ElCheapo PSU

[autir]

My first circuit.... isn't she a beauty?

[autir]

Your silence implies that nobody knows where to put a fuse in a voltage regulator circuit?  ???

[Staigen]

Hi there Autir

Well, i dont belive she is a beauty, but if it is your first circuit it will do :)
But you will have to do a few changes first >:(!

But first to the questions about fuses. First fuse should be in the mains section of the transformer, its a 9.6 VA rated one, about 80 or 100 mAT(slowblow) will do here, than a second fuse in the secondary, between the transformer and the rectifier, 800 mAT(also slowblow) is the choice here. Thats the fuses.

Then to the changes, first, the rectifier diodes should be parallelled with capacitors(4.7 to 10 nF will do here), next, remove the resistors in the outputs from the IC:s, next, a diode, connected backwards over the regulators will save them, if the input voltage to the regulators is lower than the output from them, 1N4001 will do fine here, next, a 10 uF ellyt parallelled with the 100 nF condensors on the outputs from the regulators should be fine too. Done this, this will be a nice little PSU, but it will surely eat fuses for you, every time you overload the transformer, this transformer is too small for this circuit. Also the output from the rectifier is not enough to the 7812 regulator, when loaded! Dont forget to heatsink the regulators properly! Good luck!

//Staigen

[autir]

Thank you for your reply.

-> Regarding fuses: Since the transformer is 9.6 VA, the input current will be 9.6/220 = 43.6 mA. I understand that the input current will be higher than this because no device is 100% efficient, but isn't 80 mA big? In general, how do we calculate a value that is neither too small (many burnt fuses) nor too big (lack of protection)?

-> Why do we place capacitors in parallel with the rectifier diodes? Why 4.7 to 10 nF? If I use a bridge rectifier IC will I need capacitors?

-> There are no resistors connected to the outputs of the LMs, this was just a test circuit for simulation  ;D

-> You mean a 1N4001 with its P end connected to the IC's output and the N end connected to the IC's input? I have seen such a technique in the LM317 datasheet, and it was supposed to protect the IC in case there was a short-circuit in the input area. A similar pattern is the Figure 33 in the LM7800 series datasheet from ST. In general, how can the input voltage be lower than the output? If we have a short circuit, for example, won't the output also drop to zero?

-> why do we need a 10 uF cap in parallel with the 100 nF in the output?

-> The transformer is indeed too small - but just for the 12V area of the circuit. In the LM7800 datasheet if you study the IC current vs. input-output voltage difference (Figure 5) it shows that for 500 mA the Vi-Vo can be lower than 3 Volt. But this isn't the case, as both in simulation and in a breadboard  prototype I've constructed the 12V value slowly starts to lower above 280 mAh.
What would be the ideal transformer for this circuit? If I'm not mistaken, it should be rated as 0.5*squareroot(2)=0.7 A and [12+3(for the Vi-Vo)+2(for the bridges)]*squareroot(2)/2=12 Volt. What have I calculated incorrectly?

Thanks again  :D

[Staigen]

Hi there again Autir :)

Oops, this was much, shall we take the fuses first. The transformer is rated at 9.6 VA at the output, but the efficiency is only about 75 to 80 %, so the current is going to be at least 55 mA on the input, but with a fuse that are rated at 55 mA it surely is going to blow bcause of the inrush current, also, a fuse that are rated for 55 mA is not a standard value, nearest is 63 mA, but this value is also a little bit on the low side, next standard value is 80 mA, and that i suggested, hopefully this will not blow, but dont be too sure about that, the next standard value is 100 mA, if that one will blow there is an error somewere.

Then we have the fuse on the secondary side, not much to say about that, the transformer is rated for 800 mA so the fuse will also be 800 mA!

Then the capacitors in parallell with the rectifier diodes, it was something about dynamic high frequency ground, it is a site out on the internet somewere, i dont remember where, explaining this. All i know is that they should be there, this was known already in the 1920:ies, i have seen it in old receivers from that era. The value i think is empirically found.

Ok, if there is no resistors connected to the output.

Then the diodes over the regulators. I mean a diode connected with the anode to the output and the catode connected to the input of the regulators. If you have a capacitor connected to the output and the fuse blow or you shut down the PSU, and a load connected to another output, the rectifier tank capacitor is exhausted very fast, then you have the voltages over the regulator reversed.

