Cold Alignment Engine To GearBox
Showing posts with label to. Show all posts
Showing posts with label to. Show all posts
Wednesday, 11 September 2013
You sometimes need to drive a white LED from one 1.5V battery. Unfortunately, the forward voltage of a white LED is 3 to 4V. So, you would need a dc/dc converter to drive the LED from one battery. Using the simple circuit in Figure 1, you can drive one white LED or two series-connected green LEDs, using only a few components. The circuit is a voltage-to-current converter, which converts the battery voltage to a current that passes through the LED.
You can adjust this current and, thus, the brightness of the LED, by varying resistor R3. If you turn on switch S1, resistor R2 feeds base current to transistor Q2. Q2 turns on, and its collector current, via R3, turns on Q1. Now, the current through inductor L1 increases. The slope of the increase is a function of the value of L1 and the battery voltage. The current through L1 increases until it reaches a maximum value, which depends on the gain of Q1. Because the value of R3 sets the base current drawn from Q1, Q1s collector current is also limited.
White LEDs Circuit Diagram
You can adjust this current and, thus, the brightness of the LED, by varying resistor R3. If you turn on switch S1, resistor R2 feeds base current to transistor Q2. Q2 turns on, and its collector current, via R3, turns on Q1. Now, the current through inductor L1 increases. The slope of the increase is a function of the value of L1 and the battery voltage. The current through L1 increases until it reaches a maximum value, which depends on the gain of Q1. Because the value of R3 sets the base current drawn from Q1, Q1s collector current is also limited.
White LEDs Circuit Diagram
Once the current through L1 reaches its maximum value, the slope of the current through L1 changes. At that instant, the voltage on L1 switches to a negative polarity forced by the changed slope. This negative voltage traverses capacitor C1 and turns off Q2, which in turn turns off Q1. The negative voltage on L1 increases until it reaches the forward voltage of the LED. The peak current through inductor L1 now flows through the LED and decreases to zero. Now, Q2 switches on again, via the current through R2, and the cycle starts again.
By adjusting resistor R3, you can set the peak current through L1 and the peak current through the LED. The brightness of an LED is a linear function of the current through the LED. So, adjusting the value of R3 also adjusts the brightness of the LED.
It doesnt matter which LED you use; the forward voltage on the LED always increases until the peak current through L1 flows through the LED. Different forward voltages of the LEDs yield different on-times (duty cycles) but the same peak current through the LED. With the values shown in Figure 1, the circuit oscillates at a frequency of approximately 30 kHz and delivers a 20-mA peak current through the LED.
The duty cycle depends on the ratio of the battery voltage to the forward voltage of the LED. One advantage of this circuit is that it requires no series-limiting resistor for the LED. The peak current through the LED is a function of the value of R3 and the gain of Q1.[via]
Monday, 29 July 2013
In this article we discuss an interesting circuit which can be used for controlling our mains AC line fluctuations and for producing very accurate stabilized voltage outputs for our domestic electrical appliances.
More innovative circuits HERE.
The circuit concept is rather very simple. It uses discrete op amps wired up as comparators to sense the voltage levels.
As can be seen in the diagram, each op amps inverting inputs are provided with sequentially incrementing voltage refrence levels through a series of presets which drops a certain amount of voltage across itself.
Each op amp compares this voltage with the common sample mains AC voltage level supplied to the op amps non inverting inputs.
As long as this sample voltage is below the refrence level the respective op amps keep their outputs low and the subsequent transistor relay stages remain inactive, however in case the voltage levels tends to shift from its normal range, the relevant relays trigger and toggle the transformer taps so that the output is appropriately equalized and corrected.
For example if the input AC voltage tends to fall, the upper relays may get triggered connecting the relevant higher voltage taps with the output and vice versa in case the voltage shoots upwards.
Here the op amp output inter-connections makes sure that only one optocoupler and therefore only one relay gets activated at a time.
Parts List
P1---P8 = 10 K Preset,
A1---A8 = IC 324 (2 Nos)
R1---R8 = 1 K,
All diodes = 1N4007,
All relays = 12 volts, 400 Ohms, SPDT,
Opto Couplers are all = MCT2E or equivalent,
Transformer = Pink Tap is normal voltage tap, the upper taps are in the decrementing order of 25 Volts, while the lower taps are in the incremental order of 25 volts.
