大頻寬轉阻放大器 (中英文對照)
大部分的工程師都知道，只需一個容量夠大的電阻就可使轉阻放大器(Transimpedance Amplifier)將輸入電流轉成合理的輸出電壓範圍，但若要穩定電路，便需有一個夠大的電容與回饋電阻(Feedback Resistor)並列放置，本文將討論如何計算所需的回饋電阻值，以確保設計能具有最大可能的頻寬，同時亦可保持穩定。
回饋係數可利用修改電路推導
對於工程師而言，要計算運算放大器的回饋係數(Feedback Factor)，也就是把電流轉換成電壓的設定可能有點困難。然而，只要能計算出轉阻放大器的轉換函數(Transfer Function)，並採用電壓運算放大器，便可較易於掌握電流到電壓的轉換。本文以廠商推出的運算放大器做為轉阻放大器設計範例，圖1所示為基本的轉阻放大器配置。
圖1 基本的轉阻放大器配置
圖1中只省略顯示電源耦合電容器(Power Supply Decoupling Capacitor)。在大多數情形下，選用光電二極管(Photo Diode)使設計人員可在VBias和+V採用同一個電源，但若採用分離式電源，便可把運算放大器的反向輸入保持在虛擬接地，為了得出回饋係數，須檢查圖2中的光電二極管等效電路(Equivalent Circuit)。
圖2 光電二極管等效電路
在上述等效電路中的二極管是個理想的二極管。由於光電二極管須經過反向偏壓才可正常操作，因此該理想二極管並不會包括在回饋係數的計算內。CJ是出現在二極管空乏區(Depletion Region)的電容並且已包括在光電二極管的規格內，而IPH則是來自光電二極管活動區的電流，至於電流源的阻抗則是光電二極管的串列電阻，以RSH表示，該串列電阻最小需為10百萬歐姆(MΩ)，而一般來說會較大，通常會大於100MΩ，由於RSH處於非常高的值，因此當計算回饋係數時，可以把電流源考慮成一個完美的理想值。此外，接觸電阻(Contact Resistance)和非空乏矽晶片(Undepleted Silcon)的電阻會以RS表示，在正常情況下，RS會小於100歐姆(Ω)故可以忽略，雖然RS可在此分析中被忽略，但這電阻在某些二極體中仍可能超出1千歐姆(kΩ)，這時就足以影響轉阻電路的回饋係數。另外，運算放大器還擁有一個輸入電阻，圖3表示用來計算回饋係數的電路，其中CCM代表運算放大器的輸入共模電容值。
圖3 等效AC電路
在這電路分析中，+V是AC接地，透過這個電路可輕易計算轉阻放大器的移動函數。假設這邊的運算放大器為完美的系統，因此反向輸入可處於虛擬接地，而由於CJ和CCM兩個節點都在接地，因此不會對這個電路的移動函數造成任何的影響，計算公式如下所示。
或
轉阻放大器的設計人員對上述算式並不陌生。對於低頻VOUT=-RF和極低頻VOUT=-1/sCF的情況，這個電路會在fP=1/2πRFCF時插入一個極(Pole)，而這個極可穩定電路。為了得出回饋係數，圖3被修改成圖4。
圖4 用來計算回饋係數的修改電路
正如本文一開始時所述，運算放大器是電壓放大器而不是電流放大器，圖4中的電流源擁有極大的電阻，因此不會對回饋係數有任何影響。至於該兩個輸入電容器則可透過CIN=CJ+CCM的設定組合在一起。因此，當電源流不會帶來影響時，圖4中的電路僅是一個微分器電路(Differentiator Circuit)，當中的輸入電容通往電路(CIN)的接地。即使轉阻放大器是把電流轉換成電壓的轉換器，但回饋係數仍是基於電壓放大器的配置，也就是基於微分器電路。
回饋係數的概念很簡單，就是從運算放大器的輸出回饋到輸入的程度。在計算的時候，須要假設通往運算放大器輸入的節點並沒有連接到運算放大器的輸入，之後再計算出輸入電壓和輸出電壓的比例，即VIN/VOUT，圖5所示為用來計算回饋係數的電路。
圖5 用來計算回饋係數的電路
CF和RF的並聯阻抗如下公式：
上述的算式亦可用來計算移動函數，把計算轉變成一個分壓器網路，給出的算式為：
透過進行一些代數計算，回饋係數F的算式便為：
轉阻放大器的回饋係數基本上會與微分器電路的回饋係數相同，唯一的不同是微分器電路的CIN被加上運算放大器的反向輸入上，但在轉阻放大器中，CIN只是光電二極管的電容值和運算放大器的輸入電容值之總和，且須注意在低頻時F=1。
透過過補償計算CF最佳數值
