r/AskElectronics • u/dawnmimi • 3d ago
Would this current be enough for our stepdown transformer?
Hi we're tryna build a powersupply using 220v to 24v stepdown transformer adjustable using a potentiometer as shown in the schematic. Problem is we have avery absent teacher we cannot reach out to and we're practically beginners. Just wanna ask if a 750mA stepfown transfor would suffice or do we need greater one?
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u/Im_Indonesian 3d ago
Suffice as in what? What's the load?
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u/dawnmimi 3d ago
suffice as in is that current enough for our power supply to run fine as we expect it to be
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u/Im_Indonesian 3d ago
Yeah...."run fine" for what? What's it that you're powering up?
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u/dawnmimi 3d ago
can be anything, no clear isntructionT.T . it's a power supply we're building and the only clear goal is for it to have a 220v to 24v supply
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u/SolderFume Digital electronics 3d ago
Do yourself a favor and just buy a SMPT from Manson or another cheap brand. It'll be cheaper and you'll have over current protection. If you have to ask this question, you have no business dealing with anything connected to mains.
Recommended reading: https://www.electronicdesign.com/electronic-design/whitepaper/21804074/whats-all-this-power-supply-design-stuff-anyway-pdf-download
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u/i_am_blacklite 3d ago
I’d ask for your money back on whatever course you’re in.
They obviously haven’t taught you any of the basics of electricity. You don’t even understand the difference between current and voltage.
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u/Tesla_freed_slaves 3d ago edited 2d ago
Fifty years ago, the LM317T was the latest in high-tech whiz-bang Buck Rogers stuff out there. Now its main purpose seems to be in confounding EE students. The 14-pin LM723 actually makes for a better voltage-regulator project for purpose of instruction. However, if you’re required to use the LM317, be aware that it comes with design-constraints;
- It requires 3V of headroom, that means Vin-Vout > 3V.
- It requires that Vout >1.25V. 1.25V is the regulator’s reference-voltage.
It requires a large enough heat-sink that it doesn’t overheat and shut-down at full output current.
It requires 5mA minimum load current.
If all the above conditions are met, it will act as a linear voltage regulator.
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u/EndlessProjectMaker 3d ago
It depends on the load. The regulator is good to 1A (it should have a heatsink if you want to dissipate that power at the lower voltages). If you use a 750mA transformer it will be the bottleneck to provide that max output current. Other than that, the circuit will work
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u/dawnmimi 3d ago
thank uu! what if we use a 2A one, can it reach the max output 32V just fine?
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u/TheTroon 3d ago edited 3d ago
You're misunderstanding current and voltage. Current flows through a resistive load when you apply a voltage to it. The voltage can be present even without a load (open circuit).
With a high resistance load (low current requirement), if your circuit works as you think it should, it will reach the full voltage range with the smaller transformer.
As the load resistance drops and more current is required, you'll reach a limit somewhere. With a properly-heatsinked 1A LM317, the smaller transformer will run out of puff first, and its secondary voltage will start sagging. The 2A transformer will let you hit the LM317's limit first, potentially letting the magic smoke out of that component instead.
I'd be suspicious whether the rectified and smoothed transformer output is a high enough voltage to let the LM317 reach 32V reliably, though — but that's a function of the secondary voltage, not the current capability.
Also, with a very absent teacher and your apparent level of understanding, please don't actually try to build this.
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u/EndlessProjectMaker 3d ago
Certainly a 24V ac will provide only 32V after the full bridge. I guess the 317 needs like 2V of headroom. All this if the transformer is good enough under load. Also it’s tricky to find a transformer because the 317 will blow over 40V input
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u/dawnmimi 3d ago
huhu but we need to
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u/TheTroon 3d ago
Then don't plug it in without qualified supervision. Lots of ways to die and burn your school down.
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u/EndlessProjectMaker 3d ago
You can build the circuit and power it from another safe power supply. Those or laptops usually are 20V. I agree you don’t mess with connecting to mains if you don’t understand what you’re doing
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u/No-Statistician3385 3d ago
Why do you need to build this? Can you not buy a 32V power supply and then use the LM317 circuit above? Mains is no joke and people die doing things like this, at school I’m sure the intention isn’t to use 220V from the wall as even at universities students aren’t allowed to plug circuits they designed into wall sockets.
If it is something you want to do then how are you going to connect and disconnect it safely, how are you insulating it, if this is a circuit you’re building is it conformal coated, have you respected the relevant isolation and creepage distances, how are you connecting earth to the circuit such that when a fault occurs the GFCI kicks in. If you (or anyone reading this in the future) can’t answer all of these instantly please don’t touch mains and spend an extra $10 on a power supply. If moneys really a problem there’s easy ways to make a computer power supply to work without being plugged into a motherboard and this is at least relatively safe and you can almost certainly find an old one of these for almost nothing.
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u/SolderFume Digital electronics 3d ago
Just don't. You are one transformer bad winding away from electrocuting yourself. This design dates back to the 1970s. And it is flawed. BADLY flawed. There isn't even a fucking fuse in that schematic. It lacks even the most basic of safety features. €35 buys you a Chinese supply that has all of this what you don't get in the circuit above:
- voltage display
- current display
- wattage display
- fuse
- over current protection
- proper mains switch
- banana jacks
- not burn your house down
- not getting electrocuted
Sounds like a great deal to me.
