The rectified existing will be "ironed" by C1, whose capacity will need to not be less than 40.000uF, (a golden rule of close to 2000uF/A), but we suggest fifty.000uF. This capacity could be constructed up by many smaller capacitors in parallel. The base of this design is a basic 12V regulator (7812). The output voltage might be introduced to ideal value (here thirteen.8V) by two exterior resistors (R5 and R6) using this formula: U= 12(1+R5/R6). The very low currents (here 15mA) will maintain the 7812 in its normal perform. As quickly because the present rises over 15ma, the voltage drop on R4 will "open" the Q3, basically handling the high output present. This is actually a PNP transistor (Ic > 25) and current amplification factor of a minimum of twenty. The one which has been tested and confirmed right here is the 2N5683. The existing limiting resistance RL, for that maximum output of twenty Amps must be 0.03 Ohms, rated a minimum of 15W. You possibly can make use of the resistance wire or swap numerous resistors in parallel, totaling the resistance/power values. Values for other currents could be calculated by the rule: RL=0.7/Imax
The RL and Q2 (3A PNP which include BD330) form a brief circuit automated fuse. As quickly as the maximum current reaches 20Amps, the voltage drop more than the resistor RL will open Q2, and thus restrict the B-E Existing of Q3. Parallel to Q2 is Q1, which lights the LED one when the present limiting circuit is active. Once the fuse is active, the Q2 bridges the R3, so the full present would circulation by way of the IC1, and harm it. As a result the R4 is inserted, as to limit the IC1 existing to 15mA. This may make it feasible to run the IC1 without having any cooling aid. The LED two will light up every single time the PSU is switched on.
There is an adjustable current limiter in parallel for the fixed output, thus supplying adjustable present source for scaled-down currents. This circuit is incredibly straightforward also. You might discover that there is no current sensing resistor. But it is definitely there, inside a type of the Rds-on resistance of the N-channel FET, which basically handles the load cutoff from the supply. The perform of the FET is proven in the diagram 2. When the existing Id is rising, the pressure Uds over the resistance Rds rises especially gradually in the starting, but especially quick immediately after the knick. This means, that just before the knick the FET behaves as being a resistor but right after it, works as continuous existing source. The D2, R3 and B-E connection of the Q4 will feeling the Uds voltage with the FET1. When the voltage rises sufficient, the Q4 will shortcut the FET1 gate to mass, and reduce the current circulation via the FET 1 off.
Nonetheless, to allow the FET1 to open, there is particular gate voltage required, which with this case is brought up through the voltage divider consisting of R8, Z1, P1 and R9. So the optimum Gate voltage will probably be the one with the Z1, along with the minimal is going to be around 3V6. The Z1 voltage (Uz1) will therefore establish the max current flowing via the FET one. The diagram two will indicate that for 5 Amps the Uz1 really should be 5V6, and for 20Amps close to 9V6.The Capacitor C4 will determine the “velocity” or the reaction time with the limiter. a hundred uF will make the response time to be about 100ms, and 1n will allow it to be 1us. Inside the created limits, the P1 will restrict the existing output within the array of 15mA to 20A.
You possibly can use each output simultaneously, but the complete output current will be restricted by the value of the RL. This PSU can be built also for greater outputs, as long as the transformer will deal with the present needs, and also you provide adequate cooling for your Q3.
This boost regulator is for those times when you have a 28v relay, but want to use it with a 12v source. The circuit is built around the National Semiconductor LM2585, and uses the energy stored in an inductor to boost the 12 to 28v. Output voltage can be varied by adjusting the ratio of resistor values on the feedback pin.
The voltage regulator circuit does it’s switching around 100 Khz, but generates no noise if SMT components are used. Output is good for about half an amp continuous, enough to power two or three large microwave relays. The board measures 1.5″x2″.
It is important to note at least these three cautions before powering up the board:
- A short-circuit on the output will kill U1 and D1. Always use a 1 ohm 5w resistor, or a 2.5A fast fuse on the 12v input lead.
