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advantage and disadvatnge of AC to DC converter circuit
- Thread starter pharaon
- Start date
1. It outputs 9V not ±9V so the 12V fan will run slow.
2. The output is not isolated so will give a shock if touched.
3. Very reliable high voltage components will be necessary. The input capacitor should be 1000V of the type which fails open circuit, not short circuit.
4. It is small, light and cheap.
2. The output is not isolated so will give a shock if touched.
3. Very reliable high voltage components will be necessary. The input capacitor should be 1000V of the type which fails open circuit, not short circuit.
4. It is small, light and cheap.
That's can be controlled by the zener diode
Sure I'll isolate it
Can you give me link for such one
The one that used ia ceramic capacitor 400V that run on 220 ac source
You would want a Class Y safety capacitor. Example: https://www.digikey.com/product-detail/en/kemet/PHE850ED6100MD18R06L2/399-5422-ND/1927367 .
The output voltage is going to be almost totally dependent on the load, including the load supplied by the Zener diode. With the input capacitive reactance, and the 10K resistor, you have more of a current source than a voltage source.
With the components shown the circuit can probably source around 10 mA.
The output voltage is going to be almost totally dependent on the load, including the load supplied by the Zener diode. With the input capacitive reactance, and the 10K resistor, you have more of a current source than a voltage source.
With the components shown the circuit can probably source around 10 mA.
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The input capacitor will define the input current. Use LTspice to determine this.
The current is divided between the output and the zener diode. The output voltage will limited by the 9V zener and will drop if the output current demand is more than the input current.
If you are going to isolate with a transformer as suggested, then get one with the appropriate voltage, An AC transformer and a bridge rectifier should be enough. A smoothig capacitor would only be gilding the lilly.
The current is divided between the output and the zener diode. The output voltage will limited by the 9V zener and will drop if the output current demand is more than the input current.
If you are going to isolate with a transformer as suggested, then get one with the appropriate voltage, An AC transformer and a bridge rectifier should be enough. A smoothig capacitor would only be gilding the lilly.
i'm gonna use 12 dc fan one of those like pc fan inside my old fridge to make good coolant , and will connect it to the thermostat to run and stop as the fridge work so i need something small
And just how do you intend to do this..?
I'm not certain you understand what an isolated supply comprises.
small board to solder the component and then heat shrink
why?
1) Do not, never, use this type of "power supply". This is a cheap technique intended to be used in fully isolated housings at best. As soon as there is the slightest risk of exposure of one of the output leads to human touch there is also a risk of a potentially deadly shock. You obviously lack the experience to judge this circuit. Therefore it is mandatory that you use a save power supply.
2) Once you have gained some more experience and still want to use this design (in an isolated, closed housing, of course), consider the impedance of the high voltage capacitor (as per X = 1/(2 × f × C) as a current limiting resistor. This will allow you to make the necessary calculations.
Why do I think you lack the experience? You're getting your units all wrong:
2) Once you have gained some more experience and still want to use this design (in an isolated, closed housing, of course), consider the impedance of the high voltage capacitor (as per X = 1/(2 × f × C) as a current limiting resistor. This will allow you to make the necessary calculations.
Why do I think you lack the experience? You're getting your units all wrong:
- 1 mΩ is a short circuit. The original design you copied surely calls for 1 MΩ - case matters: m = milli, M = mega.
- The unit of capacitance is F, so it is not 220 µf but 220 µF.
- The output is + 9 V vs. 0 V or 0 V vs. -9 V, but not +. 9 V (as others have mentioned before)
thanks for your post, and yes i'm still learning
can you explain this in numbers, for example what is C
We already do have another thread dealing with the same kind of circuit.
Reactance of 0.1 uF capacitor at 60 Hz
Xc = 1/(2piFC) = 1/(2pi*60*0.1e-6) = 26526 ohms
That 10K resistor is also in series, so total source impedance:
Zs = √(10,000² + 26526²) = 28348 ohms.
With a 220 vac source, short circuit current will be:
I = E/Z = 220/28348 = 7.76 mA
Any load voltage will reduce that maximum current capability. With a 12 volt load:
I = E/Z = 208/28348 = 7.33 mA
It doesn’t matter how many times I see the formula and math. It still goes WHOOOSH..
Martin
Martin
pi stands for?
F = farad
C stand for? what is e-6?
so which one of those should i use?
also what should be the AC voltage when connect the capacitor in series ?
pi = pi. pi is the ratio of a circles circumference to its diameter. pi = 3.1415926535.....
F = Frequency in Hz. For the US line that would be 60 Hz.
C = Capacitance in Farads.
e-6 is a shorthand way of saying '10^-6'.
Neither. Select a more appropriate and safer circuit to power the fan.