Then the electrolytics at the output, they reduce the dynamic output resistance(impedance).

Then it was the ideal transformer for this PSU, i think i take that in a later reply, i have to rush, the store is shutting down!

//Staigen

[audioguru]

Hi Autir,
I have seen capacitors across rectifiers in old RF and audio circuits but I have never used them and new circuits don't have them. It was to stop the buzz caused by the rectifiers switching very quickly and spraying high frequency transients around.

If you look closely at the datasheet for the 7812, it is spec'd to provide only 5mA to 500mA output with a supply voltage no lower than 14.5V. It can give much more current but its spec's are not guaranteed. Its minimum dropout voltage is 2V with a 1A load but its voltage regulation is very poor at dropout. Many of its spec's are guaranteed with a 19V input. ;D

[autir]

Then the electrolytics at the output, they reduce the dynamic output resistance(impedance).

Why/how?

Also: Can anyone spare some time to explain how we calculate the values of the proper transformer for this (and a similar) circuit?

[audioguru]

Hi Autir,
The 78xx series of voltage regulators have an opamp inside with its high frequency response rolled-off for compensation. Therefore for a quickly changing input voltage or load current the output voltage will have a transient change. An output capacitor filters the transient.

To obtain the correct voltage rating for a transformer,

[Staigen]

Hello, im back!

Well, Audioguru explained it wery good! Thanks Adioguru, then i dont have to write so much, i'm not so good in the english language!

//Staigen

[audioguru]

Hi Staigen,
Often we say exactly the same thing at the same time. ;D

[autir]

audioguru,

What would be the ideal transformer for this circuit? If I'm not mistaken, it should be rated as 0.5*squareroot(2)=0.7 A and [12+3(for the Vi-Vo)+2(for the bridges)]*squareroot(2)/2=12 Volt. What have I calculated incorrectly?

[audioguru]

Hi Autir,
You have three 1/2A supplies so the transformer's current rating should be 2.1A.
The 12V regulator needs a minimum of 15V input and the ripple on the main filter cap might be 1V. The rectifier bridge needs up to 2.5V (the rectifier current might be 20A pulses) so the total minimum DC voltage needed is 18.5V. A 13V transformer will do it but is a non-standard value. I would use a 24V center-tapped transformer (12V-0V-12V) and only two rectifier diodes. ;D

[MP]

Your silence implies that nobody knows where to put a fuse in a voltage regulator circuit?

[autir]

A little snippy there? Sometimes people are doing other things. Be patient.

Relax. I didn't bite

[audioguru]

Hi Autir,
As you can see by the low ripple voltage at the filter capacitor, the rectifiers conduct to charge it for only a very short period of time for each half-cycle of the mains. But in order to add the same amount of charge during charging that the 0.7A load removed during continuous discharging, then the rectifier current is massive for its short time.
The 1N4001 rectifier diode has a max forward voltage drop of 1.1V with a current of 1A. The datasheet has a curve for typical voltages at much higher currents but does not include guaranteed max voltages. So i think the recifier current pulses will be about 7A with your 0.7A continuous load, and each rectifier in series could easily have a 1.25V forward voltage drop. The 1N4001 rectifier diode is rated for 30A max.

[autir]

!

Won't this charging current blow the transformer's two fuses? Even if they are slowblow?

[audioguru]

Even a fast-blow fuse takes a fairly long time to blow so passes 7A pulses for about one milli-second with no problem. A fuse reacts to the average current which is low in your circuit.
High-power audio amplifiers have a huge filter capacitor in their power supply. It takes many half-cycles of the mains for it to fully charge so the amplifiers include current-limiting circuitry so the fuse or breaker doesn't blow when they are 1st turned on. ;D

[autir]

Something is wrong

[audioguru]

Hi Autir,
You shouldn't use a breadboard when tesing a power supply circuit. The resistance of its wires and contacts messes-up the voltages.
The 0V point at the IC should be soldered to its 0V pin and a heavy copper trace on its pcb should be a short distance from the main filter cap and to the 0V connection for the load.
Its heatsink tab should measure very close to 0V but its heatsink makes poor electrical contact through the thermal grease.
The output of the regulator should be soldered to its output pin. A heavy copper trace on its pcb should be a short distance to the connection for the load.
You are measuring the voltage drop across the resistance of your test circuit's poor wiring.
Connect the 0V wire of your voltmeter to the 0V pin of the IC and you can measure the voltage drops.

