Regulator current Source Circuit Diagram

The circuit powers the load via the regulator`s input instead of its output. Because the regulator`s output sees constant dummy load R1, it tries to consume a constant amount of current, no matter what the voltage across the actual load really is. Hence, the regulator`s input serves as a constant-current source for the actual load. 

Power the circuit with any one of the commonly available ± 15 or ±12 V supplies. The voltage dropped across the regulator and dummy load decreased the total compliance voltage of the~circuit. You set the load`s current with Rl. The current equals 1.25 A/`J x Rl.

Regulator current Source Circuit Diagram

Regulator current Source Circuit Diagram

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High Current Low Dropout Voltage Regulator

This circuit was designed to allow a laptop computer to be powered from a solar power setup. The computer requires 12V at 3.3A. The circuit is a linear regulator with Mosfet Q4 as the series pass device. A 100kO resistor provides Q4 with a positive gate-source voltage. Any tendency for the output voltage to exceed ZD1s voltage causes Q2 to turn on. This turns on Q3 which reduces Q4s gate voltage and thus reduces the output voltage. Note that Q2s base-emitter voltage stabilizes at about 0.35V. This combined with the zener voltage gives an output of 12.4V. If a more precise output is required, first select ZD1 so that its voltage rating is at least 0.4V less than the required output voltage.

You can then "trim" to the required output voltage by installing a resistor in series with ZD1. Q2s base-emitter voltage and the 680W base resistor set the current through ZD1 to 0.5mA. This means that the output voltage will be boosted by 0.1V for each 200O of resistance in series with ZD1. Zener diode ZD2 ensures that Q4s maximum rated gate-source voltage is not exceeded. Mosfet Q1 provides reverse polarity protection. Note that Q4 requires a heatsink since it will dissipate about 10W under worst-case conditions. No heatsink is required for Q1. At 3.3A, the regulator reduces the output voltage by just 0.2V. This can be further reduced by paralleling Q1 & Q4 with additional Mosfets.

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Fast Voltage Driven Current Source

The current source in the diagram, which react very fast to changes in the input signal, may be used, for instance, in certain measurements. Differential amplifier IC1 ensures that the potential across R2 is equal to the input voltage: Iout =Uin/R2. The bandwidth, B, is given by B=R2 f /RL, where f=80 MHz, and the load impedance RL≥ R2 (both in ohms). The input is terminated into R1 to give the usual 50Ω impedance required by measuring instruments. At the same time, this resistor sets the d.c. operating point. If the link to the driving signal source is short and d.c. coupled, R1 may be omitted. The peak voltage between pins 1 and 2 of the IC is limited to 2.1 V to prevent too large a current at the output. Therefore, the peak output current is 2.1/100=21 mA.

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Ampere or Current Booster

Small and very useful circuit, Can be built on a veroboard

Volt regulators such as the LM708, and LM317 series (and others) sometimes need to provide a little bit more current then they actually can handle. If that is the case, this little circuit can help out. A power transistor such as the 2N3772 or similar can be used.

The power transistor is used to boost the extra needed current above the maximum allowable current provided via the regulator.
Current up to 1500mA(1.5amp) will flow through the regulator, anything above that makes the regulator conduct and adding the extra needed current to the output load. It is no problem stacking power transistors for even more current. (see diagram). Both regulator and power transistor must be mounted on an adequate heatsink.

Circuit diagram:

Ampere or Current Booster Circuit Diagram

Parts:

R1 = 1R-2W
R2 = 10R-2W
C1 = 35v-470uF
C2 = 35v-470uF
Q1 = TIP2955
IC1 = 78xx Regulator

Source : www.extremecircuits.net

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