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Build a Bootstrapped Amp Current Source Circuit Diagram. This circuit responds to the difference between Vj and V2. Rq on sets gain. Resistors XR2 and (1 -X) R2 produce the bootstrap effect. These two resistors convert the circuit`s output voltage to a current. IC1 and IC2 are Burr-Brown OPA2107 or equal.

This is Simple linear Regulator Circuit Diagram. Intended for extreme temperature, radiation-hardened environments, this linear supply is capable of supplying 28 Vdc at 125 A from an ac-driven power unit. In operation, power supply output voltage is sensed by the voltage divider consisting of R24 to R28 and fed to one input of a discrete differential amplifier composed of Q13 through Q16. 

The other input of the amplifier is connected to a radiation-hardened zener diode, Dl. Local feedback using R21 and Cl produces gain to phase shift that are independent of individual component parameters, which provides stable operation into the required loads.

This dual-polarity, multivoltage power supply circuit diagram can be built for a very small investment. The circuit is built around 78XX and 79XX series i-A voltage regulators, four 3-A diodes, a 24-30-V 2-6-A transformer, and eight filter capacitors.

The circuit behaves like a zener diode over a large range of voltages. The current passing through the voltage divider R1-R2 is substantially larger than the transistor base current and is in the region of 8 mA. The stabilizing voltage is adjustable over the range 5-45 V by changing the value of R2. The total current drawn by the circuit is variable over the range 15 mA to 50 mA.This value is determined by the maximum dissipation of the zener diode. In the case of a 250 mW device, this is of the order of 50 mA.

How to Build a Efficient Negative Voltage Regulator Circuit Diagram. This simple Efficient Negative Voltage Regulator Circuit Diagram is One v;ay to provide good negative-voltage regulation is with a low-dropout positive-voltage regulator operating from a well-isolated secondary winding of switch-mode circuit transformer. The technique works with any positive-voltage regulator, although highest efficiency occurs with low-dropout types. 

Under all loading conditions, the minimum voltage difference between the regulator ViN and VoUT pins must be at least 1.5 V, the LT1086`s low-dropout voltage. Efficient-negative-voltage-regulator Rating: 7.00/10 (23Votes cast)Category: Power Supply Circuits / AC to DC & DC to DCViews: 3Rank: 5One v;ay to provide good negative-voltage regulation is with a low-dropout positive-voltage regulator operating from a well-isolated secondary winding of switch-mode circuit transformer. 

The technique works with any positive-voltage regulator, although highest efficiency occurs with low-dropout types. Under all loading conditions, the minimum voltage difference between the regulator ViN and VoUT pins must be at least 1.5 V, the LT1086`s low-dropout voltage.Efficient-negative-voltage-regulatorIf this requirement isn`t met, the output falls out of regulation. 1vo programming resistors, R1 and R2, set the output voltage to 12 V, and the LT1086`s servo the voltage between the output and its adjusting (ADJ) terminals to 1.25 V. 

Capacitor C1 improves ripple rejection, and protection diode D1 eliminates common-load problems. Since a secondary winding is galvanically isolated, a regulator`s 12 V output can be referenced to ground. Therefore, in the case of a negative-voltage output, the positive-voltage terminal of the regulator connects to ground, and the -12 V output comes off the anode of Dl. The ViN terminal floats at 1.5 V or more above ground.

This circuit uses two comparator pairs from an LM339N quad comparator; one pair drives the yellow positive (+)and negative (-)indicators, the other jointly drives the red warn LED3. The circuit draws its power from the unregulated portion of the power supply. The four comparators get their switching inputs from two parallel resistor-divider strings. Both~strings have their ends tied between the power supply`s positive and negative output terminals. 

The first string, consisting of R4, R5, and R6, divides the input voltage in half, with output taps at 0.5%. The other string, made up of R7, R8, and R9, also divides the input voltage in half, with taps at + lO%. The 0.5% R4/R5/R6 string drives the two comparators controlling the positive and negative indicators (LEDl and LED2). Their inputs are crossed so that LED2 does not fire until the positive supply is at least 0.5% higher than the negative; the positive indicator does not go off until the negative supply is at least 0.5% higher than the positive-in relative levels. 

That overlap permits both LEDs to be on when the two supplies are in 1 % or better balance. The +lOT R7/R8/R9 string drives the other two comparators, which control the warn indicator. If either side of the supply is lO% or more higher than the other, one of the two comparators will switch its output low and light the redLED3the LM339N has opened-collector outputs, allowing such wired OR connections. The inputs are not crossed, as with the other comparator pair, so there is a band in the middle where neither comparators output is low and the LED remains off.

Simple Efficient Supply Splitter Circuit Diagram. In this application, the Vl-7660 is connected as a voltage splitter. Note that the normal output pin is connected to ground and the normal ground pin is used as the output. The switches that allow the charge pumping are bidirectional; therefore, charge transfer can be performed in reverse. 

The 1-MO resistor is used to avoid start-up problems by forcing the internal regulator on. An application for this circuit would be driving low-voltage, ± 7.5 Vdc, circuits from ± 15 Vdc supplies, or low-voltage logic from 9 to 12 V batteries.

How to Build a Regulated Charge Pump Circuit Diagram. The dc-dc converter substitutes a voltage triplet in place of the external inductor and the diode that`s typically associated with the switching regulator, IC1. Inverting and non inverting amplifiers in the MOS-FET-driver (IC2) activate a diode-capacitor tripling network (D1 through D3, CI through C3). 

A 50-kHz oscillator residing within IC1 produces the EXT signal (pin 6), IC2 converts this signal into drive signals (180° out of phase) for the tripler. The resulting charge-discharge action in the capacitors recharges C3 toward 10 V every 20 The ferrite bead limits output ripple to about 20-mVpp for a 50-mA load. Conversion efficiency is about 70% for the 5-V input, 10-V output configuration.

Using a single 7812 IC voltage regulator and multiple outboard pass transistors, this power supply can deliver output load currents of up to 30 amps. The design is shown below:

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:

Calculations:
This circuit is a fine example of Kirchoffs current and voltage laws. To summarise, 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 Kirchoffs 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.

This is a Motorcycle Alarm Circuit Diagram. Any number of normally open switches may be used. Fit the mercury switches so that they close when the steering is moved or when the bike is lifted off its side-stand or pushed forward off its center-stand. Use micro-switches to protect removable panels and the lids of panniers etc. While at least one switch remains closed, the siren will sound. 

About tw1o minutes after the switches have been opened again, the alarm will reset. How long it takes to switch off depends on the characteristics of the actual components used. But, up to a point, you can adjust the time to suit your requirements by changing the value of C1.The circuit board and switches must be protected from the elements. Dampness or condensation will cause malfunction.Without its terminal blocks, the board is small. Ideally, you should try to find a siren with enough spare space inside to accommodate it. Fit a 1-amp in-line fuse close to the power source. 

This protects the wiring. Instead of using a key-switch you can use a hidden switch; or you could use the normally closed contacts of a small relay. Wire the relay coil so that it is energized while the ignition is on. Then every time you turn the ignition off, the alarm will set itself.When it`s not sounding, the circuit uses virtually no current. This should make it useful in other circumstances. For example, powered by dry batteries and with the relay and siren voltages to suit, it could be fitted inside a computer or any thing else that`s in danger of being picked up and carried away. The low standby current and automatic reset means that for this sort of application an external on/off switch may not be necessary.