運算放大器的雜訊增益為1/F。在微分器電路中，CIN會在1/F內插入一個零點，導致微分器電路長期不穩定，由於轉阻放大器本身就是一個微分器電路，所以本來就不太穩定，因此，必須用一個CF限制電路的頻寬才能穩定下來。為了確保穩定性，設計人員可考慮使用一個相對較大的回饋電容器，以過補償這個電路，不過，使用較大的CF或是求取CF最佳值的同時，將會增大電路的頻寬，使放大器的頻寬超出所需。
圖6所示為一條具備有最佳CF值的1/F曲線。1/F的極點位於A曲線上，因此CF的最佳值是在A=1/F或A*F=1。加入一個極點fP可給出一個超前-滯後補償，其在兩條曲線交點上只有135度的相移，從而為轉阻放大器提供一個45度的相位邊際。若使用的是一個沒有補償的放大器，1/F和A的交點便一定會在運算放大器的第二個極點之前出現。
圖6 開放迴路增益A和雜訊增益1/F關係圖
從計算F的算式中，可知1/F的極點為：
而1/F的零點則為：
在fZ加入零點會令1/F之斜率出現由0dB至+20dB的變化。為了達到穩定，1/F的斜率必須返回0dB，而方法就是加入CF以產生出一個極點。此外，圖6也代表一個過補償的例子，當中的CF值大於其最佳值。通過加大CF，1/F的極點和零點將可出現在較低的頻率處。這樣，CIN對零點值的影響便很小，因此極點和零點的頻率便會同時下降，並且隨著CF在過補償時的增加而逐漸接近。當使用非補償運算放大器時，便可能須使用過補償，此時1/F和A的相交點會出現在接近A的第二個極點。A的算式其實是一個增益和頻寬的乘積：
在算式中，fGBW是一個頻率，在該處增益為一的穩定運算放大器的開放迴路增益等於0dB(1V/V)。對於沒有補償的零組件，假如-20dB的斜率被擴展到0dB的線上，在這點上的頻率便等於fGBW的數值。使用算式A×F=1，可解出CF為：
由於A在1/F的極點處與1/F相交，便得出以下兩種關係：
利用這兩條公式解答A×F的算式，而透過將實質與想像零組件的平方相加起來(s=j ω)樣便可得出以下的算式：
將以上的結果簡化成一條可用來求解CF的算式，得出：
這條公式很難用來解出CF，所以最好設立一個包含有所有數值的Excel試算表，並透過重複計算來解出CF。在幾乎所有的應用中，都將會採用CIN>CF這個假設。配合這個假設，上述的公式便可被簡化，從而輕易計算出CF的數值：
這便完成了對一個轉阻放大器的理論分析和計算，而上述便是用來計算CF最佳值的公式。假如計算出來的CF值為電路帶來太大的振鈴(Ringing)，便可透過過補償減低振鈴，但當然過補償亦會同時減少轉阻放大器的頻寬。
本文設計實例採用美國國家半導體(NS)運算放大器，為速度中等的零組件，其增益頻寬乘積為88MHz，同時也是非補償式的運算放大器，因此可探究轉阻放大器中的非補償性運算放大器的操作情況。元件的輸入電容為15皮法(pF)，並選用OSI Optoelectronics的PIN-HR040光電二極管作為檢測器，該零組件在此電路的條件下之頻寬約為300MHz， 這個程度的頻寬可防止由光電二極管導致的額外頻寬限制，同時，二極管的電容亦很小，在這個電路操作的條件下只有7皮法。雷射二極管被用來作為光源，而雷射二極管的上升和下降時間在5奈秒以下，遠低於運算放大器的上升和下降時間，本例選用的是紅色雷射二極管，以便可看到光線束，以確保設置的安全性。
首先，一個高增益會被選用並將RF設定成100kΩ，圖7所示為用作轉阻放大器的電路。當光電二極管被光線照射時，電流便會流向運算放大器的反向節點，若設計的目標是要達到-20毫伏特的最低輸出電壓，從光電二極管流出來的電流便必須為0.2微安培，至於從光電二極管出來的電流大小，可以憑光源的預期光功率中計算。以PIN-HR040為例，使用的雷射二極管的光波長為630奈米，其所給出的光回應度為0.38A/W，因此假如電流是0.2微安培，光電二極管的光功率便為：
求解W(光功率以瓦特作單位單算)：
圖7 轉阻放大器示意圖
若光源的所有能量均集中在光電二極管的活動區，只須進行上述的計算便足夠，本例中光電二極管的活動區面積為0.77平方毫米，且呈圓形狀。光纖電纜的切面面積比光電二極管大，且其光功率一般都以W/cm2作為量度單位，若現在把光電二極管的活動區面積以平方釐米表示，結果便為7.7×10-3平方公分。對於相同的0.2微安培輸出電流光源，但以W/cm2作量度單位的話，所需的功率便為：
