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10档。我们在测量时，要求探头对被测的电路影响必须达到最小，并希望对测量值保持足够的信号保真度。</div><div align="left">先来看看示波器的探头的基本原理：</div><div align="left">一般示波器的垂直偏转灵敏度是固定的，5mV／格 5V／格，荧光屏垂直方向共8格，也就是说，满屏最大也只能看到40V，为扩大可观信号的幅度范围，就要用探头，它可使示波器的垂直偏转灵敏度在很大范围内调节，如用探头上的×10档，可观测的幅度就能达到400V了。而对于探头的要求是输入电阻要高，同时在示波器的整个通频带内衰减的分压比均匀不变。要达到这个要求，仅用简单的电阻分压是达不到目的的。因为在下一级的输入及引线部分都存在分布电容，这个分布电容的存在，对于被测信号的高频分量有严重的衰减，同时还会造</div><div align="left">成信号的高频分量的失真(如：脉冲上升时间变慢)。为此，一般</div><div align="left">探头采用的是阻容补偿分压器。</div><div align="left">则分压比为：可Uo= zo= Ro (1 +j toC iR可1) 可式3</div><div align="left">当调整探头电路中的可变电容元件c ，使R ×C。=R。XC。</div><div align="left">在式3中，由于l+j‘^)C。R =1+j∞coRo则式3中分压比变</div><div align="left">⋯为： 可U o1 =击Z 。+Zo =丽Ro(1+ j(I)C I R l R 。(1+j(．) cono ) ：丽R0n+oRl ⋯式4</div><div align="left">此时，分压比在整个通频带内是均匀的，与频率无 1。</div><div align="left">这种推导，<a href="http://theory.oecr.com" target="_blank" class="relatedlink">理论</a>上是在被测信号的内阻为零的情况下成</div><div align="left">立，然而在实际操作过程中是有很大出入的，特别是高频电路</div><div align="left">的测量。因为高频信号电路一般都有内阻，5O欧至几千欧，就</div><div align="left">是带功率输出的高频信号发生器电阻也有50欧或75欧(但一</div><div align="left">般不进行高频信号发生器本身的定量分析，也无测试的必要)，</div><div align="left">大量的测试是在各种有高频信号的一般放大电路中进行。这种</div><div align="left">高频放大电路，其频率上了2MH 以上，内阻常在1千欧至5</div><div align="left">千欧左右，此时用带宽为20 MH 的普通示波器很难对这种高</div><div align="left">频放大电路的幅频特性进行准确的定量测量。</div><div align="left">让我们从示波器的参数性能指标来说明，以模拟示波器</div><div align="left">CA8020技术参数为例</div><div align="left">垂直系统：</div><div align="left">带宽(-3dB)DC：0—10MHz，AC：10Hz一20MHz</div><div align="left">输入阻抗(1：1)1MQ±3％，25pF±5pF。</div><div align="left">经10：1探头：10MQ 4-5％，15pF4-5pF</div><div align="left">将以上参数中的输入阻抗1：1(1MQ，25pV)和经10：1探</div><div align="left">头(10Mn，15pF)分别画出等效电路，如图3、图4所示。并分</div><div align="left">别用OrCAD Capture<a href="http://opticaldesign.oecr.com" target="_blank" class="relatedlink">软件</a>做仿真幅频分析(设定Ri=IK，</div><div align="left">Us=IV，f=lMHz～100MHz)。如图5、图6所示。</div><div align="left">I 示波器等效输入电路</div><div align="left">【作者简介】赵珊(1960一)，女，安徽寿县人，工程师，从事电类基础实验实训教学工作。</div><div align="left">63</div><div align="left">Ri</div><div align="left">被测电路等效</div><div align="left">输出电路</div><div align="left">图3探头1：1的等效电路</div><div align="left">放 I 路</div><div align="left">效辁．1II 躁L</div><div align="left">被器等效掬入}乜强</div><div align="left">图4探头10：1等效电路</div><div align="left">图5中，屏幕正中的提示框内：A2表示lv的信号源幅度</div><div align="left">信号，A1表示下降3db处的带宽。频率从1H 至100MH 。信</div><div align="left">号幅度降至0．707V，此时的频率是6．35MH 。即U。=Us(D U。</div><div align="left">(Rj：1K．R0．1M，co )=us(D，频带1 H -6．35MHz，当用1：1探头测量</div><div align="left">频率超过6．3 5MH 的被测电路，误差将大大增加。</div><div align="left">图6中，A1表示下降3db处的带宽，A2表示1V的信号</div><div align="left">源幅度信号，频率从1Hz至100MHz；信号幅度降至0．707V，</div><div align="left">此时的频率是1 0．59MHz。即U0(HI o o=15p)=Us∞，频带</div><div align="left">1Hz一10．59MHz，当用10：1探头测量频率超过10．59MHz的被</div><div align="left">测电路，误差将大大增加。</div><div align="left">图5 U0(RII1KR0：1M 嗡。)=us(o</div><div align="left">图6 U0(Ri：1 ㈣o ll5 =us(0</div><div align="left">那么按照CA8020模拟示波器技术参数，要求被测电路的</div><div align="left">内阻不小于多少才是准确的呢?</div><div align="left">经仿真幅频分析，得出的结论是：</div><div align="left">用探头X 1档时，被测电路的等效阻抗值是315欧，小于</div><div align="left">或等于此值，它的可测带频才达到示波器说明书中20MHz的</div><div align="left">带宽；</div><div align="left">用探头×10档时，被测电路的等效内阻边缘值是530欧，</div><div align="left">小于或等于此值，它的可测带频才达到示波器说明书中</div><div align="left">2OMHz的带宽；</div><div align="left">以上的仿真分析，在实验室的实际电路测试中也得到了证</div><div align="left">实。所以使用示波器测量时，尤其是在高频电路的测量中，应注</div><div align="left">意探头的分布电容参数对被测电路的影响，同时尽可能地使用</div><div align="left">10：1的探头。在选用示波器时，一定要选用输入分布电容参数</div><div align="left">比较小的，否则测量误差会比仪器说明书中的要大很多。</div><font face="宋体"><font style="font-size:14pt"><div align="left">以上信息由西安安泰测试设备有限公司整理，希望对您能有所帮助。西安安泰测试设备有限公司是一家专注于电子测试及工业测试领域仪器销售与仪器维修的专业公司。为客户提供“丰富的测试<a href="http://b.oecr.com/sell" target="_blank" class="relatedlink">产品</a>选择、完整的系统测试解决方案、专业的测试软件开发、全面的技术支持及售后服务”。目前公司主要进行电子测试仪器的销售及维修，自研产品的开发，已经成为西北地区规模最大的仪器综合服务商之一。</div><div align="left"> </div><div align="left"> 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