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上拉电阻的阻值是不是精确计算出来的

作者:zlucas 栏目:新手园地
上拉电阻的阻值是不是精确计算出来的
如果上拉电阻选的不当会出现什么情况
我一般都选在3~5k的电阻,不知道有没有精确的计算公式

2楼: >>参与讨论
a12345678
是大致估算的
 
3楼: >>参与讨论
iC921
不同的情况有不同的要求

下面就有正规、精确的、大致的两种情形:
研讨翻译 RS-422 and RS-485 Application Note 第2章 -2-

Biasing an RS-485 Network


RS485网络的偏置

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When an RS-485 network is in an idle state, all nodes are in listen (receive) mode. Under this condition there are no active drivers on the network, all drivers are tristated. Without anything driving the network, the state of the LINE is UNKNOWN. If the voltage level at the RECEIVER’s A and B inputs is less than ±200 mV the LOGIC level at the OUTPUT of the RECEIVERs will be the VALUE of the last bit received. In order to maintain the proper idle voltage state, bias RESISTORs must be applied to force the data LINEs to the idle condition. Bias RESISTORs are nothing more than a pullup RESISTOR on the data B LINE (typically to 5 volts) and a pulldown (to ground) on the data A LINE. Figure 2.4 illustrates the placement of bias RESISTORs on a transceiver in a two-wire configuration. Note that in an RS-485 four-wire configuration, the bias RESISTORs should be placed on the RECEIVER LINEs. The VALUE of the bias RESISTORs is dependent on termination and NUMBER of nodes in the SYSTEM. The goal is to generate enough DC bias current in the network to maintain a MINIMUM of 200 mV between the B and A data LINE. Consider the following two examples of bias RESISTOR calculation.

RS-485网络处于空闲状态时,所有的节点都成为听模式(接收模式)。在这种情况下,所有发送器都是第三态,网络上没有任何工作的驱动器。没有任何(驱动器)驱动网络,则线路状态不明(未知)。如果接收器输入端A和B间的电平低于±200mV,接收器输出的逻辑电平将被当作末位值接收起来。为保持适当的空闲电压状态,就必须使用偏置电阻强迫数据线维持线路(处于正确的)空闲条件。偏置电阻不过是数据线B(典型为+5V)的上拉电阻和数据线A(接到“地”)的下拉电阻,如图2.4所示的双线系统,详细示出了收发器的偏置电阻。但要注意,对于4线系统,偏置电阻应当放在接收器一端。偏置电阻的值,取决于是不是在终端(termination)和由系统节点数(的多少)来确定。使用终端电阻的目的,是让网络发生足够的直流偏置电流,以维持数据线A和B间的最小电压在200mV(以上)。参见以下两个有关偏置电阻计算的例子。
[译者语]The VALUE of the bias RESISTORs is dependent on termination and NUMBER of nodes in the SYSTEM一句似乎很有E文特色,就着它“译”出来。只是其中关键是“termination”一词如何能与“VALUE”有关?


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Figure 2.4 - Transceiver with Bias Resistors

Example 1. 10 node, RS-485 network with two 120 W termination RESISTORs


例1 含两个120Ω终端电阻的10节点RS-485网络

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Each RS-485 node has a load impedance of 12KΩ. 10 nodes in parallel give a load of 1200 Ω. Additionally, the two 120 Ω termination RESISTORs result in another 60 W load, for a total load of 57 Ω. Clearly the termination RESISTORs are responsible for a majority of the loading. In order to maintain at least 200mV between the B and A LINE, we need a bias current of 3.5 mA to flow through the load. To create this bias from a 5V supply a total series resistance of 1428 Ω or less is required. Subtract the 57 Ω that is already PART of the load, and we are left with 1371 Ω. Placing half of this VALUE as a pullup to 5V and half as a pulldown to ground gives a maximum bias RESISTOR VALUE of 685Ω for each of the two biasing RESISTORs.