圖7中，供電源用的旁路電容器並沒有展示，圖中每一個電源都擁有一個6.8微法(μF)的鉭電容器來旁路較低的頻率，同時每個電源並設有一個0.1微法的陶瓷電容器以旁路高頻。當中的陶瓷旁路電容器須儘量放近運算放大器的電源接腳，由於光電二極管的CJ=7皮法和運算放大器的CCM=15皮法，因此CIN=22皮法，採用計算CF的公式可得出CF=0.53皮法，在此可察覺到不只是CF遠比CIN小，而且其高增益配置的數值已達到標準可用值的極限。這個實驗用的電容器是0.5皮法，但度量出來的結果是0.64皮法，因此轉阻放大器經過一些過補償。轉阻放大器的頻寬可以從RFCF時間常數中計算出來，或大約是2.5MHz，圖8所示為電路的測試結果。
圖8 電路測試結果
對於大約為108奈秒的上升和下降時間，轉阻放大器頻寬約為3.2MHz，這數值與用RFCF時間常數計算出來的2.5MHz很接近，這種過衝無疑展現一個大於45度的相角裕度，而這可能是來自由CF的最佳值，圖8的照片經過波形均化後拍攝，因此可輕易看到放大器對小型訊號的回應。
求解W(光功率以瓦特作單位單算)：
圖7 轉阻放大器示意圖
若光源的所有能量均集中在光電二極管的活動區，只須進行上述的計算便足夠，本例中光電二極管的活動區面積為0.77平方毫米，且呈圓形狀。光纖電纜的切面面積比光電二極管大，且其光功率一般都以W/cm2作為量度單位，若現在把光電二極管的活動區面積以平方釐米表示，結果便為7.7×10-3平方公分。對於相同的0.2微安培輸出電流光源，但以W/cm2作量度單位的話，所需的功率便為：
圖7中，供電源用的旁路電容器並沒有展示，圖中每一個電源都擁有一個6.8微法(μF)的鉭電容器來旁路較低的頻率，同時每個電源並設有一個0.1微法的陶瓷電容器以旁路高頻。當中的陶瓷旁路電容器須儘量放近運算放大器的電源接腳，由於光電二極管的CJ=7皮法和運算放大器的CCM=15皮法，因此CIN=22皮法，採用計算CF的公式可得出CF=0.53皮法，在此可察覺到不只是CF遠比CIN小，而且其高增益配置的數值已達到標準可用值的極限。這個實驗用的電容器是0.5皮法，但度量出來的結果是0.64皮法，因此轉阻放大器經過一些過補償。轉阻放大器的頻寬可以從RFCF時間常數中計算出來，或大約是2.5MHz，圖8所示為電路的測試結果。
圖8 電路測試結果
對於大約為108奈秒的上升和下降時間，轉阻放大器頻寬約為3.2MHz，這數值與用RFCF時間常數計算出來的2.5MHz很接近，這種過衝無疑展現一個大於45度的相角裕度，而這可能是來自由CF的最佳值，圖8的照片經過波形均化後拍攝，因此可輕易看到放大器對小型訊號的回應。
接著觀察一個具有較低增益的轉阻放大器，以圖7電路圖為例，若RF現在減少到10kΩ，放大器的增益便僅僅為第一個例子的十分之一。雖然，這個較低的RF電阻器可給出一個較廣闊的頻寬，可是對於同一個輸出級數而言，雷射二極管的光輸出須比轉阻二極體高十倍。然而，當雷射二極管的輸出級數增加，便會在光輸出產生出一個熱梯度(圖9)。現在，CF的理想值為1.7皮法，因此可使用1.8皮法的標準值，且須檢查極點的位置以確保穩定性。極點出現在8.8MHz，而雜訊增益(1/F)則等於10，這是本例元件的最小增益值，雖然符合理想兩極系統的穩定性要求，但實驗結果顯示有過大的振鈴，因此這個設計須加入過補償達到可接受的穩定程度。由於有其他的運算放大器內部極點和零點接近第二個極點，使設置與理想的兩極系統相距甚遠，因此須進行過補償，現在，CF的最終值為2.7皮法。圖9分別顯示轉阻放大器的回應，其RF=10k歐姆和CF=2.7皮法。
圖9 轉阻放大器的回應
較高增益的配置擁有大約33奈秒的上升和外降時間，提供了一個約為10.6MHz的頻寬，當CF等於2.7皮法時，極點便出現在5.9MHz。本文印證在穩定性的計算上，轉阻放大器與微分放大器相同，唯一的分別在於光電二極管，其為附帶在運算放大器輸入節點上的電流源，只要把二極管的電容加到微分放大器的總輸入電容上，光電二極管便不會對穩定性的計算有任何影響。這項實驗採用兩個不同增益的配置印證理論與測試電路可完全吻合，計算CF的算式可以使用在任何的微分放大器設計上，即使微分放大器與轉阻放大器的移動函數不相同，但用來計算穩定性的回饋係數則相同。
(本文作者任職於美國國家半導體)
上升和下降時間測量請參考最後補充說明.