Friday, 12 April 2013
A 12 V car battery can be used as the 12V source.
Use the POT R1 to set the output frequency to50Hz.
For the transformer get a 9-0-9 V , 10A step down transformer.But here the 9-0-9 V winding will be the primary and 220V winding will be the secondary.
If you could not get a 10A rated transformer , don’t worry a 5A one will be just enough. But the allowed out put power will be reduced to 60W.
Use a 10 A fuse in series with the battery as shown in circuit.
Mount the IC on an IC holder.
Remember,this circuit is nothing when compared to advanced PWM inverters.This is a low cost circuit meant for low scale applications.
Inverter Design Tips.
The maximum allowed output power of an inverter depends on two factors.The maximum current rating of the transformer primary and the current rating of the driving transistors.
For example ,to get a 100 Watt output using 12 V car battery the primary current will be ~8A ,(100/12) because P=VxI.So the primary of transformer must be rated above 8A.
Also here ,each final driver transistors must be rated above 4A. Here two will be conducting parallel in each half cycle, so I=8/2 = 4A .
source: circuitstoday.com
Use the POT R1 to set the output frequency to50Hz.
For the transformer get a 9-0-9 V , 10A step down transformer.But here the 9-0-9 V winding will be the primary and 220V winding will be the secondary.
If you could not get a 10A rated transformer , don’t worry a 5A one will be just enough. But the allowed out put power will be reduced to 60W.
Use a 10 A fuse in series with the battery as shown in circuit.
Mount the IC on an IC holder.
Remember,this circuit is nothing when compared to advanced PWM inverters.This is a low cost circuit meant for low scale applications.
Inverter Design Tips.
The maximum allowed output power of an inverter depends on two factors.The maximum current rating of the transformer primary and the current rating of the driving transistors.
For example ,to get a 100 Watt output using 12 V car battery the primary current will be ~8A ,(100/12) because P=VxI.So the primary of transformer must be rated above 8A.
Also here ,each final driver transistors must be rated above 4A. Here two will be conducting parallel in each half cycle, so I=8/2 = 4A .
source: circuitstoday.com
Thursday, 11 April 2013
A very efficient 6V to 15V DC to DC converter using LM2585 is shown here. LM2585 is a monolithic integrated voltage converter IC that can be used in various applications like flyback converters, boost converters, forward converters, multiple output converters etc. The circuit requires minimum number of external components and the IC can source up to 3A output current.
Circuit diagram :
6 to 15V DC to DC Converter Circuit Diagram
Here the IC is wired as a boost converter where resistors R1 and R2 are used to set the output voltage .The junction of R1 and R2 is connected to the feedback pin of IC1. Capacitor C4 is the input filter while capacitor C1 the filter for output. Network comprising of resistor R1 and capacitor C2 is meant for frequency compensation. Inductor L1 stores the energy for acquiring boost conversion.
Notes:
- Assemble the circuit on a good quality PCB.
- LM2585 requires a heatsink.
- Output voltage is according to the equation Vout =( (R1/R2)+1) x 1.23.
- Capacitors other than C4 and C1 are ceramic capacitors.
- Maximum output current LM2585 can source is 3A.
Source : Circuitstoday
Circuit diagram.
Seetharaman’s words about the circuit:
I am sending you a table lamp made from defunct energy saver lamp with
broken tubes. CFL converted into LED lamp. Most of the components will
be available in ones scrap box. Few components available in the CFL PCB
also can be used.
Procedure1. Carefully remove the broken glasses
2. Open the assembly carefully
3. Remove electronics and discard
4. Assemble the circuit in dot matrix PC or on a 1mm laminate sheet.
5. Cut a round laminate sheet with (scissor)
6. Mark the position of the 6 round holes on the sheet
7. Drill the holes to suit the LEDs to flush fit in the six holes
8. Use a dab of adhesive to keep the LED assembly in position
9. Close the assembly
10. Ensure the internal wiring does not touch each other
11. Now test on 230Volt AC
Your nice compact table lamp / puja room lamp / passage lamp is ready for use.
