江苏春晨电缆有限公司生产YFFB扁电缆电为啥要分交流、直流？
继电器的应用，相信大家都知道，在电路中只要给它供电、断电也就可以工作了。然而，它的应用细节，不知道大家有没注意 。下面谈谈我的观点
The application of the relay, I believe everyone knows, in the circuit as long as it gives it power supply, power failure can work. However, it is not aware of the details of its application. Let's talk about my point of view
一、现在流行的接法，如图。
A popular connection, such as a picture.
图中，继电器的线圈经过Q1作为开关，使其导通与断开。D1作为续流，消耗线圈中的能量。
In the diagram, the coil of the relay passes through the Q1 as a switch to make it open and disconnect. D1, as a continuous flow, c***umes the energy in the coil.
二、继电器的特点;
Two, the characteristics of the relay;
1、吸合电流大于释放电流
1. The absorption current is greater than the release current
2、保持电流小于吸合电流、大于释放电流。
2, the current is less than the absorption current and greater than the release current.
以上两点均为继电器的“通病”，大家可以做一下实验，或看一看说明书。
The above two points are the "common disease" of the relay. You can do the experiment, or look at the instructi***.
三、流行电路的优点及缺点
Three. The advantages and disadvantages of the popular circuit
大家知道，继电器的线圈相当于电感，它的电流不能突变。在释放时，Q1截止瞬间，线圈将仍保持原来的电流大小，如果不接入D1这个二极管，产生的电压-----理论上是无穷大的(在外电路负载为无穷大时)，流行电路中的D1的接入，给线圈中的能量提供了释放的通道。
As you know, the coil of a relay is equivalent to an inductor, and its current can not be mutated. In the release, Q1 by the moment, will still maintain the current size of the original coil, if you do not access D1 the diode, the voltage generated, is theoretically infinite (in the external circuit load is infinite), access popular in circuit D1, provides a channel to release the energy in the coil.
然而，假如(理论上)二极管为理想的，即它只单向导通而没有任何功率消耗，那么，在继电器释放时，线圈中的电流将一直保持吸合时的较大电流(同时假如线圈为理想的)，这种情况将使继电器无法释放。
However, if (theoretically) diode is ideal, it is only a one-way conduction without any power c***umption, so, when the relay is released, the current in the coil will remain larger when (and if the current pickup coil is ideal), this situation will cause the relay can not be released.
实际中的二极管及线圈都不是理想的，所以，它是可以释放的。继电器的吸合到释放是由线圈中的电流决定的，如果二极管及线圈的等效电阻(直流)很小，那么它的释放时间将很长，反之，则较短。
The actual diodes and coils are not ideal, so it can be released. The attachment of the relay to the release is determined by the current in the coil. If the equivalent resistance (DC) of the diode and the coil is very ***all, the release time will be very long, otherwise, it will be shorter.
由此看，流行电路的优点是提供了Q1截止时的能量释放通道;其缺点是，释放时间还有进一步缩短的可能。
As a result, the advantage of the popular circuit is to provide the Q1 cut-off energy release channel; its disadvantage is that the release time may be further shortened.
四、其它接法。
Four. Other connecti***.
曾见过象下图中电路，也曾见过象下图中没有二极管的接法，这些接法都考虑到了***开关Q1截止时的反向电压，但没有考虑到释放时间问题。YFFB扁电缆电为啥要分交流、直流？
I h***e seen that the circuit in the following picture has also seen the connection of diodes without the diode in the following picture. All these methods take into account the suppression of the reverse voltage when the switch Q1 is off, but the time of release is not c***idered.
五、建议接法。
Five.
1、加入电阻R1，使能量释放快一些。
1, the resistance R1 is added to make the energy release faster.
上图中，线圈在Q1关断时，能量主要消耗在R1上，使继电器可以快速降到释放电流。
In the above picture, when the coil is turned off at Q1, the energy is mainly c***umed on the R1, so that the relay can quickly drop to the release current.
R1的选择，由Q1的较高反压、线圈工作电流两者决定，电阻越大，释放时间越短。---- 计算就不说了吧。
The choice of R1 is determined by both the high reverse voltage of Q1 and the working current of the coil. The greater the resistance, the shorter the release time. The calculation will not be said.
2、减小继电器保持时的功耗。YFFB扁电缆电为啥要分交流、直流？
2, reduce the power c***umption of the relay.
大家知道，继电器吸合时需要较大的电流，而保持吸合状态则不需要和吸合时的电流一样。
It is known that the relay needs a larger current when the relay is sucking, while maintaining the absorption state does not require the same current as the suction.
下图接入R1及C1将明显减小继电器的保持功耗。在继电器吸合前，C1已充电至供电电压，吸合的瞬间将由C1为
The access to R1 and C1 below will significantly reduce the maintenance of the relay. Before the relay is absorbed, the C1 has been charged to the power supply voltage, and the moment of the absorption will be C1.
继电器供电，以保障吸合所需的大电流。当吸合后，供给线圈的电流来自R1，它将电流限制到较小状态
The relay is powered to ensure the high current required to absorb. When it is absorbed, the current of the supply coil comes from R1, which limits the current to a ***aller state.