slackjack Posted May 20, 2006 Report Share Posted May 20, 2006 I need to build a power supply to power my projetcs I am going to build. It should have an adjustable range from 0V - 32V. But I can settle for a range up to say 20V. Most of the schematics I found on the net have an input of 5V DC. Where I am suppost to get a constant 5V DC signal from ??? Its to get this 5V thats why I need to build a power supply.Now the second part of my question, is there any AC to DC power supply projetcs that take in 120Vac and meets all the requirements I stated above?--thank you :) Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 20, 2006 Report Share Posted May 20, 2006 Make a variable voltage regulated power supply from an LM317 IC. Its voltage varies from 1.25V to a max of 35V but a 24V transformer will allow it to have a max output of about 30V and a current of 1.0A. Below 15V output the current will be less.Our project has a few problems so read the topic linked in it:http://www.electronics-lab.com/projects/power/010/index.htmlA regulated 5.0V supply is made with a 6.3VAC transformer, a bridge rectifier, a big filter capacitor and a 7805 regulator IC. Connect a small capacitor at the output for stability.The regulator ICs are smart. If the load current is too high then they limit the current.If they get too hot then they shut-down until they are cool enough. Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 20, 2006 Author Report Share Posted May 20, 2006 Thank you. ;DDoes the LM317 amplify the input voltage in any way? If I decreased the transformer output voltage, then the max output volatge from the LM317 decreases proportionally?Another question - can I use a half wave rectifier instead? Since the LM317 is a voltage regulator, it should eliminate the extra ripple caused by the half wave rectifier right? Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 20, 2006 Report Share Posted May 20, 2006 Does the LM317 amplify the input voltage in any way? If I decreased the transformer output voltage, then the max output volatge from the LM317 decreases proportionally?It does not boost the input voltage. It won't regulate the output voltage if its input voltage is less than about 2V more than its output voltage. If the resistance of the pot is too high then you will not have regulation at the highest setting.Another question - can I use a half wave rectifier instead? Since the LM317 is a voltage regulator, it should eliminate the extra ripple caused by the half wave rectifier right?Definately not, it is not a generator! The input voltage must always be at least about 2V higher than its output voltage. Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 20, 2006 Author Report Share Posted May 20, 2006 Just another question regarding the schematic. What is purpose of putting the 2200uF and the 100nF capacitor in parallel? Do they act as the filter? Why two of them? Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 20, 2006 Report Share Posted May 20, 2006 Just another question regarding the schematic. What is purpose of putting the 2200uF and the 100nF capacitor in parallel? Do they act as the filter? Why two of them?The 2200uF electrolytic capacitor is a good filter and a low impedance at low frequencies, but since it is large and is made with its layers wound in a circle, it has inductance which causes its impedance to rise at high frequencies.The 0.1uF ceramic disc capacitor is small and flat so has low impedance at high frequencies.A regulator has a high gain amplifier inside that might oscillate without the ceramic disc capacitor at its input, and another at its output. Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 20, 2006 Author Report Share Posted May 20, 2006 I know that the 4k7 is a 5K-Ohm Linear Potentiometer from radio shack (part number 271-1714 as seen on http://www.uoguelph.ca/~antoon/gadgets/rs.htm ) Berfore I build my projects I always simulate using multisim or spice. Well on multisim I determined that a 5k pot only gives a max output voltage of around 14V-15V. However with a 10K pot, the max output is approx 25V. Mayb the schematic needs to be changed because I'm pretty sure I did the simulation correctly. I have determined that a 10K pot is the best alternative to the 5k pot. First graph is 5k pot and the second is 10k pot.I also noticed that multisim has LM317AH, H, K, LM, LZ. I tried the LM317LZ and LM317LM and both proved the same results. WHats the difference between all these LM chips stated above?I tried uploading the multisim file, but that file type is not allowed so I've uploaded as a JPEG.thanks again. Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 20, 2006 Report Share Posted May 20, 2006 I know that the 4k7 is a 5K-Ohm Linear Potentiometer from radio shack (part number 271-1714 as seen on http://www.uoguelph.ca/~antoon/gadgets/rs.htm ) Berfore I build my projects I always simulate using multisim or spice. Well on multisim I determined that a 5k pot only gives a max output voltage of around 14V-15V. However with a 10K pot, the max output is approx 25V. Mayb the schematic needs to be changed because I'm pretty sure I did the simulation correctly. I have determined that a 10K pot is the best alternative to the 5k pot. First graph is 5k pot and the second is 10k pot.Your pots are set to half. Therefore their max voltage setting is nearly double.Your 28VAC transformer's voltage is too high, it produces an unregulated voltage of 38VDC. A 24VAC transformer will produce 32VDC.I also noticed that multisim has LM317AH, H, K, LM, LZ. I tried the LM317LZ and LM317LM and both proved the same results. WHats the difference between all these LM chips stated above?Look on the datasheet! 