You shouldn't use an ammeter in series with the load. The ammeter has a voltage drop which reduces the voltage across the load (and therefore reduces the current through the load). Just calculate the current by measuring the voltage across the load's known resistance.

[autir]

a heavy copper trace on its pcb

Can't I build it on stripboard?  :-\

Its heatsink tab should measure very close to 0V but its heatsink makes poor electrical contact through the thermal grease.

Why very close to and not exactly 0 V? I have not used thermal paste yet.

You shouldn't use an ammeter in series with the load.

I have seen PSUs which had ammeters and voltmeters. How do they do it?
I thought I should install an ammeter in order to monitor the power consumption of my circuits. It is supposed to be connected between the 200uF's "+" and the 330nF's "live" connections, is this the correct location?

The ammeter has a voltage drop which reduces the voltage across the load (and therefore reduces the current through the load).

No offence, but isn't the ammeter's resistance supposed to be too small to produce more than the slightest voltage drop?

[audioguru]

Can't I build it on stripboard?

Use very short spacing between the IC's 0V pin and the 0V of the main filter cap and the load's 0V connection. Solder a heavy wire on the stripboard trace.
Use very short spacing between the IC's output pin and the connection for the load. Solder a heavy wire on the stripboard trace.

Why very close to and not exactly 0 V? I have not used thermal paste yet.

The IC has a small resistance in the metal of the tab and between its ground pin and the tab.

I have seen PSUs which had ammeters and voltmeters. How do they do it?

The current sensing resistor could be inside the regulator's negative feedback loop where its resistance doesn't matter.
Or you can buy cheap LCD meters with 200mV full-scale. You could add an opamp circuit to get as low a voltage drop as you want.

I thought I should install an ammeter in order to monitor the power consumption of my circuits. It is supposed to be connected between the 200uF's "+" and the 330nF's "live" connections, is this the correct location?

It is better between the rectifier and the 200uF (200uF is too small for much output current), then the input voltage of the IC will be more steady when the load changes.

No offence, but isn't the ammeter's resistance supposed to be too small to produce more than the slightest voltage drop?

My cheap DVM is made to measure car battery and appliance currents where a voltage drop doesn't matter. It has 1.25 ohms which drops 0.5V at 400mA.
My good DVM has 0.03 ohms and drops only 12mV at 400mA. Its connecting leads drop more voltage. ;D

[autir]

It worked!
I attached the power resistor directly to the breadboard and measured the voltage across its ends. It was 4.87V, dropping as the IC began to heat up and became stabilised at 4.78V.

In the lab I've attended we were used to work with external and/or PSU ammeters + voltmeters, plus a net of cables with banana clips. So much for old habits  ;)

200uF is too small for much output current

Typo. It is 2200uF  :)

The IC has a small resistance in the metal of the tab and between its ground pin and the tab.

Can I attach many ICs in the same heatsink without problems such as short-circuit? (with all tabs being ground, of cource.)

inside the regulator's negative feedback loop

What/where is that?  :)

It is better between the rectifier and the 200uF (200uF is too small for much output current), then the input voltage of the IC will be more steady when the load changes.

Won't the analog ammeter's coil be damaged by the unregulated DC passing through?

[Ldanielrosa]

p.s.: Could you explain this:

Quote
(the rectifier current might be 20A pulses)
?

The rectifier is only conducting when the input voltage is higher than the filter capacitor voltage.  This is probably a very low duty cycle so the instantaneous current is much higher than the average current.

[audioguru]

It worked!

Great!

200uF is too small for much output current

Typo. It is 2200uF

[autir]

There isn't a problem attaching ICs or transistors without insulators to a heatsink when their cases or tabs have the same voltage. Each one should be wired separately because they make a lousy electrical connection to the heatsink.

What do you mean "wired seperately"?
If there is a resistor an the tab of the TO-220s and the heatsink is soldered to the neutral of the board, won't we have unwanted current flowing through the heatsink?

So you propose that the best place for an ammeter in this circuit is the one in the picture?