12V Power Supply 30A

This is high current 12V power supply. Power supply uses LM7812 IC and can deliver up to 30A to the load by the help of the TIP2955 pass transistors. Each transistor can handle up to 5A and six of them result an total output current of 30A. You can increase or reduce the number of TIP2955s to get higher or lower current outputs. In this design the IC delivers about 800mA. A 1 amp fuse is connected after the LM7812 to protect the IC against high current transients. The transistors and the 12V regulator IC both require adequate heatsinking. When the load current is high, the power dissipation of each transistor also increases so excess heat may cause the transistors to fail. Then you will need a very large heatsink or fan cooling. 100Ω resistors are used for stability and prevent current swamping as the tolerances of dc current gain will be different for each transistor. The bridge rectifier diodes must be capable of passing at least 100 amps.

Notes
The input transformer is likely to be the most expensive part of the entire project. As an alternative, a couple of 12 Volt car batteries could be used. The input voltage to the regulator must be at least several volts higher than the output voltage (12V) so that the regulator can maintain its output. If a transformer is used, then the rectifier diodes must be capable of passing a very high peak forward current, typically 100amps or more. The 7812 IC will only pass 1 amp or less of the output current, the remainder being supplied by the outboard pass transistors. As the circuit is designed to handle loads of up to 30 amps, then six TIP2955 are wired in parallel to meet this demand. The dissipation in each power transistor is one sixth of the total load, but adequate heat sinking is still required. Maximum load current will generate maximum dissipation, so a very large heat sink is required. In considering a heat sink, it may be a good idea to look for either a fan or water cooled heat sink. In the event that the power transistors should fail, then the regulator would have to supply full load current and would fail with catastrophic results. A 1 amp fuse in the regulators output prevents a safeguard. The 400mohm load is for test purposes only and should not be included in the final circuit. A simulated performance is shown below:

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12V Power Supply 30A

Calculations
This circuit is a fine example of Kirchhoff's current and voltage laws. To summarize, the sum of the currents entering a junction, must equal the current leaving the junction, and the voltages around a loop must equal zero. For example, in the diagram above, the input voltage is 24 volts. 4 volts is dropped across R7 and 20 volts across the regulator input, 24 -4 -20 =0. At the output :- the total load current is 30 amps, the regulator supplies 0.866 A and the 6 transistors 4.855 Amp each , 30 = 6 * 4.855 + 0.866. Each power transistor contributes around 4.86 A to the load. The base current is about 138 mA per transistor. A DC current gain of 35 at a collector current of 6 amp is required. This is well within the limits of the TIP2955. Resistors R1 to R6 are included for stability and prevent current swamping as the manufacturing tolerances of dc current gain will be different for each transistor. Resistor R7 is 100 ohms and develops 4 Volts with maximun load. Power dissipation is hence (4^2)/200 or about 160 mW. I recommend using a 0.5 Watt resistor for R7. The input current to the regulator is fed via the emitter resistor and base emitter junctions of the power transistors. Once again using Kirchhoff's current laws, the 871 mA regulator input current is derived from the base chain and the 40.3 mA flowing through the 100 Ohm resistor. 871.18 = 40.3 + 830. 88. The current from the regulator itself cannot be greater than the input current. As can be seen the regulator only draws about 5 mA and should run cold.

Initial Testing and Faulting
For the initial test, do not connect a load. First use a voltmeter across the output terminals, you should measure 12 Volts, or very close to it. Then connect a 100 ohm, 3 Watt resistor or other small load. The reading on the voltmeter should not change. If you do not see 12 Volt, power off and check all connections.

I have heard from one reader whose supply was 35 Volt, not the regulated 12 Volts. This was caused by a short circuited power transistor. Should a short in any of the output transistors, occur, all 6 need to be un-soldered. Check with a multimeter set to resistance and measure between collector and emitter terminals. Power transistors usually fail short circuit so should be easy to find the faulty one.

A Finished Project
I've recently heard from Ryan Laurenciana in the Philippines who has built himself a 12V 30A power supply. Below are images from Ryans power supply.

Stable Filament Supply

 

Valves are enjoying increasing popularity in audio systems. With the European ‘E’ series of valves, such as the ECC83 (12AX7) and EL84 (6BQ5), the filament voltage is 6.3 V. Depending on how the circuit is wired, the ECC 81–83 series of twin triodes can also be used with a filament voltage of 12.6 V. In earlier times, the filament voltage was usually taken directly from a separate transformer winding, which (in part) was responsible for the well known ‘valve hum’. With regard to the signal path, current valve circuits have hardly experienced any fundamental changes. In high-quality valve equipment, though, it is relatively common to find a stabilised anode supply.

Mains hum can have a measurable and audible effect on input stages whose filaments are heated by an ac voltage. The remedy described here is a stabilised and precisely regulated dc filament voltage. The slow rise of the filament voltage after switching on is also beneficial. The exact setting of the voltage level and the soft start have a positive effect on the useful life of the valves. Diagram shows a voltage regulator meeting these requirements that is built from discrete components. The two sets of component values are for a voltage of 6.3 V (upper) and 12.6 V (lower).

Thanks to the fact that the supply works with a constant load, it can do without special protective circuits and the additional complexity of optimum regulation characteristics for dynamic loads. The circuit in Figure 1 consists of a power MOSFET configured as a series-pass regulator and a conventional control amplifier. Zener diode (D5) sets the reference potential. A constant voltage is thus present at the emitter of the BC547 control amplifier (T3). The current through D5 is set to approximately 4–5 mA by series resistor R5. The output voltage UO (the controlled variable) acts on the base of the control amplifier (T3) via voltage divider R6/R7. If the output voltage drops, the collector current of T3 also decreases, and with it the voltage drop across load resistors R1 and R2.


