Showing posts with label 30. Show all posts
Showing posts with label 30. Show all posts
Thursday, December 26, 2013
LDR Based 12V White LED Driver for up to 30 LEDs
While we have now published quite a few LED driver circuits, to date we have not published a design to drive a bunch of high-brightness white LEDs. Such a circuit is now quite desirable as the price of white LEDs has fallen and you can have a handful for not a lot of dollars. However, white LEDs do present a problem because they need a higher drive voltage than monochromatic types such as red, green, orange etc.
Instead of around 1.8V to 2V or thereabouts, they normally require more than 3V to produce their rated brightness. In fact, if you are driving a bunch of them you need to drive them all at constant current otherwise their individual brightness tends to vary markedly. However, if you only have a 12V supply available, you can only put two or maybe three LEDs in series together with a constant current source and this leads to poor efficiency.
The approach in this circuit is to boost the 12V supply to something around 21V and this means that we can have groups of five LEDs, each in series with their own current source transistors. The result is a single PC board with the drive circuitry and 30 white LEDs. It can be used for lighting in caravans and recreational vehicles, emergency lighting or whatever application you can think of. Current drain is around 190mA at 12V.
Now let’s have a look at the circuit of Fig.1. It uses just one IC (a 4093 quad NAND Schmitt trigger gate package), a few transistors and diodes, 30 white LEDs and not much else. So where is the familiar boost converter circuit? Answer: there isn’t one or least not one with an inductor switched by a Mosfet. Instead, there is a charge pump inverter, comprising IC1c, transistors Q2 & Q3, Schottky diodes D1 & D2 and a few capacitors. It works as follows:
IC1c is connected as an inverter oscillator and its running frequency of about 30kHz is determined mainly by the 6.8kΩ resistor between pins 8 & 10 together with the 4.7nF capacitor at pin 8. This produces a rectangular waveform (not quite square but pretty close) at pin 10 to drive complementary switching transistors Q2 & Q3.
The waveform at their commoned emitters drives a diode pump consisting of two 100μF capacitors and Schottky diodes D1 & D2. The waveform generated by the circuit can be seen in the scope photo of Fig.2.
Oscillator IC1c is controlled by an RS (Reset/Set) flipflop comprising the two NAND gates IC1a & IC1b and this is controlled by pushbutton switches S1 and S2. Normally, this has its pin 4 low and pins 1 & 6 are pulled high via 470kΩ resistors. Momentarily closing S1 (ON) pulls pin 6 low, causing the flipflop to change state so that pin 4 now goes high to enable IC1c which now oscillates at 30kHz.
The 30kHz waveform produced by transistors Q2 & Q3 drives the diode pump referred to earlier and this develops about 21V to drive the LED columns. Each column of five white LEDs is driven by its own current source transistor which has a 33Ω emitter resistor. The bases of all six current source transistors (Q4-Q9) are driven from pin 4 of IC1b via a 6.8kΩ resistor and clamped to a maximum of +1.2V by diodes D3 & D4.
Subtract the 0.6V between the base and emitter of each transistor and you are left with 0.6V across each 33Ω resistor, thus setting the LED drive current to 18mA. Switching the circuit off is accomplished by pushing the OFF switch, S2. This momentarily pulls pin 1 low to toggle the RS flipflop, thus causing pin 4 to go low. This disables IC1c, Q2 & Q3 and also turns off the current source transistors.
Note that there is an interesting wrinkle to this drive circuit, because there is no On/Off switch. This means that the current source transistors must be turned off otherwise they would continue to draw current from the 12V supply even when the circuit is nominally off. The current path may not be obvious but it is via the boost circuit’s diodes, D1 & D2.
As well as using the pushbutton switches S1 & S2 to turn the circuit on and off, there is also a facility to automatically turn the circuit on and off depending on ambient light levels. Links L1 & L2 can be used to provide Auto On and Auto Off respectively and these features can be used separately or together.
An LDR (light dependent resistor) is used to monitor the ambient light level. When light falls upon it, it pulls the base of Q1 low, causing pins 12 & 11 of IC1d to go low and its pin 11 to go high. When darkness falls (or the room lights go out), the process is reversed. Depending on whether you have one or both links connected, you can use the pushbuttons to turn the circuit on and off and have it turn on and/off automatically as well.
Q1 also drives a red high brightness LED (LED1) at very low current, via a 470kΩ resistor. This is a bit of a gimmick but it does have the benefit of showing that this part of the circuit is working, if you have to trouble-shoot it.
Pins 1 & 2, 5 & 6 and 8 & 9 of IC1 on the circuit are all swapped. The PC board overlay diagram is correct.
