This is generally common Unregulated DC power supply circuit. when we use electrical appliances with AC-line we need to use AC adapter instead of a battery. Because It’s convenient and economical. Which inside most low cost AC/DC adapter has this circuit. Thus it need to has low ripple or low noise similar the normally battery. Although it is a very simple circuit. But it is very important. Because it is a first step of power source. If it does not have enough energy. It will result in our electrical appliances malfunction. Let’s learn to use them.
The working of circuit
In the example circuit above, the output about 8 volt(No load 8.4V) power supply at 100mA. When we apply the power cord to AC line, the 120VAC/220VAC/230VAC fed to a step down transformer-T1 via the on-off switch-S1 and the 500mA fuse-F1. It will blow if the current through the output terminals is too high.
The 6V AC output (approximately) from the transformer is rectified by a bridge rectifier, which We want to save money with not use the normal Bridge Rectifier, we can easily use the 4 general purpose 1N4007 rectifier diode D1 through D4. Now they rectified AC (Alternating Current) to DC (Direct Current).
This pulsating DC output is filtered via the 1000uF capacitor-C1, it acts as a storage capacitor. Now the pulsating DC smooth look like DC voltage from the battery.
How to choose right components
The simple full wave rectified or unregulated power supply above. We have to choose right components to load works best performance.
The its output has low RIPPLE or Vrip
The transformer is chosen according to the desired load. For example, if the load requires 12V at 1amp current, then a 12V, 1 amp rated transformer would do. However, when designing power supplies or most electronic circuits, you should always plan for a worst case scenario. With this in mind, for a load current of 1 amp a wise choice would be a transformer with a secondary current rating of 1.5 amp or even 2 amps. Allowing for a load of 50% higher than the needed value is a good rule of thumb. The primary winding is always matched to the value of the local electricity supply.
An approximate formula for determining the amount of ripple on an unregulated supply is:
Vrip = Iload * 0.007 / C
where I load is the DC current measured through the load in amps and C is the value of the capacitor in uF.The diagram below shows an example with a load current of 0.1 amp and a smoothing capacitor value of 1000uF.
The calculated value of ripple is (0.1 * 0.007) / 1000e-6 = 0.7 volts or 700mV. The value of peak-peak ripple measured from the graph is 628mV. Therefor, the equation is a good rule of thumb guide for choosing the correct value for a smoothing capacitor in a power supply.
This is a basic AC line operated 9 volt power supply. For low ripple (superimposed AC at Vout),use large value for C1.
add one or more capacitors in parallel with C1 for more capacitance
Capacitors must have a DC working voltage (WVDC) of at least 12 volts.
Rectifier bridge B1 must have peak inverse voltage (PIV) of at least 12 volts.
T1 and B1 must have adequate power and current ratings. (Use ohms law)
Caution: You must insulate or enclose all exposed AC line connections!
The power cord must be unplugged when you assemble or service the circuit.
Up to $20 shipping discount on first order now: https://jlcpcb.com/quote