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Talkin' PSUs |
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Ashleip 
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Joined: 08 Jan 2021 Status: Offline Points: 23 |
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 Posted: 22 Jan 2021 at 9:02am |
 need a good map to navigate the marketing campaigns! Good to hear that for something like this EI is perfectly good and no need to spend 3x as much for a heavy doughnut.
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Graham Slee
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Retired Joined: 11 Jan 2008 Location: South Yorkshire Status: Offline Points: 16298 |
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Posted: 22 Jan 2021 at 9:16am |
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So much for the interest in the design process. The significant bit about 40.05 volts wholly ignored because the topic must descend into a debate about the incredible beauty of toroidal transformers. And there was me thinking you'd like to learn something. You know it all already!
But I will continue for the sakes of those who might eventually come along with unprejudiced minds. |
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Ashleip
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Joined: 08 Jan 2021 Status: Offline Points: 23 |
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Posted: 22 Jan 2021 at 9:24am |
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I had noticed this, but got sidetracked... What is it about transformers eh?!
As this is only slightly above 40, can we change diodes to increase the drop slightly or is that a stupid idea? Do we need to go looking for a transformer with <1.1 voltage regulation, or is that a pretty standard value? (Edited this post because I hadn't properly read the previous post
)Edited by Ashleip - 22 Jan 2021 at 9:37am |
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Graham Slee
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Posted: 22 Jan 2021 at 9:48am |
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The point was 40.05 volts.
The sustaining voltage - the absolute maximum on the datasheet for the 7824 voltage regulator, is stated as being 40V DC. We're supposed to learn the difficulty in designing a mains power supply, having a regulated output, for operation on all 230V voltages. I started this topic because of the number of members who have asked me about a PSU1 in kit form. Then there is another group who wonder if a PSU1 can be used to power other items, and if not, what can I do to assist them? Rather than pushing all the bits into a bag and selling it as a kit - a lethal instrument due to its mains connection - perhaps some education into the hows and whys might result in something the DIY'er can make better use of. So, back to the 40.05 volts calculated earlier. That is the maximum DC input on the voltage regulator we can squeeze it down to for a mains supply at the top of the specified grid voltage range while ensuring it doesn't drop out of regulation at the bottom of the specified grid voltage range. The crucial questions here should be about how we can ensure reliability. The reader could be in some European country where the mains power demands take the voltage right down or somewhere in industrial England where a large substation now supplies a smaller domestic demand. The latter can be subjected to 255V AC. The DIY enthusiast cannot easily obtain a transformer of the exact voltage required (which is around 23V) and must work with what is available. We should be asking, will 40.05V damage a 7824 voltage regulator? |
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Graham Slee
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Posted: 22 Jan 2021 at 10:19am |
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A diode is NOT a stupid idea. (
 )Edited by Graham Slee - 22 Jan 2021 at 10:20am |
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Graham Slee
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Posted: 22 Jan 2021 at 7:50pm |
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If we put a forward-biased diode just before the 7824 input, the off-load voltage drop is roughly 0.6V, and that adds 1.5% into the "component tolerance pool." It might still reach 40V, but we've made extra allowances for inaccuracies, which inevitably creep in everywhere.
We also have to put some faith in the electricity grid. The datasheet doesn't tell us the conditions of the absolute maximum voltage, whether it's on-load or off-load. If it said on-load, then we know that off-load, the input voltage will be higher. Unfortunately, it doesn't say. The datasheet also notes, "Absolute maximum ratings are those values beyond which damage to the device may occur. Functional operation under these conditions is not implied." So, it says not implied, so better not conjecture. At the low mains voltage end, we had about 0.7V going spare, plus the 26.5V minimum 7824 input. Theoretically, the 7824 input can go as low as 26V, but take a look at the output tolerances on any 7824 datasheet. Every little helps! The diode has made it so the rectifier circuit centres-up with the 7824 input range for all expected grid conditions. |
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Graham Slee
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Posted: 23 Jan 2021 at 8:23pm |
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The non-technical layman understands electricity due to it being a greater part of modern living.
He understands Voltage that, in most cases, it appears between two points. The live and neutral of a mains plug and the pointed end and the blunt end of a battery are two examples. In a mains derived DC power supply, both types of electricity exist. The input is from the live and neutral of a mains plug. The output mimics that of a battery. A battery contains charge, and in use, the charge flows in the circuit it powers, and that flow is called current. As the charge is consumed, the current flow diminishes, and eventually, all the charge has gone. The smoother or reservoir capacitor of a mains derived DC power supply acts similarly to a battery. It doesn't have the long term charge of a battery. Instead, a rectifier charges it, and the load discharges it. The rectifier is constantly applying charge, and the load is continuously discharging it.  (image copied from Morgan Jones) The transformed and rectified waveform is like a series of upward going loops. The top of each loop is the peak value of the AC voltage. VDC = √2 x Vac The charge is used up in current flow in the circuit powered by the power supply. The smoother capacitor voltage decays, only to be met by the next loop. These loops are 0.01 seconds apart. Rather than the DC voltage being a straight line over time, it ventures up and down by an amount called ripple voltage. The ripple voltage is directly related to the current being drawn from the power supply. If we don't draw any current, then this DC voltage is a straight line over time, but as soon as we draw current, the Voltage starts to ripple. The letter used for charge is Q, and it stands for coulombs. Charge is related to Capacitance (the storage of charge) and Voltage. Q = CV Charge is used up by current draw over time, so Q = IT (I stands for current in amps, T stands for time in seconds) Therefore Q is related to both, so Q = CV = IT If we want to find Capacitance and know what ripple voltage is acceptable and the current the load draws, we also understand that the rectified waveform repeats every 0.01 seconds. We can use CV = IT Rearranging it to obtain C on its own involves dividing both sides of the equation by V. The V's of CV/V cancel to become C. On the other side, we have IT/V. C = IT/V If we want to know the ripple voltage, then we divide both sides by C V = IT/C So now, even the layman should have a better idea of designing power supplies. |
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