補充說明:
1)數位器的頻寬最好比要測量的訊號中的最高頻率高 3 ~ 5 倍.
2)Tr=2.2RC ==>RF=10k歐姆和CF=2.7皮法 ==> Tr=59.4ns
==> BW=5.892255Mz(已知RC常數值求BW)
圖9 轉阻放大器的回應
較高增益的配置擁有大約33奈秒的上升和外降時間，提供了一個約為10.6MHz的頻寬，當CF等於2.7皮法時，極點便出現在5.9MHz。本文印證在穩定性的計算上，轉阻放大器與微分放大器相同，唯一的分別在於光電二極管，其為附帶在運算放大器輸入節點上的電流源，只要把二極管的電容加到微分放大器的總輸入電容上，光電二極管便不會對穩定性的計算有任何影響。這項實驗採用兩個不同增益的配置印證理論與測試電路可完全吻合，計算CF的算式可以使用在任何的微分放大器設計上，即使微分放大器與轉阻放大器的移動函數不相同，但用來計算穩定性的回饋係數則相同。
(本文作者任職於美國國家半導體)
上升和下降時間測量請參考最後補充說明.
補充說明:
1)數位器的頻寬最好比要測量的訊號中的最高頻率高 3 ~ 5 倍.
2)Tr=2.2RC ==>RF=10k歐姆和CF=2.7皮法 ==> Tr=59.4ns
==> BW=5.892255Mz(已知RC常數值求BW)
3) Tr=33ns BW=0.35/33ns=10.606060MZ (由示波器測出Tr值求BW)
4)Tr=108奈秒的上升和下降時間，BW=0.35/108ns=3.240740Mhz頻寬約為3.2MHz.
EX: RF=100k歐姆和CF=0.64皮法 Tr=2.2RC = 140.8ns,
BW=0.35/140.8ns=2.485795Mhz==>頻寬約為2.4MHz.
5)另一個和頻寬有關的重要主題是上升時間 (Rise time)。輸入訊號的上升時間是指訊號從最大訊號振幅的 10% 轉換到 90% 的時間，而且與頻寬成反向相關，由以下公式呈現。此公式採用單極模型，R-C 限制輸入反應為基礎。
圖5
這表示100 MHz數位器的輸入途徑的上升時間是3.5 ns。我們建議數位器輸入途徑的上升時間為受測訊號上升時間的 1/3 到 1/5，才能在上升時間誤差最低的情況下測量訊號。測得之上升時間的理論值 (Trm) 可以利用數位器的上升時間 (Trd) 和輸入訊號的實際上升時間 (Trs) 計算而得。
圖6
舉例來說，在使用 100 MHz 高速數位器測量上升時間為 12 ns 的訊號時，測得的上升時間約為 12.5 ns=sqr(3.5^2+12^2)-->(由示波器測出Tr值求真實Tr值,即去掉示波器誤差值)。
英文原文
Most engineers know that to design a transimpedance amplifier circuit, they just need a large-enough resistor to convert the input current to a reasonable output voltage range. To stabilize this circuit, a large enough capacitor must be placed in parallel with the feedback resistor. This article will show how to calculate the value for the feedback capacitor to ensure that the design has the largest possible bandwidth, and will still be stable.
Feedback factor calculation
Calculating the feedback factor for an op amp that is set up for current-to-voltage conversion may be a bit of a mystery to many engineers. By deriving the transfer function for a transimpedance amplifier and using a voltage amplifier op amp, the conversion will be easy to understand. Here we use the LMV793 op amp as an example for a transimpedance amplifier design. A basic transimpedance amplifier configuration is shown in Figure 1.
Figure 1: Basic transimpedance amplifier configuration
(Click to enlarge image)
The figure shows the complete transimpedance amplifier; only the power supply decoupling capacitors are not shown. In most cases, the selection of the photodiode will allow the designer to use the same supply for VBias and +V. Using split supplies keeps the inverting input of the op amp at virtual ground. To derive the feedback factor, it is necessary to examine the equivalent circuit of the photodiode, Figure 2.
Figure 2: Photodiode equivalent circuit
(Click to enlarge image)
The diode is an ideal diode in the equivalent circuit. Since the photodiode must be back-biased for proper operation, the ideal diode is not included in the feedback factor calculations. CJ is the capacitance that occurs from the depletion region of the diode and it is included in the photodiode specifications. IPH is the current that occurs from the photodiode action. The impedance of the current source is the series resistance of the photo diode, shown as RSH. The series resistance is at least 10 MΩ and typically much higher, usually over 100 MΩ.
