Types of AC/DC Power Supply Adapter Circuits and How to Use Them

Power supply is one of the main element in every electronic circuit, and the right AD/DC adapter circuit should be chosen if we want our device to accept power line voltage as the main supply. There several types of AC/DC adapter circuits, transformer-less (non-isolated), step-down transformer, and switched-mode power supply adapter. The schematic diagram of the circuits and how to implement them for our design is presented in this article.

Transformer-less Power Supply Adapter Circuit

Warning! This circuit is not isolated from the power-line voltage and causes hazardous electrical shock when get touched!

Before discussing this type of circuit, please be aware about this caution: Warning! This circuit is not isolated from the power-line voltage and causes hazardous electrical shock when get touched!

There two types of this non-isolated power supply adapter circuits: half-wave rectified and full-wave rectified circuits. The main application of this types of adapter is where a small power is needed by an isolated device. A schematic diagram of a half-wave rectified transformer-less ac/dc adapter circuit is shown in the Figure 1.A, and the full-wave rectified one is shown in the Figure 1.B.

Figure 1. Transformer-less Power Supply Circuit A) Half-Wave Rectified; B) Full-Wave Rectified

Half-Wave Rectified Transformer-less Power Supply Circuit

The circuit depicted in the schematic diagram shown in the Figure 1.A uses only one high voltage diode, one zener diode, one resistor, and one capacitor. This circuit is suitable only for devices that draws very small current. Practically, this power supply circuit can be used for devices requiring not more that 10 mA. The reason behind this limitation is the use of resistant current limiter (R1) that waste to much power when supplying higher current. There are 3 component’s values that should be selected when designing with this circuit.

Half-Wave Rectified Transformer-less Adapter Circuit Application Examples

Because of limited current specification, we might find that this circuit is only used in very limited application. Please be aware that the connected circuit should be isolated from any user’s touch since the adapter is not isolated from the power-line supply.

  • Isolated blinking live indicator for power-line network
  • Isolated live charger and supply for low power devices backed with small rechargeable coin battery.
  • Isolated part of low power micro-controller
  • Isolated driving section of electronic circuitry

The Formula on How to Design The Half-Wave Rectified Transformer-less Adapter Circuit

To design this simple transformer-less AC/DC adapter circuit, you have to know the parameter of the power-line input voltage (Vin) and its frequency (f), the expected (average) voltage output or the voltage at the load (vL), the acceptable voltage ripple at the output (vRipp), and the load resistance (RLoad). If the resistance of the load is unknown as static resistance, then you can conduct a simple measurement. To do this, you can supply the load with a general power supply at its operating voltage (vL), then measure the load current (iL) using an Ampere meter, and deduct the RLoad = vL/iL. Below is the step-by-step design procedure:

  1. Compute the average load current iL = vL/RLoad.
  2. Compute the R1 value, where il = Vin/(R1+RLoad), so R1 = (Vin/iL)-RLoad
  3. Make sure the rating for R1 wattage >= power dissipation Pd, Pd = iL*(Vin-vL)
  4. Half the R1 to compensate the half-wave rectification, R1 = R1/2
  5. Select vL as the zener diode’s voltage value, with minimum wattage rating Wr = (vL x iL)
  6. Select the C1 capacitance value to minimize the peak-to-peak ripple voltage Vrip, C1 = iL/(Vrip* f),
  7. Use voltage rating of equal or higher than 1.5 of vL for the capacitor C1, VC = 1.5*vL

When the resistance of the load is really fixed, then the zener diode can be omitted, and the voltage will be swinging around the vL between the peak-to-peak of the ripple voltage. For possible dynamic load resistance, the zener diode should be installed. As the effect, the average final voltage vL will be shifted down dynamically (depending on the load Current). At the rated load current iL, this voltage shift will be about the half of the ripple voltage.

