Showing posts with label current. Show all posts
Showing posts with label current. Show all posts
Friday, January 10, 2014
Build a Bootstrapped Amp Current Source Circuit Diagram
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.
Bootstrapped Amp Current Source Circuit Diagram

Build a Bootstrapped Amp Current Source Circuit Diagram
Wednesday, April 10, 2013
High Current Regulated Supply Circuit Using LM317
The high current regulator circuit is built uses an additional winding or a separate transformer to supply power for the LM317 regulator so that the pass transistors can operate closer to saturation and improve efficiency. For good efficiency the voltage at the collectors of the two parallel 2N3055 pass transistors should be close to the output voltage. The operation of this circuit is explained like this.

The LM317 requires a couple extra volts on the input side, plus the emitter/base drop of the 3055s, plus whatever is lost across the (0.1 ohm) equalizing resistors (1volt at 10 amps), so a separate transformer and rectifier/filter circuit is used that is a few volts higher than the output voltage. The LM317 will provide over 1 amp of current to drive the bases of the pass transistors and assumption a gain of 10 the combination should deliver 15 amps or more.
The LM317 always operates with a voltage difference of 1.2 between the output terminal and adjustment terminal and requires a minimum load of 10mA, so a 75 ohm resistor was chosen which will draw (1.2/75 = 16mA). This same current flows through the emitter resistor of the 2N3904 which produces about a 1 volt drop across the 62 ohm resistor and 1.7 volts at the base. The output voltage is set with the voltage divider (1K/560) so that 1.7 volts is applied to the 3904 base when the output is 5 volts. For 13 volt operation, the 1K resistor could be adjusted to around 3.6K. The regulator has no output short circuit protection so the output probably should be fused.

The LM317 requires a couple extra volts on the input side, plus the emitter/base drop of the 3055s, plus whatever is lost across the (0.1 ohm) equalizing resistors (1volt at 10 amps), so a separate transformer and rectifier/filter circuit is used that is a few volts higher than the output voltage. The LM317 will provide over 1 amp of current to drive the bases of the pass transistors and assumption a gain of 10 the combination should deliver 15 amps or more.
The LM317 always operates with a voltage difference of 1.2 between the output terminal and adjustment terminal and requires a minimum load of 10mA, so a 75 ohm resistor was chosen which will draw (1.2/75 = 16mA). This same current flows through the emitter resistor of the 2N3904 which produces about a 1 volt drop across the 62 ohm resistor and 1.7 volts at the base. The output voltage is set with the voltage divider (1K/560) so that 1.7 volts is applied to the 3904 base when the output is 5 volts. For 13 volt operation, the 1K resistor could be adjusted to around 3.6K. The regulator has no output short circuit protection so the output probably should be fused.
Thursday, April 4, 2013
Non Switching Current Drive Amplifiers
Non-Switching Amplifiers
Most of the distortion in Class-B is crossover distortion, and results from gain changes in the output stage as the power devices turn on and off. Several researchers have attempted to avoid this by ensuring that each device is clamped to pass a certain minimum current at all times. This approach has certainly been exploited commercially, but few technical details have been published. It is not intuitively obvious (to me, anyway) that stopping the diminishing device current in its tracks will give less crossover distortion .
Current-Drive Amplifiers
Almost all power amplifiers aspire to be voltage sources of zero output impedance. This minimizes frequency-response variations caused by the peaks and dips of the impedance curve, and gives a universal amplifier that can drive any loudspeaker directly.
The opposite approach is an amplifier with a suffi ciently high output impedance to act as a constant-current source. This eliminates some problems – such as rising voice-coil resistance with heat dissipation – but introduces others such as control of the cone resonance. Current amplifiers therefore appear to be only of use with active crossovers and velocity feedback from the cone . It is relatively simple to design an amplifier with any desired output impedance (even a negative one), and so any compromise between voltage and current drive is attainable. The snag is that loudspeakers are universally designed to be driven by voltage sources, and higher amplifier impedances demand tailoring to specifi c speaker types
Tuesday, March 19, 2013
How to Make 1 A Constant Current LED Driver Circuit
The article explains a simple circuit using the IC MBI6651 from MACROBLOCK. The IC has been specifically designed for operating high power LEDs safely by providing a constant current output. The circuit includes very few external components and therefore becomes very easy to assemble at home.
About the IC MBI6651
The IC MBI6651 is a high efficiency, step down DC to DC converter chip capable of driving high power LEDs at a safe 1 Amp constant current.
The IC requires just four passive external components for making it functional.
The output current of the IC can be externally set by selecting the appropriate resistor value.
The IC also features a PWM controlled dimming control of the connected LEDs.
Some of the other outstanding features of this IC includes UVLO meaning under voltage lockout, over temperature shut down, LED open circuit protection and LED short circuit protection, all these ensure complete safety to the IC from wrongly configured output loads.
Typical Application of this device are:
Automotive decoration and illumination
LED flood lights using high intensity, high power LED.
The IC also can be used as a constant current source in particular circuit applications.
Setting the output Current
The output current of the IC is fixed through an external resistor Rsen. The output current Iout and the adjustment resistor Rsen has the following relation:
Given Vsen=0.1V
Rsen=(Vsen/Iout)=(0.1V/Iout)
Where Rsen is the value of the external resistor. This resistor is connected across the pin outs SEN and Vsen of the IC.
The optimum current with Rsen 0.1 Ohms is 1000 mA or 1 Amp.
Optimizing External Component Selection
Inductor: Two issues specify the inductor type, the switching frequency and the ripple current. The involved calculation can be written as:
L1>{Vin - Vout - Vsen - (Rds(on) * Iout)} * D/fsw * delta.IL
where, Rds(on) is the on-resistance of the ICs internal MOSFET. The value is typically around 0.45 at 12V
D is the duty cycle of the IC, given as D = Vout/Vin
fsw is the switching frequency of the IC
While designing the inductor for the given circuit, along with the inductance the saturation current must also be taken into account,because these are two basic factors which typically affects the overall performance of the circuit.
The rule of thumb, the saturation current of the inductor should be selected 1.5 times greater than the LED current.
Moreover, selecting high values for the inductance provides better line and load regulation.
Refer circuit diagram
Selecting the Schottky diode
The diode D1 shown in the circuit diagram basically acts as the flywheel diode for nullifying the inductor back emf during the periods when the LED is switched OFF.
The diode must be selected with the following couple of important characteristics:
It should have a low forward voltage rating and maximum possible reverse voltage tolerance.
Selecting the capacitor
The general rule is always to select a capacitor value with a voltage tolerance 1.5 times higher than the supply voltage.
Preferably, a tantalum capacitor should be selected because these have high capacitance and low ESR characteristics.
The proposed circuit of 1 Amp constant current LED driver circuit is given below:

