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Power Supply

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On the dwg I posted you will see the EARTH is used, connect the E wire to the bolt/screw which fixes the transformer to the psu chassis.

Have you got the mating IEC wired cable socket to suit the chassis mount connector you have bought.?

On the back of the chassis connector 'N' goes to one of the wires on the transformer primary, the 'L' on the connector goes to the other transformer primary winding.

As its a double pole switch I would have thought the 'N' would be internally switched on the connector.

The only thing connected internally is the fuse to 'L'. The switch is just in the same casing. You could think of it as being completely separate (as if I bought it as a separate component).

I don't have a complete chassis yet, as I want to try it on the breadboard first. Should I wire the Earth to the metal casing of the transformer for now?

I don't understand what you mean about the 'mating IEC wired cable socket', etc...
 
I don't understand what you mean about the 'mating IEC wired cable socket', etc...

Its the mains lead socketed cable that connects into the chassis mounted plug.

EDIT:

This IEC cable.

**broken link removed**
 
Last edited:
I assembled this circuit this morning, and it mostly works!

However, I only seem to get 6.81v output, which I don't understand.

The rectifier is a L7809CV.
I'm using the 12.5v connectors on the transformer (although it's actually producing a little over 13v).

Does the type of smoothing capacitor matter? It's definitely 0.33uF, but could I have the wrong type?

The resistor and LED do cause a bit of voltage drop. If I remove them I get an output of about 7v.

How can I get a nice steady 9v output?
 
I have found my answer, though I don't understand it...
The smoothing cap I was using was 0.33uF, which is what they were using in the datasheet.
This was far too small. I swapped it with a 47uF cap, and all is now good.
Why is this happening? Is there a formula to determine the correct smoothing cap?

Thanks
 
I have often wondered where the over-voltage goes when using voltage regulators. A 7809 allows 9volts to pass to the curcuit side and dumps some excess voltage as heat . Does the rest just go to ground until a bigger current draw is made?

This seems like a waste and I have seen 7805 s fed with 15 Volts in some circuits which seemed silly.Where does the energy go?
 
I assembled this circuit this morning, and it mostly works!

However, I only seem to get 6.81v output, which I don't understand.

The rectifier is a L7809CV.
I'm using the 12.5v connectors on the transformer (although it's actually producing a little over 13v).

Does the type of smoothing capacitor matter? It's definitely 0.33uF, but could I have the wrong type?

The resistor and LED do cause a bit of voltage drop. If I remove them I get an output of about 7v.

How can I get a nice steady 9v output?

Mark,
Your cct, which I edited for you, clearly shows 2 electrolytic capacitors, which are essential for correct psu operation.

It will not work correctly with a 0.33uF.!!1
 
Mark,
Look at this image of your psu operation.

The left image is with a 0.33uF and the centre image a 1000uF, with a load current of 0.5Amp.
You can see on the left image the 0.33uF cap has no smoothing effect and is applying a full wave rectified voltage to the 7809 input.

The rightside image is when the load current is 1Amp, using a 1000uF.

Dont remove the 0.33uF, add the 1000uF in parallel as shown in the edited dwg.
 

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I have often wondered where the over-voltage goes when using voltage regulators. A 7809 allows 9volts to pass to the curcuit side and dumps some excess voltage as heat . Does the rest just go to ground until a bigger current draw is made?

This seems like a waste and I have seen 7805 s fed with 15 Volts in some circuits which seemed silly.Where does the energy go?

The excess voltage is a voltage drop across the regulator.

Take a 12 V circuit, with a battery, a bulb and a switch. The switch is on or off. When the switch in on there is no voltage drop across the switch, and the bulb has 12 V across it and it lights. When the switch is off, the voltage across the switch is 12 V, so none is left across the bulb.

Now put a variable resistor instead of the switch. When there is a lot of resistance it is like a switch that is off, so there is 12 V across the resistor and none across the bulb. When there is no resistance it like a switch that is on, with no voltage across the resistor and the 12 V across the bulb.

So far, it is just like a switch. However, if you put some resistance in the way, there will be some voltage across the resistor and some across the bulb. You could adjust the resistor to get 9 V on the bulb, and there will be 3 V across the resistor. You can change the resistor to get any voltage you want on the bulb.

That is all that a linear regulator is doing, just automatically. It adjusts its resistance to give you the correct output voltage.

If you have a 7805 fed from a 15 V supply, there will be 10 V drop across the regulator.

The power loss in the regulator is the 10 V drop multiplied by the load current. If there is no load, no power is lost. If there is a 0.1 A load, the power loss is 10 V * 0.1 A = 1 Watt which is lost as heat in the regulator.
 
The excess voltage is a voltage drop across the regulator.

Take a 12 V circuit, with a battery, a bulb and a switch. The switch is on or off. When the switch in on there is no voltage drop across the switch, and the bulb has 12 V across it and it lights. When the switch is off, the voltage across the switch is 12 V, so none is left across the bulb.

Now put a variable resistor instead of the switch. When there is a lot of resistance it is like a switch that is off, so there is 12 V across the resistor and none across the bulb. When there is no resistance it like a switch that is on, with no voltage across the resistor and the 12 V across the bulb.

So far, it is just like a switch. However, if you put some resistance in the way, there will be some voltage across the resistor and some across the bulb. You could adjust the resistor to get 9 V on the bulb, and there will be 3 V across the resistor. You can change the resistor to get any voltage you want on the bulb.

That is all that a linear regulator is doing, just automatically. It adjusts its resistance to give you the correct output voltage.

If you have a 7805 fed from a 15 V supply, there will be 10 V drop across the regulator.

The power loss in the regulator is the 10 V drop multiplied by the load current. If there is no load, no power is lost. If there is a 0.1 A load, the power loss is 10 V * 0.1 A = 1 Watt which is lost as heat in the regulator.

Yes OK I think I understand that.
The resistance is heated and the energy goes as heat in the drop across the resistor example .
Thats one thing . But the 7809 has a third connection which goes to ground. Does any of the excess voltage go straight to ground?
 
Yes OK I think I understand that.
The resistance is heated and the energy goes as heat in the drop across the resistor example .
Thats one thing . But the 7809 has a third connection which goes to ground. Does any of the excess voltage go straight to ground?

You are confusing voltage and current.

Ground is, by definition, zero volts.

In any linear regulator there will be some ground current. The ground current is usually very small and does not add significantly to the power dissipated in the regulator.

For example, a 7805 will use about 5 mA. If it is supplied at 12 V, and the load is 600 mA, the voltage and current at the connections are:-

Input. 12 V, 605 mA in
Output. 5 V, 600 mA out
Ground 0 v, 5 mA out.

The power dissipated will be:-

(12 - 5) * 0.6 + 12 * 0.005 = 7 * 0.6 + 12 * 0.005 = 4.2 W + 0.06 W = 4.26 W

As you can see, it is only a tiny fraction of the power comes from the ground current.
 
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