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        Home>Services>Product Knowledge>Sine Wave Inverter schematic

        Sine Wave Inverter schematic

        Date: 2010-7-25      Source: Zhejiang Tress Electronic Technology Co.,Ltd.     Views: 5787

        Sine wave inverter schematics, well-wave output and the sine wave output. Square-wave output inverter, high efficiency, but are designed for the sine wave electrical power, using always worried, though can be applied to many appliances, but some appliances do not apply, or electrical indicators up will change. Sine wave inverter output will not have the shortcomings, but there is low efficiency. To this end I designed a highly efficient sinewave inverter, the circuit shown in figure1.

        The circuit with the 12V battery. First with a double voltage module Voltage supply for the op amp.Can select the ICL7660 or MAX1044. 50HZ sine wave generated op amp 1 as the reference signal. Op-Amp 2 as inverter. Op amp amplifier 3 and 4 as a hysteresis comparator. In fact, the op amp 3 and switch 1 is composed of the proportion of switching power supply. Op amp 4 and switch 2 are the same. Its switching frequency instability. In the op-amp output signal is phase 1, the op amp 3 and switch work. Then the output op amp 2 is the negative phase. Operational amplifier 4 is then input potential(constant 0) is better than the negative input of the high potential, so the op amp output constant 4 to 1, switch closure. In an op-amp output signal is negative phase, then the opposite. This realization of the work of the two alternating.

        The following diacussion about how the switch works. When the reference signal than the detection signal, that is the op amp 3 or 4 of the negative input signal than the positive input signal high for a small value to compare the output 0, the switch to open, resulting detection signal is rapidly increasing, when the detection signal higher than the reference signal a small value, the comparator output 1, swith off . Here we note that when the comparator circuit turn have a positive feedback process, which is characterized by hysteresis comparator. For example, detection of signals in the reference signal lower than the premise, as they continue to close the difference, in the miment they are equal, reference signal detection signal immediately higher than a certain value. This "certain value" affect the switching grequency. The greater the frequency , the lower it. Select here it 0.1~0.2V.

        C3,C4's role is to allow continued flow of high frequency switching current passing through, whereas the lower frequency of 50HZ signals have a greater resistance. C5 by the formula:50=calculated. L generally 70H, measuring about makng the best time. This is 0.15μ C about. R4 and R3 should be strictly equal to the ratio of 0.5, big wave distrotion significally, smaller and it does not have vibration, but rather larger, not smaller. The maxium switch current is:I==25A .

        More detailed discussion here about L1,L2 selection value. Electricity equivalent to the load back to the input transformer, the circuit in Figure 2.

        Taking into account the switching frequency much higher than 50HZ, the switch from open to related processes, the voltage transformer as constant. The power output of energy through the L:

        W=∫Uccdt=t

        Ignore all the loss is not ideal, this energy should be equal to the load enegy consumption. On the type of average power: P==t

        We hope Ucc-U is close to a small value, the battery can be a higher switching frequency and meet the requirements to the transformer. That " a small value" here take 0.5V, frequency of taking 5KHZ.. When Ucc-U <0.5V, switch to open a long time (this is a relative concept). If you need the power of the maximum output power is 150KW, then the load resistance 322.7Ω, converted to the transformer input side:0.48Ω.

        ∴instantaneous power load at this time: P==276W

        ∴P=×=276

        ∴L=2.2μH

        Can see that L is very small, on the switch negative, and the output are clipping. L value can increase production, but the maximum output power will decrease. The best way to solve this problem is with the 16V  power supply, also with a 8.5V TRANSFORMER (peak to 12V), and the reference signal 12V peak, but this time the circuit needs to be changed, not discussed here.

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