Half-Wave Rectified Transformer-less Adapter Design Example

Using the previously described procedure and formulas, now we can easily implement such adapter in our circuit design. Suppose we a device that behaves like a fixed resistance load of 2k. It needs to be powered by 12V DC source with acceptable 1V(peak-to-peak) ripple voltage using the half-wave rectified ac/dc adapter circuit. The power line voltage is 220 V with 50Hz AC frequency. Let’s find the correct specification of the R1 resistor, C1 capacitor, and the Zener diode.

Let’s list the known parameters,

  • vL = 12 V
  • Vin = 220 V
  • f = 50 Hz
  • RLoad = 2000 Ohm
  • Vrip= 1 V

Next, let’s compute the specification:

  • Using formula #1, iL = vL/RLoad = 12V/2000 kOhm = 6 mA (=0.006A)
  • Using formula #2, R1 = Vin/iL – RLoad = 220V/0.006A – 2000 = 36667 – 2000 = 34667 kOhm
  • Using formula #3, Min R1 wattage = Pd = iL*(Vin-vL) = 0,006A * (220-12)V = 1.25 Watt
  • Using formula #4, Half the R1, R1 = 34667/2 = 17333 Ohm
  • Using formula #5, Zener voltage = vL = 12V, WR= vL*iL = 12V * 6mA = 72 mW
  • Using formula #6, C1 = iL/Vripp * f = 0.006 / (1*50Hz) = 0,000120F = 120uF
  • Using formula #7, C1 voltage min = 1.5*vL = 1.5*12 = 18V

Using slightly higher specs with the available components found in the market, we can use 15k/2W resistor for R1, 12V/250mW for the Zener diode, and 120uF/25V for C1. You can also use 220uF/25 capacitor since it might be more widely available, and you’ll get smaller voltage ripple as the bonus. As the final note for this non-isolated half-wave rectified ac/dc adapter circuit, please keep in mind that the formula is only approximation based on simple practical component models and assumptions. You have to build and test the circuit yourself to validate the design.

Full-Wave Rectified Transformer-less Power Supply Circuit

The circuit of full-wave rectified version shown in the Figure 1.B is constructed by replacing the single diode half-wave rectifier with full-wave rectifier diodes, and replacing the R1 resistant current limiter with the reactant current limiter C1. The position of the current limiter is also moved from “after the rectifier” to “before the rectifier”. This placement should be done since the reactant component C1 will only work in alternating-current only, so it won’t work if we place it after the rectification. Because the reactant current limiter doesn’t dissipate any power (in ideal condition), the efficiency of the overall circuit will be much better that the half-wave rectified version. Up to 50 mA of output current is practically feasible with this transformer-less AC/DC adapter circuit design.

Full-Wave Rectified Transformer-less Adapter Circuit Application Examples

The higher efficiency feature of this circuit make it possible for wider range of application. Please be aware that this power adapter circuit are incorporated in an isolated system where it is safe from any user’s touch. It’s definitely not for general purpose power supply adapter where we can plug-and unplug the load from the adapter.

  • LED lighting, this is probably the most popular application
  • Isolated intercom system
  • Isolated door bell system

The Formula on How to Design The Full-Wave Rectified Transformer-less Adapter Circuit

In the same way with the half-wafe rectified version, we have to know some parameters to design the full-wave rectified version of the transformer-less adapter circuit:

  • Power-line input (rms) voltage (Vin)
  • The frequency of the power-line input voltage (f)
  • The output or load voltage (vL)
  • The acceptable (peak-to-peak) ripple of the output voltage (vRipp)
  • The load resistance (RLoad)

Now we can compute some parameters of the components specs

  1. Compute the load current iL = vL/RLoad
  2. Compute the reactance of the current limiter RC1 = (Vin/iL)-RLoad
  3. Compute the capacitance C1 = 1/(6.284*f*RC1)
  4. Compute the voltage rating of C1 capacitor, VC1 = 1.5 * (Vin-vL)
  5. Compute the wattage rating of the Zener diode, WR = vL*iL
  6. Compute the C2 capacitance, based on the ripple voltage Vrip, C2 = iL/(Vrip* 2 * f)
  7. Compute the voltage rating spec of C2, VC2 = 1.5 vL