Monday, 8 April 2013
This is an
audio amplifier that can be used with a small 9 volt Battery
Operated,Current use as little as 5 milliamps.And amplification up to
500 mW.
Which is sufficient to expand the sound from a sound about or the CD Walk Man out to the small speakers clearly.
When entering the power supply 9-volt circuit IC1 number LM386 amplifier IC size is 300-800 mW, Depending on the power supply circuit with,This is from 4-15 volts.
Once entered into the input pin 3,The non inverting pin to amplifier
non-return phase.C1 will be served cut out the noise input to ground.And
C2 increases the rate of amplifier,C2 is to add more value.But if the
C2 Too much distortion (the C2 should not exceed 100uF).The output of
IC1 is out of the pin 5 through C4 coupling audio signals to better and
DC block and not passed to the speaker.For the audio portion will also
be fed back through R2 and C3 to the high frequency response better.
++++++++++++++++++++++++++++++
Next circuit ideals that use LM386 IC.
circuit number LM386 IC is used as the IC, which is popular is that it
has. It is a simple circuit. Less equipment items. Suitable for use or
used in small trials.
The properties of the IC can be used from
4V-12V power supply for low current at 50 mA only. And the frequency
response from 40Hz – 100 kHz rate of expansion of 46 dB and distortion.
Less than 1%.
When entering the power supply light LED1 circuit to tell the
operating environment of the circuit. C6 and C7, with a page filter to
smooth then be entered through one input sound signal.
C1 coupling
signal protection dc voltage noise in circuit to the sound signal that
is transmitted through VR1 for controller level reputation of sound
signal and then sent to the input to pin 3 of IC1.
boost up output at
pin 5 through C5 for protection dc voltage and meet the low frequency
better and send out put speakers. The C4 and R1 is acting eliminate
noise signal out and the pin 1 of IC will have a jumper for. to access
the C3 to boost up rate increase in case the circuit to be used to
boost up signal is very small.
circuit, will help expand the sound small. such as radio, sound out,
including CD WALKMAN. To the more powerful sound.The amplifier circuit
can be up to 7 watts. Enough to use a good listening room.And most
importantly is a very simple circuit.
Operation of the circuit. be Input signal is coupling with the
C1.Passed into the pin 1 of IC1 number LM383. Amplifier output from
the pin 4.Through C3 increase in low frequency stability better, before
leaving to the speaker. The R2 and R3 set the gain 100 times. Which is
calculated from the (R3/R4) +1.The C2 functions help in frequency
response.
audio amplifier that can be used with a small 9 volt Battery
Operated,Current use as little as 5 milliamps.And amplification up to
500 mW.
Which is sufficient to expand the sound from a sound about or the CD Walk Man out to the small speakers clearly.
When entering the power supply 9-volt circuit IC1 number LM386 amplifier IC size is 300-800 mW, Depending on the power supply circuit with,This is from 4-15 volts.
Once entered into the input pin 3,The non inverting pin to amplifier
non-return phase.C1 will be served cut out the noise input to ground.And
C2 increases the rate of amplifier,C2 is to add more value.But if the
C2 Too much distortion (the C2 should not exceed 100uF).The output of
IC1 is out of the pin 5 through C4 coupling audio signals to better and
DC block and not passed to the speaker.For the audio portion will also
be fed back through R2 and C3 to the high frequency response better.
++++++++++++++++++++++++++++++
Next circuit ideals that use LM386 IC.
Amplifier 500 mW with ic LM386N
Thiscircuit number LM386 IC is used as the IC, which is popular is that it
has. It is a simple circuit. Less equipment items. Suitable for use or
used in small trials.
The properties of the IC can be used from
4V-12V power supply for low current at 50 mA only. And the frequency
response from 40Hz – 100 kHz rate of expansion of 46 dB and distortion.
Less than 1%.
When entering the power supply light LED1 circuit to tell the
operating environment of the circuit. C6 and C7, with a page filter to
smooth then be entered through one input sound signal.