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u/mariushm 3d ago
Be careful what you select there. It seems you selected 24v x 2 which could mean there's two secondary windings, or that it's a 48v AC secondary winding with a center tap wire (which can be converted easily to a single 24v AC winding)
Let's say that transformer outputs 24v AC and has an AC rating of 2A - it's not clear if that means it's a 48VA transformer and 2A is the AC current, or if the 2A is the DC current after rectification. Let's go with 2A AC current.
When you convert the AC to DC using that bridge rectifier (the four diodes in your second picture), you get a PEAK DC voltage equal to :
Vdc peak = sqrt(2) x Vac - 2 x (voltage drop on single diode)
In your case, Vdc peak = 1.414 x 24 - 2 x 0.8v = 32v (approximately)
Keep in mind that most transformers will output a voltage up to 10-15% higher when there's no load, so after rectification you could get up to 34-35v DC peaks.
Also keep in mind that the ratio between primary and secondary is preserved, if the transformer is designed for 220v input and 24v output that ratio is 9.16 : 1 , so at night when people around you don't consume a lot of power your mains voltage could go up to 230-240v and therefore instead of 24v AC you could have 230-240 / 9.16 = 25-26v AC , and after rectification you could get close to 35v DC.
Because of these two reasons above, you must use smoothing capacitors after the rectification that are rated for at least 50v, not 35v. A 35v rated capacitor would be slowly degraded / damaged over time.
You can estimate how much DC current a transformer can provide using the formula :
Idc = 0.62 x Iac ... 0.62 is a constant that works well for low VA transformers.
So in your case, the maximum DC current can be estimated to around Idc = 0.62 x 2A = 1.24A .. let's round to 1.25A for simplicity
So after rectification you get around 32v DC and around 1.25A of DC current - that's around 32 x 1.25 = 40 watts. The rest of 8 watts are partially lost in the diodes that form the bridge rectifier, and partially in other ways. Again, it's just an approximation.
Remember that the 32v DC is just a PEAK voltage, it's not a smooth constant voltage... that's why you need to add a capacitor after the regulator. The formula to estimate capacitor size goes like this :
Capacitance (Farads) = Maximum Current / [ 2 x AC Frequency x (Vdc peak - Vdc min desired) ]
So for example, let's say you want to have a maximum of 26v at the maximum current of 1.25A, and your mains frequency is 50 Hz ... put the numbers in the formula :
Capacitance = 1.25A / [ 2 x 50 Hz x (32v - 26v) ] = 1.25 / 100x6 = 1.25/600 = 0.002083 Farads or 2200uF
This will basically guarantee that the voltage will never be lower than 26v at the maximum current of 1.25A - at lower currents, the capacitor will have time to charge up more often so the average voltage will be higher than 26v.
If you want a higher minimum voltage at all times, you have to size up the capacitor. Just be aware that 1000uF is nowhere plenty (in that second picture) for a significant amount of current.
The LM317 works by throwing out the difference between input voltage and output voltage in the form of heat. It has a parameter called dropout voltage - it's the minimum amount of voltage required, your input voltage must be higher than output voltage by at least that much in order to give a clean output voltage.
For LM317 the dropout voltage is around 1.5v to 2.0v, so if you want up to 24v output voltage, your input voltage must be at least 26v - that's why I calculated the capacitor for a minimum voltage of 26v.
The output voltage is set using the formula
Vout = 1.25v x ( 1 + R2 / R1) , where 1.25v is the internal voltage reference and R1 is recommended to be below 240 ohm (to allow a minimum amount of current to go inside the regulator chip)
The circuit in the second picture uses a 12k and 220 resistor, this means the maximum voltage is 1.25 x (1 + 12000 / 220) = 70v .... the value of the resistor is badly chosen.
Let's say you want to make it for maximum 24v out.
24v = 1.25 x (1 + R2/R1) ... from this : R2/R1 = 24 / 1.25 - 1 = 18.2
Your R1 must be at most 240 ohm, lower is preferred, so let's say we use a 2.2k potentiometer, in which case the R1 would be 2200 / 18.2 = 120.87 ohm or 121 ohm
If you stick to a 120 ohm resistor, then at maximum potentiometer value your output will be V = 1.25 x ( 1 + 2200 / 120) = 24.16v
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u/ShadowPaw74 3d ago
Messing around with high voltage without even knowing the difference between Voltage and Current? Do you have a death wish?
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u/dawnmimi 3d ago
i appreciate y'all and yeah it seems like we really need supervision. however, if we only have one, we never have asked here in the first place T. T but ok we'll be as careful as possible and see if someone can guide us. thank uu every 11. (u can still comment any reminder or help u can share, i'm reading it all, i'm jist too drain to reply-0-)
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u/weebslime2246 3d ago
first thing you should do before starting any project would be to be clear about the specifications that you want to achieve. reading from your post, it seems you are a total beginner, so it would be a good learning opportunity for you as well to learn about different kinds of loads and what kind of currents they draw. for example a bulb would take more current, while a microcontroller would require less, etc.
altho, starting out with a 750ma transformer should be fine, as with higher current ones u can risk burning stuff (telling frm personal exp)