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Stable Filament Supply

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The voltage on the gate of the MOSFET thus increases. This closes the control loop. The values of the resistors forming the voltage divider are chosen for the usual tolerances of Zener diodes, but they must be adjusted if the diode is out of spec (which can happen). The load resistance of the control amplifier is divided between R1 and R2. The current through the load resistance and the collector current of T3 are practically the same, since the MOSFET draws almost no gate current. Filter capacitor C2 is connected to the junction of R1 and R2 to reduce residual hum. Electrolytic capacitor C4 and power supply filter capacitor C1 serve the same purpose. The hum voltage also depends on the magnitude of the load current.

The voltage drop over the series-pass regulator is nearly the same for an output voltage of 6.3 V or 12.6 V. With a BUZ11 and a load of 1 A at 6.3 V, for instance, the average voltage across the source–drain channel is approximately 7V. The power dissipation of 7 W requires a corresponding heat sink. The slow rise of the output voltage is due to the presence of timing network R3/C3 and T1. When power is switched on, T1 holds the gate of the MOSFET at nearly ground level. As C3 charges, T1 conducts increasingly less current, so ultimately only the control transistor affects the gate voltage.

The mains transformer must be selected according to the required load current. The required value of the input voltage can be read from the chart. The transformer should have a power rating at least 30 % greater than what is necessary based on the calculated load dissipation. Where possible, preference should be given to a filament voltage of 12.6 V. When twin triodes in the ECC81–83 series are used, for example, the power dissipation in the series pass transistor is lower with a voltage of 12.6 V.

Negative Auxiliary Voltage

Some circuits need a negative supply voltage that only has to supply a small current. Providing a separate transformer winding for this (possibly even with a rectifier and filter capacitor) would be a rather extravagant solution. It can also be done using a few gates and several passive components. The combination of gate IC1a and the other three gates (wired in parallel) forms a square-wave generator. D1 and D2 convert the ac voltage into a dc voltage. As a CMOS IC is used here, the load on the negative output is limited to a few milliampères, depending on the positive supply voltage (see chart), despite the fact that three gates are connected in parallel.


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However, as the figure shows, the negative voltage has almost the same magnitude as the positive input voltage, but with the opposite sign. If a clock signal in the range of 10–50 kHz is available, it can be connected to the input of IC1a, and R1 and C1 can then be omitted.

5V Power Supply For On-Train Radio Camera

 


A radio camera on a model railway should transmit constantly while the train is moving and continue transmitting for a few minutes after the train stops. But if the train starts up again after a relatively long halt, imagery should be transmitted immediately. Consequently, the power source for the camera cannot be rechargeable batteries (since they take too long to charge), nor can it be primary batteries (for environmental reasons). Instead, GoldCaps provide a good alternative. They can be charged in no time flat, and they assure sufficient reserve power for operating the radio camera for a few minutes. Coming from the left in the schematic shown in Figure 1, the dc voltage arrives at the supply circuit and is buffered by capacitor C1, which bridges brief power interruptions.

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5V Power Supply For On-Train Radio Camera

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The actual reserve power source consists of four GoldCaps connected in series, each rated at 22 F / 2.3 V, which yields a net capacitance of 5.5 F / 9.2 V. The maximum charging voltage must never exceed 9.2 V. This is ensured by a modern adjustable low-drop voltage regulator (LT1086), which is set to a nominal output voltage of 9.57 V by resistors R2 and R3 (since there is an 0.6-V voltage drop across D5). The LT1086 can handle a current of 1.5 A (with current limiting), so even completely empty GoldCaps can be charged in a few seconds. Whenever the dc voltage is present, the GoldCaps are charged via D2. When the dc voltage is present, the camera is not powered from the Gold-Caps, but instead directly from the track via D4.

Diode D5 prevents this voltage from reaching the bank of capacitors, and D4 prevents the GoldCaps from discharging via the track when no voltages present on it. D4 and D5 thus form a sort of OR gate. The radio camera used by the author requires 5 V and draws a current of approximately 70 mA. This means the circuit must have an output stage consisting of a ‘normal’ 7805 fixed voltage regulator and the usual capacitors (C9 C10, C11). The two low-current LEDs respectively indicate whether voltage is present on the track and whether the storage capacitors are charged. They can also be omitted. The 100-nF capacitors must be placed as close as possible to the voltage regulators.

7.2V Battery Replacement Power Supply For Camcorders

 


This circuit lets an external 12V SLA battery power a camcorder which normally has an inbuilt 7.2V battery. Such batteries can now be very difficult or expensive to obtain for earlier model camcorders. In essence, the circuit is a standard LM317 adjustable regulator with resistors R1 & R2 set to provide 7.2V (depending on the accuracy of the 1.25V internal reference). If the resulting output voltage is low, it can be increased by reducing the 130 resistor and vice versa. The circuit can be assembled on to the Eliminator PC board or the simple DC power supply PC board. The regulator should be fitted with a flag heatsink. Note that the circuit should be disconnected from the battery when not in use, otherwise its quiescent current (from the LED and regulator) will flatten the SLA battery.

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7.2V Battery Replacement Power Supply For Camcorders  

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