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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
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
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
-> 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
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
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
audioguru2
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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
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
audioguru2
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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,
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,
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audioguru2
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Hi Staigen,
Often we say exactly the same thing at the same time. ;D
Often we say exactly the same thing at the same time. ;D
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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
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
audioguru2
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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.
View attachment 37533
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.
View attachment 37533
audioguru2
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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
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
audioguru2
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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.
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.
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