Full-Wave Rectified Transformer-less Adapter Design Example

In order to have more understanding of the design procedure of this transformer-less adapter circuit, let’s try to design a power supply circuit for LED lighting. One typical bright white LED has 3.6V forward voltage drop at 30 mA current, so we can use 9 LEDs in series for our design. The load voltage vL will be 3.6V*9, or 32.4V. Let’s use 220VAC power-line input voltage at 50Hz, with 10V ripple as the specification. At vL = 32.4V and iL = 30mA, this means that the RLoad is vL/iL = 32.4/0.03 = 1080 Ohm. Let’s list the already known parameters:

  • Vin = 220V
  • iL = 30 mA
  • vL = 32.4V
  • RLoad = 1080
  • Vrip = 10V

Now let’s find some parameters and components values using the previously described procedure and formulas.

  • iL = 30 mA
  • The reactance of the current limiter, RC1 = (Vin/iL)-Rload = (220V/0,03A) – 1080 Ohm = 6253.33 Ohm
  • Find the capacitance of C1 = 1/(6.284*f*RC1) = 1/(6.284 *50Hz*6253.33Hz) = 0,00000051F = 0.51uF
  • Find the C1 voltage rating VC1 = 1.5*(Vin-vL) = 1.5*(220-32.4) = 281.4V
  • Find the wattage rating of the zener diode, WR = vL * iL = 32.4V * 30 mA = 972 mW
  • Find the C2 capacitance, C2 = iL/(Vrip*2*f) = 0.03/(10*2*50) = 0.00003 F = 30uF
  • Find the C2 voltage rating, VC2 = 1.5*vL = 1.5 * 32.4V = 48.6V

C1 and C2 capacitor. From the parameter computation, we find that C1 capacitor specification is 0.51uF/281.4V, so we can use 0.47uF/350V for the closest available component in the market, or just use two 1uF/350V back-to-back series electrolytic capacitor to create a non-polar 0.5uF/350V capacitor. For C2 capacitor, a 33uF/50V electrolytic capacitor will be more than enough for this purpose.

Removing the zener diode. Ideally, the zener diode should never be exist where the load behaves like fixed resistor in the steady state operation. When the load current is dynamically changing at the normal operation, the zener diode regulates the output voltage by shorting the voltage and wasting the power as heat. Especially for this LED lighting application, ideally, the zener diode should never be conducting to avoid wasting the power. If the zener diode is kept installed to protect the LEDs from surge current, then the selection of the proper value will be complicated. A too low break-down voltage will cause power wasting, and too high voltage will cause the LEDs unprotected. To simplify the design, we can remove the zener diode since the real protection from surge current is already provided by the C2 capacitor.

The surge current limiter resistor. Referring to the schematic diagram in the Figure 1.B, it is shown a fixed 22 ohm resistor to provide surge current protection. A surge current happens at the moment when the Vin is plugged to the power-line while the C1 charge is empty and the power-line voltage is not at zero crossing state. There will be a current rush in very brief period, the maximum value of this current glitch will be Vin(peak)/22. The peak voltage of 220V(rms) source is about 311V, so the current glitch will be 14.14A. This surge current will be safely handled by the 1N4007 diodes since they have 30A surge current (according to its data sheet). At the normal operation the dissipated power by this surge protection resistor is very small, about 20 milli Watts, a 0.5W small resistor should be enough to handle both normal and surge current.

Step-Down Transformer Power Supply Adapter Circuit

(to be continued..)

One comment

  • Mark Izatt

    I would like to ask a question. (This is difficult to Google) Regarding a small power converter for a cell phone: Is there a polarity? Does flipping the converter upside down in the wall socket have ANY effect? My guess is that there is no difference. But some people swear they get a quicker charge by doing this. Thank you for answering

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