Since RSH is such a high value, the current source will be considered ideal when deriving the feedback factor. The contact resistance and the resistance of the undepleted silicon are represented by RS. Normally RS is under 100 Ω and can be ignored. Even though RS is being ignored in this analysis, this resistance can exceed 1 kΩ in certain diodes, making it large enough to affect the transimpedance circuit feedback factor.
The op amp also has an input resistance. Figure 3 shows the circuit used in the feedback factor calculations. CCM is the input common mode capacitance of the op amp.
Figure 3: Equivalent AC circuit
(Click to enlarge image)
For the circuit analysis, +V is an AC ground. From this circuit it is easy to derive the transfer function of the transimpedance amplifier. The op amp is assumed to be ideal; therefore, the inverting input is at virtual ground. CJ and CCM have no affect on the transfer function of this circuit since both nodes are at ground. The transfer function is (Equation 1):
Or (Equation 2):
This equation is quite familiar to designers of transimpedance amplifiers. For low frequencies,
VOUT = -RF, and for very high frequencies VOUT = -1/sCF. This circuit inserts a pole at fP = ½πRFCF and this pole stabilizes the circuit. (The effect of this pole on the circuit is explained later.)
To derive the feedback factor, the circuit in Figure 3 is modified as shown in Figure 4.
Figure 4: Circuit for feedback-factor derivation
(Click to enlarge image)
As noted at the beginning of this article, the op amp is a voltage amplifier, not a current amplifier. The current source in Figure 4 has infinite impedance; therefore, it has no affect on the feedback factor! The two input capacitors have been combined by setting CIN = CJ + CCM. Therefore, with no effect from the current source, the circuit in Figure 4 is just a differentiator circuit with the input capacitance to the circuit (CIN) grounded. Even though a transimpedance amplifier is a current-to-voltage converter, the feedback factor is still based on a voltage-amplifier configuration, the differentiator circuit.
The feedback factor is simply what is fed back to the input from the output of the op amp. This is calculated by assuming that the node at the input to the op amp is not connected to the input of the op amp, then calculating the ratio of the input voltage to the output voltage, VIN/VOUT. Figure 5 shows the circuit used to calculate the feedback factor.
Figure 5: Feedback-factor derivation
(Click to enlarge image)
The parallel impedance of CF and RF is (Equation 3):
This is also used in the transfer function. Now the calculation becomes a voltage divider network giving the equation (Equation 4):
With some algebraic manipulations the equation for the feedback factor, F, becomes (Equation 5):
The feedback factor for a transimpedance amplifier is identical to the feedback factor of a differentiator circuit. The only difference is that CIN of a differentiator circuit is added to the inverting input of the op amp, but for a transimpedance amplifier CIN is just the sum of the capacitance of the photo diode and the input capacitance of the op amp. Note, that for low frequencies, F = 1.
Optimum value for CF
The noise gain for an op amp circuit is 1/F. In a differentiator circuit, CIN will insert a zero in 1/F, thus making a differentiator circuit inherently unstable. Since a transimpedance amplifier is a differentiator circuit, it is inherently unstable. CF must be added to make the circuit stable. CF limits the bandwidth of the circuit. To insure stability a designer can add a feedback capacitor that is rather large, thus overcompensating the circuit. However, finding the optimum value for CF will maximize the bandwidth of the circuit and avoid using an op amp with a wider bandwidth than necessary. Figure 6 shows the relationship of 1/F and the open loop gain for an optimum design.
Figure 6: Open loop gain, A, and noise gain, 1/F
(Click to enlarge image)
Figure 6 shows the 1/F curve with the optimum value for C. The pole of 1/F is located on the A curve. Therefore the optimum value for CF is where A=1/F, or A*F=1. Without CF there would be only a zero for the 1/F curve, giving close to a 180° phase shift at the intersection of A and 1/F. Adding the pole fP gives a lead-lag compensation with a phase shift of only 135° at the intersection of the two curves; giving the transimpedance amplifier a phase margin of 45°. If using a decompensated amplifier the intersection of 1/F and A must occur before the second pole of the op amp.
From the equation for F the pole for 1/F is (Equation 6):
And the zero for 1/F is (Equation 7):
The addition of a zero at fZ changes the slope of 1/F from 0 dB to +20 dB. For stability the slope of 1/F must be changed back to 0 dB and this is done by the pole created by the addition of CF.