各RS-485节点有负载阻抗为12kΩ,10个节点并联后得出的阻抗为1200Ω,另外,两个120Ω的线路终端电阻形成的另一个负载为60Ω,因此,总负载电阻为57Ω。明显地,终端电阻将是线路的主要负载。为保持A、B线间至少有200mV的电压,这就需要3.5mA的偏置电流流过(这57Ω的)负载。用5V电源进行偏置时,所要串联的电阻最大为1428Ω。扣除已经有的57Ω负载电阻,剩下的电阻为1371Ω。如将其值的一半接到5V电源做上拉电阻,将另外的一半拉接到地做下拉电阻,则各个节点的两边偏置电阻最大总共为685Ω。
[译者语]最后一句说的是上拉或下拉电阻的并联值而不是直接利用这个值。这一点比较容易误会,而目前看到的资料又没有对此加以说明的,因而特别强调之。

Example 2. 32 node, RS-485 network without termination


例2:没有终端电阻的32节点网络

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Each RS-485 node has a load impedance of 12KΩ. 32 nodes in parallel gives a total load of 375 Ω. In order to maintain at least 200 mV across 375Ω we need a current of 0.53 mA.  To generate this current from a 5V supply requires a total resistance of 9375W maximum. Since 375 Ω of this total is in the RECEIVER load, our bias RESISTORs must add to 9KΩ or less. Notice that very little bias current is required in SYSTEMs without termination.

各RS-485节点的负载阻抗为12kΩ,32节点并联所得的总负载为375Ω。为保持375Ω负载的电压至少为200mV,需要的电流为0.53mA。通过5V电源发生这样的电流,所需的负载电阻最大为9375Ω。因有375Ω是接收口器总负载(固有),因此,总的偏置电阻必须增加到9kΩ或更小。注意,系统没有终端电阻时所需的偏置电流非常小。


[025]

[025]

Bias RESISTORs can be placed anywhere in the network or can be split among multiple nodes. The parallel combination of all bias RESISTORs in a SYSTEM must be equal to or less than the calculated biasing requirements. B&B ELECTRONICS uses 4.7KΩ bias RESISTORs in all RS-485 products. This VALUE is adequate for most SYSTEMs without termination. The SYSTEM designer should always calculate the biasing requirements of the network. Symptoms of under biasing range from decreased noise immunity to complete data failure. Over biasing has less effect on a SYSTEM, the primary result is increased load on the drivers. Systems using PORT powered RS-232 to RS-485 converters can be sensitive to over biasing.

偏置电阻可以放在网络的任何位置或分别放到网络中的多处节点。系统中,所有偏置电阻并联后的总阻抗必须等于或小于计算出的偏置要求值。B&B Electonics公司所有的RS-485产品用4.7KΩ偏置电阻。此值适用于多数没有终端电阻的系统。系统设计师务必计算网络的偏置要求(值)。在偏置范围以下,则存在降低噪声的抗干扰性导致数据失效的symptoms。

Extending the Specification



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Some SYSTEMs require longer distances or higher NUMBERs of nodes than supPORTed by RS-422 or RS-485. Repeaters are commonly used to overcome these barriers. An RS-485 repeater such as B&B ELECTRONICS’ 485OP can be placed in a SYSTEM to divide the load into multiple segments. Each “refreshed” signal is capable of driving another 4000 feet of CABLE and an additional 31 RS-485 loads.

性能扩展
往往有一些系统需要比RS-422或RS-485所支持的有更长的距离或更多的节点数。要突破这类局限,通常要用到重复器。象B&B公司的485OP型这样的RS-485总线重复器,可以插入到系统中将负载分割成许多段,各“复原”后的信号有驱动另一段4000ft电缆和带动另外31个RS-485负载的能力。

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Another method of increasing the NUMBER of RS-485 nodes is to use low load type RS-485 RECEIVERs. These RECEIVERs use a higher input impedance to reduce the load on the RS-485 drivers to increase the total NUMBER of nodes. There are currently half and quarter load integrated CIRCUIT RECEIVERs available, extending the total allowable NUMBER of nodes to 64 and 128.
4楼: >>参与讨论
iC921
注意它也包括了上拉下拉两种情形
 
5楼: >>参与讨论
zlucas
谢谢
 
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