- Do not omit the LED (D2); It provides a visual indicator of a properly operating boost condition, but more importantly, it also provides a minimum load for the output, preventing an output “spike” which will otherwise appear when the load is disconnected abruptly.
- Keep the ratio of r2 and r3 to 22 or less to keep the output voltage within the ratings of C4 (C4 on my board is rated at 35wvdc). This ratio plus 1, multiplied times 1.25v, determines the output voltage.
One of the things you may not think about when buying a new laptop is what will happen if you lose your laptop AC adapters. Sure this does not seem like something you would lose, but it is very easy to misplace. So it is always a good idea to know where to buy another one, or even have another one handy just in case something like this should ever happen.
Also, be sure that it is a good model of laptop AC adapters. Some of them can get too hot and overheat. The best thing to do is read some online reviews of the notebook battery chargers, and see which ones work the best. From there you can choose the one that you are going to need. I think it is always a good idea to have a back up anyway. Nothing is worse than having a laptop and not being able to use it. So do not let that happen to you. Go out and get an extra one today.
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This circuit will generate a smaller DC output voltage from a larger DC input voltage.It is quick and simple to make and by changing the value of the zener diode, the circuit can be universally adapted to provide other output voltages.The circuit and all diagrams represent a DC convertor with 12V battery input and 9Volt DC output.
The output voltage is equal to the zener diode voltage less 0.7 volts, or :-
Vo = Vz - 0.7 where V z is the value of the zener diode.
With the 10V zener diode as shown in the diagram the output voltage is about 9.3 Volts DC. The supply voltage used must always be at least a few volts higher than the zener voltage. In this example I have used a 12 Volt DC battery to provide the regulated 9 Volt DC output.
The above graph shows how the output is affected by input voltage variations. This was produced with a load current of 100mA and using a 10 volt rated zener diode. Note that the circuit falls sharply out of regulation when the input voltage falls to 11.5 volt, hence the requirement for an adequate supply voltage.
Temperature stability is very good as the above graph shows. The output voltage changes by 8.5mV for every 10 degree rise in temperature. This is less than 1 mV / degree.
With a DC-DC convertor, the most important consideration is power dissipation in the output device. Power dissipation is the product of the transistors emitter current and collector-emitter voltage. With this circuit the maximum power dissipation of the BD139 or maximum collector current cannot be exceeded, otherwise the transistor will be destroyed.
With a 12 Volt supply and a 9 Volt, 100 mA load the dissipation is as follows. Using a 10 volt zener the output voltage will be about 9.3 volts DC therefore:
VCE * IC = (12 - 9.3) * 100 mA = 2.7 Watts
This is well within the maximum limits of power dissipation and collector current, which for the BD139 are 8 watts and 1 amp respectively. If higher load currents are required then the following circuit may be used.
Output dissipation is calculated in the same way, the BD131 has a maximum power dissipation of 15 watts and collector current of 3 amps. The output voltage is approximately 1.4 volts less than the zener diode voltage and supply voltage must be higher than the input voltage by at least 3 volts.
The 5 volt regulated power supply for TTL and 74LS series integrated circuits, has to be very precise and tolerant of voltage transients. These IC's are easily damaged by short voltage spikes. A fuse will blow when its current rating is exceeded, but requires several hundred milliseconds to respond. This circuit will react in a few microseconds, triggered when the output voltage exceeds the limit of the zener diode.
This circuit uses the crowbar method, where a thyristor is employed and short circuits the supply, causing the fuse to blow. This will take place in a few microseconds or less, and so offers much greater protection than an ordinary fuse. If the output voltage exceed 5.6Volt, then the zener diode will conduct, switching on the thyristor (all in a few microseconds), the output voltage is therefore reduced to 0 volts and sensitive logic IC's will be saved. The fuse will still take a few hundred milliseconds to blow but this is not important now because the supply to the circuit is already at zero volts and no damage can be done. The dc input to the regulator needs to be a few volts higher than the regulator voltage. In the case of a 5v regulator, I would recommend a transformer with secondary voltage of 8-10volts ac.