Figure 6 also shows the overcompensated case, where the value of CF is larger than its optimum value. By making CF larger, the pole and zero for 1/F will occur at lower frequencies. In addition, CIN becomes a smaller portion of the denominator for the value of the zero; therefore, the pole and zero both drop in frequency and come closer together as the value of CF is increased for over compensation. Overcompensation may be necessary when using decompensated op amps and the intersection of 1/F and A is located near the second pole of A.
The equation for A is simply the gain bandwidth product which is (Equation 8):
In this equation, fGBW is the frequency where the open loop gain is 0 dB (1 V/V) for unity gain stable op amps. For decompensated parts, if the -20 dB slope was extended to the 0 dB line, the frequency at this point would be the value for fGBW. Using the equation A*F = 1, we now solve for CF (Equation 9):
Because A intersects 1/F at the pole of 1/F, we have the following two relationships (Equations 10a and 10b):
Using these two formulas to solve the AxF equation, the magnitude is found by taking the square root of the sum of the squares (rss) of the real and imaginary parts (s = jω). This results in (Equation 11):
(Click to enlarge image)
The above result simplifies into the equation to use to solve for CF (Equation 12):
(Click to enlarge image)
This formula is difficult to solve for CF. It is best to set up an Excel spreadsheet with all the values and use iteration solve for CF. In almost all applications, the assumption that CIN >> CF will apply.
With this assumption the above formula can be simplified to easily calculate the value for CF as follows (Equation 13):
This completes the theory and calculations for a transimpedance amplifier. The above formula is the one to use for calculating the optimum value for CF. If the value calculated for CF gives too much ringing in the circuit, then some overcompensation can be used to reduce the ringing. Overcompensation will, of course, reduce the bandwidth of the transimpedance amplifier.
Design Examples
National Semiconductor's LMV793 is the op amp used for the design example. This is a medium-speed part with a gain bandwidth product of 88 MHz. It is also a decompensated op amp, which gives the opportunity to investigate the operation of a decompensated op amp used in a transimpedance amplifier.
The input capacitance of the LMV793 is 15 pF. The PIN-HR040 photodiode from OSI Optoelectronics was selected for the detector. This part has bandwidth of about 300 MHz under the conditions being used in the circuit. This bandwidth prevents additional bandwidth limitation due to the photodiode. This diode also has a low capacitance, only 7 pF for the circuit operating conditions. Laser diodes were used for the light source. A laser diode has rise and fall times under 5 nsec, well under the rise and fall times of the op amp. A red laser diode was selected so that the beam of light can be seen, ensuring the safety of the setup.
A high gain will first be selected, setting RF to be 100 kΩ. Figure 7 shows the circuit used for the transimpedance amplifier.
Figure 7: Transimpedance amplifier
(Click to enlarge image)
When the photodiode is exposed to light, the current will flow into the inverting node of the op amp. The direction of the current flow is shown in the equivalent circuit in Figure 2. If the design goal is to have a minimum output voltage output of -20 mV, then the current flow from the photodiode must be 0.2 μA.
The expected current from the photodiode can be calculated from the expected optical power from the light source. For the PIN-HR040, the responsivity to light is given as about 0.38 A/W for the 630 nm wavelength of light from the laser diode being used. So for 0.2 μA of current the optical power on the photo diode is (Equation 14):
Solving for W (light power in Watts), we get (Equation 15):
If the light source has all of its energy on the active area of the photo diode, then this is the only calculation required. The active area of this photo diode is 0.77 mm2 and is a circular pattern.
Fiber optical cable may have an area larger than the photo diode and normally the optical power is measured in W/cm2. Expressing the area of the photo diode in cm2, the result is 7.7x10-3 cm2. For the same 0.2 μA output current from a light source measured in W/cm2, the necessary power would be (Equation 16):
In Figure 7, the bypass capacitors for the power supplies are not shown. Each supply has a 6.8 μF tantalum capacitor for bypassing the lower frequencies. In addition, each supply has a 0.1 μF ceramic capacitor for bypassing the high frequencies. The ceramic bypass capacitor must be located as close as possible to the power pins of the op amp.
For the photodiode CJ = 7 pF. For the op amp CCM = 15 pF. Therefore CIN = 22 pF. Using the formula for CF gives CF = 0.53 pF. Note that not only is CF much smaller than CIN, but with high-gain configurations the value reaches the limits of standard available values. For this experiment, a 0.5 pF capacitor was available, which measured 0.64 pF, so the transimpedance amplifier had slight overcompensation. The bandwidth of the transimpedance amplifier is derived from the RFCF time constant, or about 2.5 MHz. Figure 8 and Figure 9 show the test results of this circuit.
Figure 8: 1 VP-P output
(Click to enlarge image)
Figure 9: 20 mVP-P output
(Click to enlarge image)
From the rise and fall time being around 108 nsec, the estimated bandwidth is 3.2 MHz, which is close to the 2.5 MHz calculated from the RFCF time constant. The overshoot does show a phase margin greater than the 45° phase margin that would come from the optimum value of CF. Waveform averaging was used for the Figure 9 photograph so the actual response of the amplifier for small signals can easily be seen.
