Check calculations project dual power supply stabilized 5V - Page 1
Forum section: electronics / beginners

author: giorgik
author_rank: Contributor
timestamp: April 28, 2015, 06:19:04 pm
content: Hello everyone, I did a project for a dual power supply stabilized by +5 / -5V using as voltage stabilizers L7805CV / L7905CV. I put in the attachment designed circuit.C3, C4, C5, C6 are defined by the product description L7805CV / L7905CV.C1, C2 are obtained from C = Iout / (2 * f * Vripple) so if Iout = 440mA,Vripple = 2V -> C = 0.0022F = 2200uF.The input voltage is given by the L7805CV Vin1 = Vout1 + Vdropout + 2V = 5V = 7V.In C1 tension appears to have Vc1min = Vin1 = 7V, Vc1max = Vc1min Vripple = + 7V + 2V = 9V.Bearing in mind that the diode Dp1 of the rectifier bridge has a voltage of 0.7V, the voltage on the first transformer secondary Vsec1 = Vc1max VDP1 = + 9V + 0.7V = 9.7V -> 10V.If the secondary of the transformer has a loss of 8%, V1 = 10V + 8% * 10V = 10.8V ->V1 rms = V1 / sqr (2) = 7.63V -> 8VIn the market find a transformer with dual secondary 9V rms, 16VA -> I = 16VA / (9V + 9V) = 0.888A -> I1 = I2 = I = 0.888AR1 = R2 = (5V - Vled)/Iled = (5V - 2V)/15mA = 200 ohm -> 220 ohm.Do you think these calculations are correct ?

author: giorgik
author_rank: Contributor
timestamp: April 29, 2015, 08:50:55 am
content: no one can tell me if the calculations are correct ?

author: codeboy2k
author_rank: Super Contributor
timestamp: April 29, 2015, 10:55:18 am
content: Calcs look ok, yes. Good work there.Some additional points:1) You might want to include a diode across the regulator from output to input. This protects the regulator if the output rises above the input, especially on turn-off when the load might still be connected and it has some capacitance that has not discharged fully. This is definitely needed for any bench supply, but might not be needed in a supply that has a fixed purpose and you control the load.2) 15 mA might be too bright for Iled, especially if it's a modern LED. You can usually run modern LEDs at 5mA or so. The LED resistor can be the next step up maybe 330 (8 mA) or 470 (6 mA) ohms. It really depends on what you like. I personally don't like bright LEDs3) consider the power dissipated in the regulators. I always use the max values for this, disregarding the ripple. So you will have about 12V max input with 5V output at 440mA, that's P=VA = (12-5) * 0.440 = 7 * .440 = 7 * .5 = 3.5W3.5W Pd.. Can it work in free air? (I know it cannot, but let's do the calculation). The datasheet says the TO-220 is 50 degC per watt Junction to Air. So 50 degC/W *3.5W = 175 degC rise above ambient. If ambient is 35-40C, then the junction will be at 40+175 = 215C. The datasheet also says the maximum junction temperature is 125C so clearly 3.5W in free air will not work. You need a heatsink.So if we say maximum ambient will be 40C and maximum junction temperature allowed is 125C , then the maximum temperature rise that you can tolerate will be 125C - 40C = 85C. So the power dissipated across the thermal resistances cannot allow a rise of more than 85C.TR = thermal resistanceTRmax= 85C / 3.5W = 24.3 C/W max.TRj-c= junction-to-case = 5 deg C/W (from datasheet)TRpaste= 0.5 deg C/W (varies, but you can use .2 to .5 C/W for paste)So TRheatsink= 24.3 - 5 - 0.5 = 18.8 C/W.So you need a heatsink with a thermal resistance of 18.8 deg C / W.This oneis 20 deg C/W, it will probably work if your ambient is not 40C as I calculated (or in an airflow, where it might be about 11 deg C/W)However, that's the absolute minimum heatsink you need, because using that one could run the regulator junction at it's max temperature of 125C. We don't like to do that, so let's try to keep the junction at 100C now.redoing the calculations:100 - 40C = maximum temperature rise of 60C60 C / 3.5W = 17 deg C/W17 - 5 - 0.5 = 11.5 deg C/W for the heatsink. The above heatsink will probably work, in an airflow (fan).However, if you are in free air, you might wantthis onewhich is 10.5 deg C/W in free air.