Continue Reading[..]
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| 30 White LEDs Driver Circuit Diagram |
Instead of around 1.8V to 2V or thereabouts, they normally require more than 3V to produce their rated brightness. In fact, if you are driving a bunch of them you need to drive them all at constant current otherwise their individual brightness tends to vary markedly. However, if you only have a 12V supply available, you can only put two or maybe three LEDs in series together with a constant current source and this leads to poor efficiency.
The approach in this circuit is to boost the 12V supply to something around 21V and this means that we can have groups of five LEDs, each in series with their own current source transistors. The result is a single PC board with the drive circuitry and 30 white LEDs. It can be used for lighting in caravans and recreational vehicles, emergency lighting or whatever application you can think of. Current drain is around 190mA at 12V.
Circuit description
Now let’s have a look at the circuit of Fig.1. It uses just one IC (a 4093 quad NAND Schmitt trigger gate package), a few transistors and diodes, 30 white LEDs and not much else. So where is the familiar boost converter circuit? Answer: there isn’t one or least not one with an inductor switched by a Mosfet. Instead, there is a charge pump inverter, comprising IC1c, transistors Q2 & Q3, Schottky diodes D1 & D2 and a few capacitors. It works as follows:
IC1c is connected as an inverter oscillator and its running frequency of about 30kHz is determined mainly by the 6.8kΩ resistor between pins 8 & 10 together with the 4.7nF capacitor at pin 8. This produces a rectangular waveform (not quite square but pretty close) at pin 10 to drive complementary switching transistors Q2 & Q3.
The waveform at their commoned emitters drives a diode pump consisting of two 100μF capacitors and Schottky diodes D1 & D2. The waveform generated by the circuit can be seen in the scope photo of Fig.2.
RS flipflop
Oscillator IC1c is controlled by an RS (Reset/Set) flipflop comprising the two NAND gates IC1a & IC1b and this is controlled by pushbutton switches S1 and S2. Normally, this has its pin 4 low and pins 1 & 6 are pulled high via 470kΩ resistors. Momentarily closing S1 (ON) pulls pin 6 low, causing the flipflop to change state so that pin 4 now goes high to enable IC1c which now oscillates at 30kHz.
The 30kHz waveform produced by transistors Q2 & Q3 drives the diode pump referred to earlier and this develops about 21V to drive the LED columns. Each column of five white LEDs is driven by its own current source transistor which has a 33Ω emitter resistor. The bases of all six current source transistors (Q4-Q9) are driven from pin 4 of IC1b via a 6.8kΩ resistor and clamped to a maximum of +1.2V by diodes D3 & D4.
Subtract the 0.6V between the base and emitter of each transistor and you are left with 0.6V across each 33Ω resistor, thus setting the LED drive current to 18mA. Switching the circuit off is accomplished by pushing the OFF switch, S2. This momentarily pulls pin 1 low to toggle the RS flipflop, thus causing pin 4 to go low. This disables IC1c, Q2 & Q3 and also turns off the current source transistors.
Note that there is an interesting wrinkle to this drive circuit, because there is no On/Off switch. This means that the current source transistors must be turned off otherwise they would continue to draw current from the 12V supply even when the circuit is nominally off. The current path may not be obvious but it is via the boost circuit’s diodes, D1 & D2.
Auto on/off
As well as using the pushbutton switches S1 & S2 to turn the circuit on and off, there is also a facility to automatically turn the circuit on and off depending on ambient light levels. Links L1 & L2 can be used to provide Auto On and Auto Off respectively and these features can be used separately or together.
An LDR (light dependent resistor) is used to monitor the ambient light level. When light falls upon it, it pulls the base of Q1 low, causing pins 12 & 11 of IC1d to go low and its pin 11 to go high. When darkness falls (or the room lights go out), the process is reversed. Depending on whether you have one or both links connected, you can use the pushbuttons to turn the circuit on and off and have it turn on and/off automatically as well.
Q1 also drives a red high brightness LED (LED1) at very low current, via a 470kΩ resistor. This is a bit of a gimmick but it does have the benefit of showing that this part of the circuit is working, if you have to trouble-shoot it.
Note:
Pins 1 & 2, 5 & 6 and 8 & 9 of IC1 on the circuit are all swapped. The PC board overlay diagram is correct.
Wednesday, December 18, 2013
Simple 12 Volt 30 Amp PSU Circuit Diagram
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:
Simple 12 Volt 30 Amp PSU Circuit Diagram
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 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:
Circuit Diagram
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 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.
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