Next let's look at a transimpedance amplifier with a lower gain. Referring to Figure 7, if RF is now reduced to 10 kΩ, then the gain of the amplifier is one-tenth of the first example. This lower RF resistor gives a wider bandwidth. However, the light output of the laser diode must be ten times brighter for the same output level from the transimpedance amplifier. The higher output level of the laser diode did create a thermal gradient in the optical output, Figure 10.
Figure 10: 1 VP-P output
(Click to enlarge image)
The ideal value for CF is now 1.7 pF, so the standard value 1.8 pF will be used.
The location of the pole must be checked for stability concerns. The pole is located at 8.8 MHz. This makes the noise gain (1/F) equal to 10; which is the minimum stable gain of the LMV793. Although this matches the requirements of an ideal two-pole system for stability, the lab results showed extensive ringing. Thus, this design requires overcompensation for acceptable stability.
There are other poles and zeros internal to the op amp that are located close to the second pole, making this setup far from the ideal two pole system, therefore creating the need for overcompensation. The final value for CF is 2.7 pF. Figure 10 and Figure 11 show the response of the transimpedance amplifier with RF = 10 kΩ and CF = 2.7 pF.
Figure 11: 20 mVP-P output
(Click to enlarge image)
The higher-bandwidth configuration has rise and fall times of about 33 nsec. This gives an estimated bandwidth of 10.6 MHz. The pole, when using the 2.7 pF capacitor for CF, is located at 5.9 MHz.
This article has shown that the transimpedance amplifier is the same as a differentiator amplifier for stability calculations. The only difference between the transimpedance amplifier and a differentiator amplifier is the photodiode, which is a current source attached to the input node of the op amp. The photodiode has no affect on the stability calculations for a differentiator amplifier, once the diode capacitance is added to the total input capacitance of the differentiator amplifier.
Two different gain configurations were used in the lab, confirming a good match between the theory and the circuits tested in the lab. The equation for calculating CF does apply for any differentiator amplifier design. Even though the transfer function for a differentiator amplifier and a transimpedance amplifier is different, the feedback factor used for stability calculations is identical.
About the author
David Westerman is a staff applications engineer for National Semiconductor Corp., Santa Clara, CA. He has worked in several different applications groups at National Semiconductor for the last 16 years. For the last three years, he has been in the amplifiers applications group, covering low-power and precision op amps. He holds a bachelor's degree from the University of California (Berkeley) and a master's degree from the Georgia Institute of Technology (Atlanta), both in electrical engineering. In his spare time, he enjoys square dancing, hiking, and photography.
補充說明:
1)數位器的頻寬最好比要測量的訊號中的最高頻率高 3 ~ 5 倍.
2)Tr=2.2RC ==>RF=10k歐姆和CF=2.7皮法 ==> Tr=59.4ns
==> BW=5.892255Mz(已知RC常數值求BW)
3) Tr=33ns BW=0.35/33ns=10.606060MZ (由示波器測出Tr值求BW)
4)Tr=108奈秒的上升和下降時間，BW=0.35/108ns=3.240740Mhz頻寬約為3.2MHz.
EX: RF=100k歐姆和CF=0.64皮法 Tr=2.2RC = 140.8ns,
BW=0.35/140.8ns=2.485795Mhz==>頻寬約為2.4MHz.
Calculating the feedback factor for an op amp that is set up for current-to-voltage conversion may be a bit of a mystery to many engineers. By deriving the transfer function for a transimpedance amplifier and using a voltage amplifier op amp, the conversion will be easy to understand. Here we use the LMV793 op amp as an example for a transimpedance amplifier design. A basic transimpedance amplifier configuration is shown in Figure 1.
Figure 1: Basic transimpedance amplifier configuration
(Click to enlarge image)
The figure shows the complete transimpedance amplifier; only the power supply decoupling capacitors are not shown. In most cases, the selection of the photodiode will allow the designer to use the same supply for VBias and +V. Using split supplies keeps the inverting input of the op amp at virtual ground. To derive the feedback factor, it is necessary to examine the equivalent circuit of the photodiode, Figure 2.
Figure 2: Photodiode equivalent circuit
(Click to enlarge image)
The diode is an ideal diode in the equivalent circuit. Since the photodiode must be back-biased for proper operation, the ideal diode is not included in the feedback factor calculations. CJ is the capacitance that occurs from the depletion region of the diode and it is included in the photodiode specifications. IPH is the current that occurs from the photodiode action. The impedance of the current source is the series resistance of the photo diode, shown as RSH. The series resistance is at least 10 MΩ and typically much higher, usually over 100 MΩ.