100mA, 500mA or 1A. Different packages and heatsink possibilities.The datasheet shows that a higher-price LM117 uses a 240 ohm resistor. An LM317 is supposed to have a 120 ohm resistor instead, or its output voltage might rise without a load. Then the pot's value will also need to be cut in half. Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 21, 2006 Author Report Share Posted May 21, 2006 So you recommend that I use a 120 ohm(for LM317) instead of the 240ohm? And with that I must change the pot to a 2.5k right? Why wouldnt the 5k pot suffice with the resistor change?Also on multisim, using a 24Vac, I get an output of about 21Vdc and not 32Vdc. ??? Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 21, 2006 Report Share Posted May 21, 2006 So you recommend that I use a 120 ohm(for LM317) instead of the 240ohm? And with that I must change the pot to a 2.5k right? Why wouldnt the 5k pot suffice with the resistor change?Look at the datasheet or at a tutorial for the LM317 to see how to calculate its resistors.A 240 ohm resistor will have 1.25V across it which makes 5.2mA. 5.2mA times 5k= 26V. The output is 26V + 1.25V= about 27.25V. If you use a 120 ohm resistor then its current is 10.4mA and the max output voltage with a 5k pot is 53.25V which is impossible. A 120 ohm resistor and a 2.5k pot make a max output of 27.25V.Also on multisim, using a 24Vac, I get an output of about 21Vdc and not 32Vdc. Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 21, 2006 Author Report Share Posted May 21, 2006 Now lets say I wanted to make a digital voltage indicator to tell me my output voltage from the power supply and I wanted to power the indicator from my power supply. If it uses 5V would it be safe to setup some sort of voltage divider across the .1uF capacitor? The digital volt meter here has a range up to 1999. This is way more than I need. Are there any other low power ones that I can use with a smaller range that fits my purpose? Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 21, 2006 Report Share Posted May 21, 2006 The digital voltmeter project cannot measure its own supply. It needs a separate supply.Its IC has a max reading of 0.1999V and measures higher voltages with a voltage divider at its input. Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 21, 2006 Author Report Share Posted May 21, 2006 The reason I was getting around 21Vdc-22Vdc on the output when using 24Vac is beacuse the source I was using was already in peak voltage and not RMS ;D Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 21, 2006 Author Report Share Posted May 21, 2006 Since I cant use a digital voltage indicator mayb I can opt for a bunch of LEDs to indicate the voltage. Mayb one likethis would do? Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 21, 2006 Author Report Share Posted May 21, 2006 I plan on using comparators instead to indicate voltage levels. For now I experiment with 3 different voltages - 5V, 12V, and 25V. I plan on using the LM741 general purpose operational amplifer. I have a problem with this circuit though:V1 is used to represent the reference voltage (connected to the non inverting input). I set up a voltage divider so that each 741 has its own reference voltage. V2 is used to represent the output voltage from the power supply. V3 supplies +-Vcc to both comparators (and is also the voltage source for the LEDs). The forward voltage drop for the LED is 1.8V and activates at 2mA. I calculated that the current limiting resistors should be 3600 ohms.What did I forget to add to this circuit (or connected wrongly) because it does not work? Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 22, 2006 Report Share Posted May 22, 2006 The problem is that the opamp's input voltages far exceed their common-mode voltage limit which is a few volts less thas the opamp's supply voltage. You should use voltage dividers to reduce the input voltages.Why not use an LM3914 bar-graph IC? It has a programmable voltage reference, 10 comparators/LED-drivers with a 10 resistors divider and a programmable constant current output to the LEDs so current-limiting resistors aren't required. Quote Link to comment Share on other sites More sharing options...
slackjack Posted May 22, 2006 Author Report Share Posted May 22, 2006 The LM3914 looks very neat. According to this website they say that it has a LM317 voltage regulator built in. Is it a good idea to use this IC for both regualtion and voltage display? Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 22, 2006 Report Share Posted May 22, 2006 The voltage regulator in an LM3914 is to provide a stable reference voltage for its comparators. Whan it is loaded then the LEDs become brighter. An emitter-follower transistor can be used to feed a regulated voltage to something else.The LEDs don't need a voltage regulator because each LED driver is a constant current source with the current determined by the load on the voltage regulator. Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 24, 2006 Report Share Posted May 24, 2006 Hi Zeppelin,The LM117 and LM317 operating current goes to the output. The LM117 has a max of 5mA and the LM317 has a max of 10mA. The operating current is the highest at high input voltages.The resistor from its output to its adj terminal, and in series with the adjustable resistor to ground is its load when there isn't another load. Therefore if the resistor is 240 ohms and the LM317 has an operating current of 10mA then the voltage across the 240 ohm resistor is 2.4V which is much higher than the 1.25V required so the regulator's output voltage will rise above what it should be. If you use 120 ohms then 10mA through it makes 1.2V which is less than the required 1.25V so the output voltage remains at the regulated voltage.The 1st page of the datasheet shows an expensive LM117 with a 240 ohm resistor. Quote Link to comment Share on other sites More sharing options...
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