The basic operating parameters are given below:

Pin Out Specs:

About the IC MBI6651
The IC MBI6651 is a high efficiency, step down DC to DC converter chip capable of driving high power LEDs at a safe 1 Amp constant current.
The IC requires just four passive external components for making it functional.
The output current of the IC can be externally set by selecting the appropriate resistor value.
The IC also features a PWM controlled dimming control of the connected LEDs.
Some of the other outstanding features of this IC includes UVLO meaning under voltage lockout, over temperature shut down, LED open circuit protection and LED short circuit protection, all these ensure complete safety to the IC from wrongly configured output loads.
Typical Application of this device are:
Automotive decoration and illumination
LED flood lights using high intensity, high power LED.
The IC also can be used as a constant current source in particular circuit applications.
Setting the output Current
The output current of the IC is fixed through an external resistor Rsen. The output current Iout and the adjustment resistor Rsen has the following relation:
Given Vsen=0.1V
Rsen=(Vsen/Iout)=(0.1V/Iout)
Where Rsen is the value of the external resistor. This resistor is connected across the pin outs SEN and Vsen of the IC.
The optimum current with Rsen 0.1 Ohms is 1000 mA or 1 Amp.
Optimizing External Component Selection
Inductor: Two issues specify the inductor type, the switching frequency and the ripple current. The involved calculation can be written as:
L1>{Vin - Vout - Vsen - (Rds(on) * Iout)} * D/fsw * delta.IL
where, Rds(on) is the on-resistance of the ICs internal MOSFET. The value is typically around 0.45 at 12V
D is the duty cycle of the IC, given as D = Vout/Vin
fsw is the switching frequency of the IC
While designing the inductor for the given circuit, along with the inductance the saturation current must also be taken into account,because these are two basic factors which typically affects the overall performance of the circuit.
The rule of thumb, the saturation current of the inductor should be selected 1.5 times greater than the LED current.
Moreover, selecting high values for the inductance provides better line and load regulation.
Refer circuit diagram
Selecting the Schottky diode
The diode D1 shown in the circuit diagram basically acts as the flywheel diode for nullifying the inductor back emf during the periods when the LED is switched OFF.
The diode must be selected with the following couple of important characteristics:
It should have a low forward voltage rating and maximum possible reverse voltage tolerance.
Selecting the capacitor
The general rule is always to select a capacitor value with a voltage tolerance 1.5 times higher than the supply voltage.
Preferably, a tantalum capacitor should be selected because these have high capacitance and low ESR characteristics.
The proposed circuit of 1 Amp constant current LED driver circuit is given below:

The basic operating parameters are given below:

Pin Out Specs:

Courtesy: http://www.ledlabs.ru/pdf/macroblock/mbi6651.pdf
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