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basic power supply design......

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p72

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Hi All,
I have made circuit diagram of simple AC to DC switch mode power supply.

AC (110V,60Hz) --> Transformer --> Resistor and Capacitor --> Bridge Rectifier --> Capacitor --> DC-DC converter --> Output DC Voltage.

I am looking for 24VDC, 2Amp (14W) as an output. I have three questions.

(Questions #1) which specific transformer and (Questions #2)Bridge Rectifier should I use?
(Question #3) To get 24VDC as an output what kind of DC-DC converter is proper?

I would like to make a very simple SMPS.
Please see attached schematic for the same.

Thank you!!!

Untitled.png
 

(Question #3) To get 24VDC as an output what kind of DC-DC converter is proper?
Why do you want to use DC-DC converter?

**broken link removed**
 

24V at 2 Amps is 48W by my calculations.

With that circuit, there is no point in using the DC-DC converter as the input to it is DC anyway. The only problem you have is the transformer. To push 2A through 10K + 2.652K (Xc) you need a voltage of 25.305 KV. I'm not saying you can't get 2A 25KV transformers but it might be a bit bigger and more expensive than you think.

Brian.
 

24V at 2 Amps is 48W by my calculations.

With that circuit, there is no point in using the DC-DC converter as the input to it is DC anyway. The only problem you have is the transformer. To push 2A through 10K + 2.652K (Xc) you need a voltage of 25.305 KV. I'm not saying you can't get 2A 25KV transformers but it might be a bit bigger and more expensive than you think.

Brian.

If I will not use DC-DC converter then how do I get exact 24 V DC?
And I did not get your second point related to 2A 25KV transformer. Could you please explain in detail?
 

Answer: you use a regulator as demonstrated in Yura717's link.

However, consider that passing 2A through a 10K (R2) resistor will drop (Ohms law V= I*R) 20,000 Volts and the capacitve reactance of C1 at 60 Hz is 1/(2*pi*60*1^-6) = 2.652K so that also drops 5,304 Volts. If you want 24V out at all, you would have to get 24 + 20,000 + 5,304 = 25,328V from the transformer. Additionally, the resistor would dissipate (W=V*I) around 40KW so think in terms of one the size of a refridgerator but with lots of cooling fans.

In other words, it isn't really a practical circuit at all.

I would go back to basics and read that link, you could improve the efficiency of your design by around 1000% !!

Brian.
 

Answer: you use a regulator as demonstrated in Yura717's link.

However, consider that passing 2A through a 10K (R2) resistor will drop (Ohms law V= I*R) 20,000 Volts and the capacitve reactance of C1 at 60 Hz is 1/(2*pi*60*1^-6) = 2.652K so that also drops 5,304 Volts. If you want 24V out at all, you would have to get 24 + 20,000 + 5,304 = 25,328V from the transformer. Additionally, the resistor would dissipate (W=V*I) around 40KW so think in terms of one the size of a refridgerator but with lots of cooling fans.

In other words, it isn't really a practical circuit at all.

I would go back to basics and read that link, you could improve the efficiency of your design by around 1000% !!

Brian.

Hi Brian,
Thanks for the details.

How about this if I put Resistor and Cap before Transformer? (I will low down my resistor value 1ohm only.) And the purpose of capacitor is to block DC in case if someone applied DC instead of AC.

AC (110V,60Hz) --> Resistor and Capacitor --> Transformer --> Bridge Rectifier --> Capacitor --> DC-DC converter --> Output DC Voltage

If the above case looks ok then what should be a proper transformer and bridge rectifier? (I am clear about DC-DC converter.) I don't want to use voltage regulator IC because they get hot very quickly.
Last question is, Can I call this supply as SMPS?

Thank you,
p72
 

I would leave the resistor and capacitor out altogether. If you need to protect it against DC just wire a fuse in line with the transformer primary. If you really do want to use a capacitor, go for one with much higher value, say 100uF or more but make sure it is non-polarized and rated at least to handle 160V AC. DO NOT use a normal electrolytic type.

A linear regulator works by converting excess power to heat so it is normal for them to warm up. The dissipation is the voltage dropped between input and output pins multiplied by the current passing through it so if you can keep the voltage overhead as low as possible it will not need a huge heat sink.

It most certainly is NOT an SMPS, they work on a different principle altogether. SMPS control the amount of output by using different proportions of time within a cycle to enable or disable the pass component, your is on all the time and would be considered to be a linear power supply.

I would suggest you change to using an SMPS design because it would be smaller and more efficient (= cooler) than a linear one. It can also produce the 24V output you need without any additional DC-DC converter. I suggest you look at the data sheets for the LM2756-ADJ which would be ideal for your purpose. Be careful to use the components mentioned in the data sheet because they are critical. Also note that it can accept a very wide range of input voltage. You still need the bridge rectifier and transformer but it's voltage becomes far less important. As long as both are rated at 24V or more and 2 Amps they will work.

Brian.
 

I would leave the resistor and capacitor out altogether. If you need to protect it against DC just wire a fuse in line with the transformer primary. If you really do want to use a capacitor, go for one with much higher value, say 100uF or more but make sure it is non-polarized and rated at least to handle 160V AC. DO NOT use a normal electrolytic type.

A linear regulator works by converting excess power to heat so it is normal for them to warm up. The dissipation is the voltage dropped between input and output pins multiplied by the current passing through it so if you can keep the voltage overhead as low as possible it will not need a huge heat sink.

It most certainly is NOT an SMPS, they work on a different principle altogether. SMPS control the amount of output by using different proportions of time within a cycle to enable or disable the pass component, your is on all the time and would be considered to be a linear power supply.

I would suggest you change to using an SMPS design because it would be smaller and more efficient (= cooler) than a linear one. It can also produce the 24V output you need without any additional DC-DC converter. I suggest you look at the data sheets for the LM2756-ADJ which would be ideal for your purpose. Be careful to use the components mentioned in the data sheet because they are critical. Also note that it can accept a very wide range of input voltage. You still need the bridge rectifier and transformer but it's voltage becomes far less important. As long as both are rated at 24V or more and 2 Amps they will work.

Brian.

LM2756-ADJ?? (Why this? It is LED driver.)
Is it true that if I connect fuse between line and transformer's primary, it blocks DC? How?
If possible can you suggest rough diagram of same power supply but with SMPS technology?

Thank you,
p72
 

The fuse doesn't block DC, it burns out if DC is applied by mistake. When DC is applied across the transformer primary, only the resistance of the copper in the wire will limit the current so it will be many times higher than if AC was applied. You can use that property to protect it with a fuse that can carry the low AC current but burns out at the higher DC current. The only component that will actually block DC is a capacitor but as I pointed out, it needs to be high value and high voltage and non-polarized so it could be big and expensive.

The schematic is simply the fuse in line with the AC, transformer rated at 2A (or more) at about 24V, a bridge rectifier and reservoir capacitor then the LM2756-ADJ schematic from the data sheet. It will run cool and provide a regulated 24V at 2 Amps. All the SMPS control is done inside the LM2756-ADJ.

Brian.
 

It's the same device but the "-ADJ" version has the ability to set the output voltage using two resistors. To be perfectly honest, I can't see why the non-adjustable version should work any differently, perhaps it's just the way they optimize the error amplifier inside the IC.

Brian.
 

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