author: giorgik
author_rank: Contributor
timestamp: April 29, 2015, 03:03:23 pm
content: Thanks codeboy2k for your answer, advice and replicas of calculations.You've been very helpful.)

author: schopi68
author_rank: Regular Contributor
timestamp: April 29, 2015, 03:32:05 pm
content: Looks a little bit as if you've used the datasheet data from the L7805 also for the L7905. The L7905 does need other capacitor values to be stable than the 7805. The 7905 has a much higher tendency to oscillate than the 7805 because of it's different internal design.The reverse-protection diode from output to input on the regulator mentioned by codeboy2k is only neccessary for the positive regulator and only if the output voltage level is greater than 6 Volts.And please add a fuse on the power line (and maybe a line EMI filter).

author: giorgik
author_rank: Contributor
timestamp: April 29, 2015, 04:58:33 pm
content: Thanks schopi68, at this point the circuit as is, can you give me the correction ?Another question: Is there another way (circuit diagram) to create a dual power supply stabilized? can you give me the diagram and possibly explain how to obtain the values of the various components ?Is there any tutorial or book to download (free) from the Internet to teach me good to design power supplies ?

author: schopi68
author_rank: Regular Contributor
timestamp: April 30, 2015, 09:43:16 am
content: Replace C4 with at least a 2,2 uF tantalum capacitor. Alternative would be an 25 uF aluminium electrolytic.And replace C6 with 1uF.C3,4,5,6 should be placed as near as possible to the regulators. (Pin 1 is GND, Pin 2 is IN, Pin 3 is OUT)Please note that the 7905 has a different pinout than the 7805.I have to correct my previous made notice about the reverse-protection diode between input and output of the 7805: There are some manufacturer differences. Some seems to need it even in the 5V-Version of the 7805. The 7905 has another internal circuit design, it does not need this diode.Another good idea would be to give some additional protection against static discharge. This can be done by replaceing D1 and D2 with a Voltage supressor diode. I.e. 1N5908. It will give the same function as D1/D2 against negative polarity during startup or shortage, but will also act as a fast Zener diode with a breakdown Voltage of 6V. So high-energy transients on the output from electrostatic discharge will be absorbed and the regulators are protected (It depends on the usage of your power supply if you need this protection). Furthermore you have some protection of the circuits behind your power supply against failure of the regulators.For fusing and EMI filtering you could simply use an fused inlet like this one (PS20/A):http://www.bulgin.co.uk/Products/Mains_Filters/Fused_Inlets.htmlFusing:16VA/220V = 72 mAtransformer loss (~ 1.2) and derating (1.5-2)72mA * 1.2 * 1.5 = 129 mA-> Fuse with 0,16A T (slow blow)A 0,1A Fuse

author: giorgik
author_rank: Contributor
timestamp: April 30, 2015, 05:29:29 pm
content: Thanks schopi68. As a protection diode for only L7805 place between input (pin 1) and output (pin 3) what kind is fine (eg. 1N4001) ?This power supply it uses it only for projects of simple pre / amplifier with LM324, TL084 ect ... .For example I wanted to do the sound-detector circuit Byron Jacquot (open hardware). Here, however, we only use the + 5V (then I will use a single power supply stabilizzatousando the only part of the circuit with L7805).I have a transformer like the one shown in my project (220Vac / 9 + 9Vac, 16VA), I can also use the power adapter for single output stabilized + 12V or + 5V stabilized ?