LM7812 input voltage

I'm building a power supply with a 2x18 Volt transformer. The output shall be above the max 35V input for the LM7812.

Would this solution be OK for feeding the LM7812 regulator?

The 12 V shall feed a small fan, a thermostat and a voltage/ampere display. Estimated consumption is max 200mA

You are really pushing your luck. 23V at 200mA will dissipate 4.6 Watts. You would be better with a switching converter unless dealing with low level audio or RF.

Current is measured in Amperes (Amps), so glad to see that you haven't used that awful made up word "Amperage". Always makes me cringe when I see that.

I suspected that. Perhaps a PWM DC-step-down would be better.

But I assume it will work if the 7812 is replaced by a DC-step-down module?

thehardwareman:
But I assume it will work if the 7812 is replaced by a DC-step-down module?

Certainly should. One rated at least 35 V to be sure.

The 1N4007 is quite unnecessary.

Paul__B:
Certainly should. One rated at least 35 V to be sure.

The 1N4007 is quite unnecessary.

Why is the 1n4007 not required? It's a DC step-down unit. It does not take AC in.

It is ALL wrong! The correct way to do the rectification is with two diodes and a ground centre tap. All of your supplies then come from the junction of the two diodes, regulated as required.

I take it that you don't actually need 50V for anything?

Its clever!

In operation one end of the trf is clamped to 0V approx by a diode. The other at +V between 0 and 36 sqrt2 .
So the mid-point is full wave rectified 18V ac. peak 23V approx. and will drive the 7812 no bother.

If on-load you have 20V and 12V out at 200mA you are dissipating 8 * 0.2 = 2W which will need a small heat sink.

If your load is roughly constant you can calculate C to reduce the voltage a bit - see here how to calculate your capacitor.

http://www.skillbank.co.uk/psu/smoothing.htm

AJLElectronics:
It is ALL wrong! The correct way to do the rectification is with two diodes and a ground centre tap. All of your supplies then come from the junction of the two diodes, regulated as required.

I take it that you don't actually need 50V for anything?

I realize that it is wrong when I see your drawing. However, the main design is as shown with the reflector bridge and the 6x 1000uF caps.

And yes, I need up to 50 Volt. It's a power supply that shall heat water pipes so that they do not freeze during winter. What the kikad drawing does not show, is that there is control unit (controlled by an arduino of cause) that regulate the heat.

In total there are 7 heating wire with heating wire around the water pipes (already installed) together with one NTC resistor for each heating wire.

To get enough heat I must have 35-40 volt out (less when warmer). The transformers (total of 3 2x18V 3.2A AC) are also "in house" (bought 30 years ago but never used). So I'm shucked with this design.

Perhaps the best is to add a single 12V supply to drive the components that require 12V, but I was hoping to avoid it.
Each transformer will drive 3 heating wires (the last one only two).

36VAC full wave rectified will be about 32.4VDC, the capacitors will charge up to peak of 1.414 times that, 45.8VDC, but load current will drag that down. What is your load current? What is the transformer's VA (Volt Amp) rating?

Why not use AC for the heater.
Leo..

Sorry hardwareman I was giving you undeserved credit.

It is ALL wrong! The correct way to do the rectification

.. is however suits your requirements effectively.

Let me explain why I said it was clever. I've redrawn the fwr bridge to make the explanation clearer.

The diagram shows one half of the cycle

cct.png

during the cycle when a is positive relative to b:

point b is clamped by D4 to a voltage very close to ground. Point a is positive with respect to ground and at the peak is + 36 * 1.4.. volts.

The centre tap is at +18 * 1.4.. volts.

On the other half of the cycle
point a is clamped by D3 to a voltage very close to ground.
Point b is positive with respect to ground and at the peak is + 36 * 1.4.. volts.

The centre tap is at +18 * 1.4.. volts.

So since the 18V load is very low compared to your heaters you CAN take a full-wave rectified 18V from the centre tap.

cct.png

johnerrington:
Sorry hardwareman I was giving you undeserved credit... is however suits your requirements effectively.

That's OK. I should be ashamed of myself (regards my first drawing). After all I'm suppose to be an electrical engineer, but after I completed 2 years education back in 1985, I never practiced - the computer came and I was lost in programming...
Just a year ago since I started with Arduino and electronics again - as a hobby.

johnerrington:
So since the 18V load is very low compared to your heaters you CAN take a full-wave rectified 18V from the centre tap.

That was on my mind when I saw your first drawing. So I'll consider both another full-wave bridge or an external 12V power supply that can give me 12 V for all three boxes. Using an external power supply (re)move the heat problem from the boxes with the transformers.
And since I might get a problem with space in the boxes, an external 12V power only require a connector, which occupy less space.

So, the original circuit is fine in regard to connecting that transformer and obtaining half the DC voltage from the centre tap - the additional rectifier is not necessary.

The "split" supply is more commonly used with the centre tap grounded so the bridge gives positive and negative supplies. You can see that a diode in the centre tap makes no sense.