Wiring circuit

Friday, October 24, 2014

Loudspeaker Protector Monitors Current

This circuit uses a 0.1O 1W resistor connected in series with the output of a power amplifier. When the amplifier is delivering 100W into an 8O load, the resistor will be dissipating 1.25W. The resulting temperature rise is sensed by a thermistor which is thermally bonded to the resistor. The thermistor is connected in series with a resistor string which is monitored by the non-inverting (+) inputs of four comparators in an LM339 quad comparator. All of the comparator inverting inputs are connected to an adjustable threshold voltage provided by trimpot VR1. As the thermistor heats up, its resistance increases, raising the voltage along the resistor ladder.

Loudspeaker Protector Monitors Current Circuit diagram:

Loudspeaker Protector Circuit Diagram

When the voltage on the non-inverting input of each comparator exceeds the voltage at its inverting input, the output switches high and illuminates the relevant LED. NOR gate latches are connected to the outputs of the third and fourth comparators. When the third comparator switches high, the first latch is set, turning on Q1 and relay 1. This switches in an attenuation network (resistors RA & RB) to reduce the power level. However, if the power level is still excessive, comparator 4 will switch, setting its latch and turning on Q2 and relay 2.

This disconnects the loudspeaker load. The thermistor then needs to cool down before normal operation will be restored. The values of R1-R4 depend on the thermistor used. For example, if a thermistor with a resistance of 1.5kO at 25°C is used, then R1 could be around 1.5kO and R2, R3 and R4 would each be 100O (depending the temperature coefficient of the thermistor). The setup procedure involves connecting a sinewave oscillator to the input of the power amplifier and using a dummy load for the output. Set the power level desired and adjust trimpot VR1 to light LED1. Then increase the power to check that the other LEDs light at satisfactory levels.

Author: David Devers - Copyright: Silicon Chip Electronics

Monday, October 20, 2014

Adjustable Voltage and Variable Current Limiter Circuit Diagram

This Mini Voltage Source with Adjustable Current Limit variable is very simple and efficient, ideal for small shippers or circuits that require a current limiter with adjustable voltage that this circuit can vary from a few mA up to 2 A and voltage 1.225v up to 12V or more, depending on the power supply.

Once the current reaches the limit set by the 100R resistor, transistor BC547 begins to act on the voltage regulator in the Adj pin and the output voltage starts to decrease. If the output is shorted, the output voltage will reduce to almost zero, as a kind of protection.

Adjustable Voltage and Variable Current Limiter Circuit Diagram

Adjustable Voltage and Variable Current Limiter Circuit Diagram

Sunday, October 19, 2014

2 36V 10A High Current Variable Voltage Regulator

Yesterday I no enhance the circuit in my blog. Today lead High Current power supply 10A and Can Adjustable Voltage or Variable Voltage Regulator 2V to 36V. By the prominent point of this circuit. Be there is the a little equipment and is simple have VR1 for control Volt output. For the equipment must use the size is appropriate. If use the a little value goes to get Current not tall on schedule keep, such as The Transformer use 10A, Transistor sizes should are the number that notes , and A capacitors should choose many values especially C1 – 18000 uF , use 10000uF values wasp parallel can. For other detail , friends sees in the circuit please yes.

Saturday, October 18, 2014

MJ2955 for IC 78xx Boosting Regulator Current Diagram Circuit

Source :: http://www.zen22142.zen.co.uk/Circuits/Power/boosti.htm

IC 78xx series of voltage regulators are available with different current outputs, you can boost the available current output with this circuit. A power transistor is used to supply extra current to the load the regulator, maintaining a constant voltage. Currents up to 650mA will flow through the regulator, above this value and the power transistor will start to conduct, supplying the extra current to the load. This should be on an adequate heat sink as it is likely to get rather hot. Suppose you use a 12v regulator, 7812. The input voltage should be a few volts higher to allow for voltage drops. Assume 20 volts. Lets also assume that the load will draw 5amps. The power dissipation in the transistor will be Vce * Ic or (20-12)*8=40watt. It may keep you warm in the Winter, but you will need a large heatsink with good thermal dissipation. If you want to Boost the output current with a negative regulator, such as the 79xx series, then the circuit is similar, but an NPN type power transistor is used instead.

Tuesday, September 23, 2014

Adjustment power supply values 1 25 15V Max current 0 5 amp circuit

This is power supply that have Voltage output fine the value has 1.25-15V. From by Output current that be valuable about 0.5 amp that 12V and 0.2 amp that 15V.
When Volt , from 220V houses reach transformer. It will modify Volt 220V to be 18VAC already to change rectifier circuit. Which D1, D2, C1 and R1 wasp be Full wave rectifier circuit. For modify DCV to ACV to a signal DCV. It make get a signal DCV that have voltage at pin 3 of IC1 be 20V. From that time DCV signal this reach fight DC Regulator circuit. Which use IC number LM317.

This circuit will perform to maintain one’s position Voltage smoothly. Which level output voltage at get this will go out the way pin 2 of IC1. By have capacitors C2 be voltage filter in order that voltage output level of the circuit is will high class modify to follow fining. The performing fee withstands VR1 there.

from:eleccircuit.com

Tuesday, September 16, 2014

AC Line Current Detector Wiring diagram Schematic

This schema diagram will detect AC line currents of about 250 mA or more without making any electrical connections to the line. Current is detected by passing one of the AC lines through an inductive pickup (L1) made with a 1 inch diameter U-bolt wound with 800 turns of #30 - #35 magnet wire. The pickup could be made from other iron type rings or transformer cores that allows enough space to pass one of the AC lines through the center.

Only one of the current carrying lines, either the line or the neutral should be put through the center of the pickup to avoid the fields cancelling. I tested the schema using a 2 wire extension cord which I had separated the twin wires a small distance with an exacto knife to allow the U-bolt to encircle only one wire.

AC Line Current Detector Circuit Diagram

The magnetic pickup (U-bolt) produces about 4 millivolts peak for a AC line current of 250 mA, or AC load of around 30 watts. The signal from the pickup is raised about 200 times at the output of the op-amp pin 1 which is then peak detected by the capacitor and diode connected to pin 1. The second op-amp is used as a comparator which detects a voltage rise greater than the diode drop. The minimum signal needed to cause the comparator stage output to switch positive is around 800 mV peak which corresponds to about a 30 watt load on the AC line.

The output 1458 op-amp will only swing within a couple volts of ground so a voltage divider (1K/470) is used to reduce the no-signal voltage to about 0.7 volts. An additional diode is added in series with the transistor base to ensure it turns off when the op-amp voltage is 2 volts. You may get a little bit of relay chatter if the AC load is close to the switching point so a larger load of 50 watts or more is recommended. The sensitivity could be increased by adding more turns to the pickup.