CC BY
8Scientific Journal Impact Factor (SJIF 2022=4.63) Passport: http://sjifactor.com/passport.php?id=22230
DETERMINATION OF THE MATHEMATICAL EXPRESSION FOR THE EFFICIENCY OF A TWO-SEMICONDUCTOR DIODE RECTIFIER (q)
M. Shamsiddinov
Doctoral student of Andijan Institute of Mechanical Engineering
A. Ismailov
Associate Professor of Andijan Mechanical Engineering Institute
ABSTRACT
This paper presents a method for determining the efficiency of two semiconductor rectifiers by a mathematical expression.
Keywords: transformer, diode, rectifier, bridge diode, VAX (Volt-Ampere characteristic).
Theoretical part. Modern electronics is based on electronic devices made of semiconductor materials and is rapidly evolving. From the smallest details of devices to integrated circuits, microcontrollers and even microprocessors, semiconductor materials are being created. Therefore, a quality AC power supply is required to ensure proper and smooth operation of these devices. Because the electricity in our networks is sinusoidal, we need two half-cycle rectifiers. Our article explains that we can rectify alternating current using semiconductor rectifiers. Semiconductors are widely used in electronics, and almost all modern electrical appliances - from computers to cell phones - are based on semiconductor technology.
Main part: There are two types of single-phase full-wave rectifier, namely, full-wave rectifiers with center-tapped transformer and bridge rectifiers. A full-wave rectifier with a center-tapped transformer is shown in Fig.1. It is clear that each diode,
U,
D1
together with the associated half of the transformer, acts as a half-wave rectifier. The outputs of the two half-wave rectifiers are combined to produce full-wave rectification in the load. As far as the transformer is concerned, the dc currents of the two halfwave rectifiers are equal and opposite, such that there is no dc current for creating a transformer core saturation problem [1].
FIGURE 1. Full-wave rectifier with
center-tapped transformer. _
UZBEKISTAN
Scientific Journal Impact Factor (SJIF 2022=4.63) Passport: http://sjifactor.com/passport.php?id=22230
inst
FIGURE 2. Voltage and c of the full-wave rectifier vt rec1 transform*
can
rectification without using a center-tapped transformer. During the positive halfcycle of the transformer secondary voltage, the current flows to the load through diodes D1 and D2. During the negative halfcycle, D3 and D4 conduct.
The voltage and current waveforms of the full-wave rectifier are shown in Fig.2. By observing diode voltage waveforms UDi and UD2 in Fig.2, it is clear that the PIV of the diodes is equal to 2Um during their blocking state[1]. Hence the VRRM rating of the diodes must be chosen to be higher than 2Um to avoid reverse breakdown. During its conducting state, each diode has a forward current which is equal to the load current, therefore the IFRM rating of these diodes must be chosen to be higher than the peak load current, Um=R, in practice[5].
J - 0 -
1 i L - u.
UL
FIGURE 3. Bridge rectifier.
Scientific Journal Impact Factor (SJIF 2022=4.63) Passport: http://sjifactor.com/passport.php?id=22230
The voltage and current waveforms of the bridge rectifier are shown in Fig. 4 As with the fullwave rectifier with center-tapped transformer, the IFRM rating of the employed diodes must be chosen to be higher than the peak load current, Um = R. However, the PIV of the diodes is reduced from 2Um to Um during their blocking state [3].
U=UmaxC0S9
U=Umaxcosrat (2)
FIGURE 4. Voltage and current waveforms of the bridge rectifier.
current to constant current. (1)
the
sin(| — a) = c o s a
cos(| — a) = s in a
n
n
From this expression we get the integral of rat and make the boundaries - - and -because sin (|) is 1 and sin(- |) is -1 then the values in the interval are [-1; 1].
n
U=¡\ i/maxCos(rat)d(rat) (3)
2
In this case, we divide the expression by because the rectifier corrects each period. We also take the Umax out of the integral [4].
n
U=Umax~ JICO S ( 0)t) d ( 0)t) (4 )
If we take the integral of the expression, it becomes
U=Umax~ Sin ( tot) (5)
We calculate the integral by setting the boundaries.
1
n
ItN
U=Umax * - (Sin(- ) - Sin(--) ) (6)
it
CENTRAL ASIAN ACADEMIC JOURNAL ISSN: 2181-2489
OF SCIENTIFIC RESEARCH VOLUME 2 I ISSUE 5 I 2022
Scientific Journal Impact Factor (SJIF 2022=4.63) Passport: http://sjifactor.com/passport.php?id=22230
U=Umax * I ( 1 + 1 ) (7) U=Umax *1 *2 (8)
U=Umax *l (9)
We now express the maximum value of the voltage by the effective value
Umax Ueff=^=- (10)
From this expression, Umax looks like this:
Umax =V2 * Ueff (11) Once we have expressed the maximum voltage, we take (9) and the expression is as follows.
U=Umax *2=2V" Ueff (12) U=2^ Ueff (13)
U=0.9 Ueff (14)
This expression means that we can use only 90% of the effective value of any voltage.
REFERENCES
1. POWER ELECTRONICS HANDBOOK DEVICES, CIRCUITS, AND APPLICATIONS Third Edition, Edited by Muhammad H. Rashid, Ph.D., Fellow IET (UK), Fellow IEEE (USA) Professor. (Muhammad H. Rashid, 2011)
2. Alijanov D.D., Topvoldiyev N.A. (2021). SOLAR TRACKER SYSTEM USING ARDUINO. Theoretical & Applied Science, 249-253.
3. Alijanov D.D., Topvoldiyev N.A. (2022). PHYSICAL AND TECHNICAL FUNDAMENTALS OF PHOTOELECTRIC SOLAR PANELS ENERGY. Theoretical & Applied Science, 501-505.
4. Шамшиддинов М.К. (2021) повышение энергоэффективности энергосберегающего светодиодного освещения.
Universum: технические науки, 74-76.
5. Shamshiddinov. M.Q, Parpiev.O.B. (2021). Use Of Electronic Keys To Increase Energy Savings Of Led Lights. The American Journal of Applied sciences, 39-46.