C1 coupling
signal protection dc voltage noise in circuit to the sound signal that
is transmitted through VR1 for controller level reputation of sound
signal and then sent to the input to pin 3 of IC1.
boost up output at
pin 5 through C5 for protection dc voltage and meet the low frequency
better and send out put speakers. The C4 and R1 is acting eliminate
noise signal out and the pin 1 of IC will have a jumper for. to access
the C3 to boost up rate increase in case the circuit to be used to
boost up signal is very small.
Small audio amplifier IC 7w using lm383
Thiscircuit, will help expand the sound small. such as radio, sound out,
including CD WALKMAN. To the more powerful sound.The amplifier circuit
can be up to 7 watts. Enough to use a good listening room.And most
importantly is a very simple circuit.
Operation of the circuit. be Input signal is coupling with the
C1.Passed into the pin 1 of IC1 number LM383. Amplifier output from
the pin 4.Through C3 increase in low frequency stability better, before
leaving to the speaker. The R2 and R3 set the gain 100 times. Which is
calculated from the (R3/R4) +1.The C2 functions help in frequency
response.
Saturday, 6 April 2013
This circuit in the figure is designed to help if you must transfer dc power and audio over a pair of copper wires. One application for such a circuit is a low-cost door-opening system with speech input. The circuit uses only one IC, the well known LM317, a low-cost power supply regulator. Using this chip, you can modulate the adjustment-pin input with the audio signal from an electrets condenser microphone, connected between the output and the adjustment terminals of the IC. The LM317 regulates the output in such a way that the voltage on the microphone is always 1.25V dc.
This circuit uses a WM34 electrets microphone, which comes in a standard 10-mm capsule from Panasonic and is common in low-cost equipment. You can use nearly any electrets capsule, because the well-regulated voltage on the microphone never exceeds 1.25V.
The principle work is every electrets capsule contains an integrated JFET based impedance converter that translates speech into a current flowing from the source to the drain terminal. This current through the microphone modulates the voltage on the variable resistor, RP. Because the output of the LM317 must follow the voltage on RP, you obtain a low-impedance audio signal riding on the output dc voltage. The microphone directly modulates the adjustment pin, so a smoothing capacitor, such as C1, for noise and hum does not influence the level of the audio signal. C1 shunts some of the audio signal to ground, but the LM317 compensates for the loss with internal gain. To avoid excessive losses in the LM317, use a capacitor with as low a value as possible. The circuit works well without a capacitor, but values as high as 47F do not present a problem. Using RP, you can adjust the dc output voltage and the gain for the microphone signal. For proper operation, the LM317 needs to deliver a minimum current of 4 mA from its output terminal. If your design uses no loudspeaker, you can connect a load resistor to sink this 4 mA. Designs using low-impedance loudspeakers must also have load resistors. You must add the ac current in the audio signal to the minimum current requirement of 4 mA. For an 8 loudspeaker, you need a minimum resistive load of 470_ to avoid distortion.
This circuit uses a WM34 electrets microphone, which comes in a standard 10-mm capsule from Panasonic and is common in low-cost equipment. You can use nearly any electrets capsule, because the well-regulated voltage on the microphone never exceeds 1.25V.
The principle work is every electrets capsule contains an integrated JFET based impedance converter that translates speech into a current flowing from the source to the drain terminal. This current through the microphone modulates the voltage on the variable resistor, RP. Because the output of the LM317 must follow the voltage on RP, you obtain a low-impedance audio signal riding on the output dc voltage. The microphone directly modulates the adjustment pin, so a smoothing capacitor, such as C1, for noise and hum does not influence the level of the audio signal. C1 shunts some of the audio signal to ground, but the LM317 compensates for the loss with internal gain. To avoid excessive losses in the LM317, use a capacitor with as low a value as possible. The circuit works well without a capacitor, but values as high as 47F do not present a problem. Using RP, you can adjust the dc output voltage and the gain for the microphone signal. For proper operation, the LM317 needs to deliver a minimum current of 4 mA from its output terminal. If your design uses no loudspeaker, you can connect a load resistor to sink this 4 mA. Designs using low-impedance loudspeakers must also have load resistors. You must add the ac current in the audio signal to the minimum current requirement of 4 mA. For an 8 loudspeaker, you need a minimum resistive load of 470_ to avoid distortion.
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