Many of you must have faced this situation when purchased imported household appliance (such as a telephone or calculator) was equipped with remote power supply for mains voltage of 120 V. a Case this, of course, a pleasant one, but the radio quite a force to make the unit work normally and from the mains voltage of 220 V.
How to finalize a 120-volt low-power remote power unit for inclusion in the network 220? This can be done in several ways. Consider them briefly.
As a rule, all the "stuffing" of the block consists of a mains transformer, rectifier and the smoothing capacitor. Therefore, the first way is to disassemble the transformer, to remove it from the frame, all the windings, to calculate them again to 220 again to reel coil and collect the transformer.
Transformer calculation is not difficult [1], but the wound will require a lot of trouble and, of course, skill.
An insurmountable obstacle when using This method may be the fact, the imported magnetic transformer often perform non-separable - plate tightly connect the narrow weld. In this case, you can only to recommend to throw the transformer and to choose a replacement with the magnetic circuit of the same (or slightly larger) section.
Those who rewind transformer seems unacceptable, another offer the obvious way is to include the network in series with the winding of the transformer ballast resistor, pre-calculating the resistance (in ohms) the formula:
Rбал= 12000/RG, where RG is the overall capacity of the transformer in watts, usually specified on the unit.
The method is very simple, but if you calculate the power that will be allocated to this resistor (and it will be approximately equal to the power transformer!), it will be it is clear that the applicability of the method is limited.
Instead of the ballast resistor can be used ballast capacitor [2]. Then with a sort of thermal power will have no problems - it is close to zero, but the capacitor need of impressive dimensions. Suffice it to say that its rated voltage should be at least 520 In!
To connect to the network of small electric appliances with constant consumption power is sometimes used another method, based on the phenomenon of resonance current. It may occur in two parallel branches of an electric circuit fed by AC voltage, if the resistance of one branch of inductive and other - capacitive (see diagram). Here Rэкв and Lekv respectively equivalent the resistance and inductance of the transformer power supply, refurbished to the power winding, and the elements R1 and C1 additionally introduced for the implementation of the resonance current.
It is easy to see that R1 is the same ballast resistor, but the capacitor C1 here compensates the inductive component of the current of the primary winding, so the power allocated to the ballast resistor, less in 30% to 50%. The amplitude of the voltage the capacitor C1 even in the moment of switching is less than 200 V.
Thus, it is only necessary to determine the values of the additional elements, and you need to know Rэкв and Lekv. On the PSU usually indicate input nominal voltage power supply UBX, full overall power of the WG, the output voltage UBыX, the load current lH and sometimes supply current lBX. The ohmmeter to measure the resistance network RI and secondary windings RII of the transformer.
The calculation starts with the determination of current consumption (if not specified):
lBX=WG/UBX. Next, calculate the active power consumed by the power supply from the network: RA=I2BX · RI+I2H · RII+IH · Uвых
(it is assumed that the load is purely resistive block and losses are not taken into account in eddy current and magnetization reversal of the magnetic circuit), and reactive power is: Px= √ WG2 - RA2
The values of active and reactive power equivalent to counting active the resistance and inductance of the transformer, refer to the power winding:
Recv=RA/|2BX; Lekv=Px/ω · I2BX, where ω is 2π · f;
f - frequency of the voltage - 50 Hz.
The capacitance of the capacitor C1 is determined from the condition of equality to zero of reactive the conductivity of the circuit formed by the parallel capacitor and transformer:
C1=Lekv/A, where A=ω2 · L2экв +R2экв.
The resistance of the ballast resistor R1 and its power PR1 is calculated by the formulas:
R1=A/Rэкв(UC/UBX-1); PR1=UBX-Rэкв(UC-UBX)/A, where UC=220 V.
The proposed method was used for finalization of the external power supply for a calculator that had the following parameters: UBX=120 V; RG=3 B-A; Uвыx=5.6 V; lH=0.2 A; the resistance of the windings is measured with an ohmmeter, RI=764 Ohms; RII=3 Ohms. On the reference values were calculated parameters of the elements: Rэкв=2748 Ohms; Lekv=12,54 GN, C1=0,54 UF; R1 =6987 Ohms; PR1=1,48 watts. Choose the capacitor MBGC capacity 0.5 UF voltage of 250 V and a resistor MLT-2 resistance of 6.8 com. The calculations showed that when turned on, the voltage on the capacitor does not exceed the value corresponding to the steady-state regime (120 In), when it is disabled exceeds by only 4 %.
In conclusion, some recommendations. The capacitance of the capacitor C1 preferably choose close as possible to the calculated ones. This is achieved by parallel the required number of capacitors (capacitance values summing). The rated voltage of the capacitor should be at least 200 V. The capacitors should be used paper (MBGC, BGP, etc.) intended to work in an AC circuit; selecting the type and nominal voltage you need to use the Handbook of electrical capacitors.
The power of the resistor R1 is chosen larger than the estimated. Sometimes adjustment is required the resistance of a resistor, which is best done when you connect to the unit power rated load. At low output voltage of the resistance should be less with increased - more.
The capacitor and resistor can be placed inside of the power supply, if there is free space (don't forget the side walls of the block to drill ventilation the holes), or in a separate housing, in the form of adapter.
Literature
Author: V. Chudnov, Ramenskoye, Moscow region.