 10.5Vac (and not 9Vac as expected), after the bridge 12.7Vdc and the current is less than 100mA and here I am confused by the diodes witch are colored gray yellow black and a large yellow but it seems are simply rectifiers. All other problems are solved but the 24/9 transformer is hard to construct and as a result I wonder if I can use the LM317 to replace it but this I have to do from 24Vac I already have in the soldering iron.

Regards, CCOAlex

You are not making much logical sense. I get the impression that you have NO idea what you are doing/need to do....

You are spending far too much time on a power supply, for something that you haven't built, why?

If you rectify the 24VAC and apply this correctly to the 317, you can have a preset resistor to allow the voltage to be set up later for what is eventually needed.....

You could build your circuit to copy the data from the original display, power it with batteries till you are sure it works, then design the needed power supply, as you may only need 5 volts......

What you need to do is to get your needs for the extra displays decided and how you want them to work/cable/cables?

We on CR4 can then help with a design that should work.....

Get your wishes down on paper, forget the power supply for the moment.....its like putting the cart in front of the horse......

First, in general terms you can use a LM317 to step down a voltage. Being a linear regulator though you will not have a very efficient conversion. Your 24/9 transformer was providing a reasonably efficient change in voltage. I still think you have other concerns that are more technically challenging than your supply, but I'll focus my answer to your concerns.

I'm not certain of the meaning of your numbers, so I will have to do a few assumptions of what you mean. I will explain these assumptions though in my analysis. (This is where a diagram reduces much confusion.)

Assuming the primary current you specify is the primary of the 24/9 transformer, then the power traveling across this transformer is 24V*25ma=0.6W. So 0.6W/12.7V=47.2ma, this is the current being drawn by your display chip to light your LEDs. The display chip can operate off of 12.7V DC. So you will first have to rectify your 24VAC to DC with some rectifying diode. Next store sufficient charge in a capacitor to provide current during diode off time. And then follow the LM317 data sheet to produce a regulated 12.7V output. My concern comes from how hot the LM317 will be, for the power dissipated by the LM317 will be (24Vac*√(2)-0.6Vdc-12.7Vdc)*47.2ma≈1W. But even if proper heat-sinking keeps this chip cool, this will mean that your primary transformer that produces 24Vac from your supply main must provide more power for the whole circuit. Hopefully it can.