Since RSH is such a high value, the current source will be considered ideal when deriving the feedback factor. The contact resistance and the resistance of the undepleted silicon are represented by RS. Normally RS is under 100 Ω and can be ignored. Even though RS is being ignored in this analysis, this resistance can exceed 1 kΩ in certain diodes, making it large enough to affect the transimpedance circuit feedback factor.
The op amp also has an input resistance. Figure 3 shows the circuit used in the feedback factor calculations. CCM is the input common mode capacitance of the op amp.
Figure 3: Equivalent AC circuit
(Click to enlarge image)
For the circuit analysis, +V is an AC ground. From this circuit it is easy to derive the transfer function of the transimpedance amplifier. The op amp is assumed to be ideal; therefore, the inverting input is at virtual ground. CJ and CCM have no affect on the transfer function of this circuit since both nodes are at ground. The transfer function is (Equation 1):
Or (Equation 2):
This equation is quite familiar to designers of transimpedance amplifiers. For low frequencies,
VOUT = -RF, and for very high frequencies VOUT = -1/sCF. This circuit inserts a pole at fP = ½πRFCF and this pole stabilizes the circuit. (The effect of this pole on the circuit is explained later.)
To derive the feedback factor, the circuit in Figure 3 is modified as shown in Figure 4.
Figure 4: Circuit for feedback-factor derivation
(Click to enlarge image)
As noted at the beginning of this article, the op amp is a voltage amplifier, not a current amplifier. The current source in Figure 4 has infinite impedance; therefore, it has no affect on the feedback factor! The two input capacitors have been combined by setting CIN = CJ + CCM. Therefore, with no effect from the current source, the circuit in Figure 4 is just a differentiator circuit with the input capacitance to the circuit (CIN) grounded. Even though a transimpedance amplifier is a current-to-voltage converter, the feedback factor is still based on a voltage-amplifier configuration, the differentiator circuit.
The feedback factor is simply what is fed back to the input from the output of the op amp. This is calculated by assuming that the node at the input to the op amp is not connected to the input of the op amp, then calculating the ratio of the input voltage to the output voltage, VIN/VOUT. Figure 5 shows the circuit used to calculate the feedback factor.
Figure 5: Feedback-factor derivation
(Click to enlarge image)
The parallel impedance of CF and RF is (Equation 3):
This is also used in the transfer function. Now the calculation becomes a voltage divider network giving the equation (Equation 4):
With some algebraic manipulations the equation for the feedback factor, F, becomes (Equation 5):
The feedback factor for a transimpedance amplifier is identical to the feedback factor of a differentiator circuit. The only difference is that CIN of a differentiator circuit is added to the inverting input of the op amp, but for a transimpedance amplifier CIN is just the sum of the capacitance of the photo diode and the input capacitance of the op amp. Note, that for low frequencies, F = 1.
Optimum value for CF
The noise gain for an op amp circuit is 1/F. In a differentiator circuit, CIN will insert a zero in 1/F, thus making a differentiator circuit inherently unstable. Since a transimpedance amplifier is a differentiator circuit, it is inherently unstable. CF must be added to make the circuit stable. CF limits the bandwidth of the circuit. To insure stability a designer can add a feedback capacitor that is rather large, thus overcompensating the circuit. However, finding the optimum value for CF will maximize the bandwidth of the circuit and avoid using an op amp with a wider bandwidth than necessary. Figure 6 shows the relationship of 1/F and the open loop gain for an optimum design.
Figure 6: Open loop gain, A, and noise gain, 1/F
(Click to enlarge image)
Figure 6 shows the 1/F curve with the optimum value for C. The pole of 1/F is located on the A curve. Therefore the optimum value for CF is where A=1/F, or A*F=1. Without CF there would be only a zero for the 1/F curve, giving close to a 180° phase shift at the intersection of A and 1/F. Adding the pole fP gives a lead-lag compensation with a phase shift of only 135° at the intersection of the two curves; giving the transimpedance amplifier a phase margin of 45°. If using a decompensated amplifier the intersection of 1/F and A must occur before the second pole of the op amp.
From the equation for F the pole for 1/F is (Equation 6):
And the zero for 1/F is (Equation 7):
The addition of a zero at fZ changes the slope of 1/F from 0 dB to +20 dB. For stability the slope of 1/F must be changed back to 0 dB and this is done by the pole created by the addition of CF.
Figure 6 also shows the overcompensated case, where the value of CF is larger than its optimum value. By making CF larger, the pole and zero for 1/F will occur at lower frequencies. In addition, CIN becomes a smaller portion of the denominator for the value of the zero; therefore, the pole and zero both drop in frequency and come closer together as the value of CF is increased for over compensation. Overcompensation may be necessary when using decompensated op amps and the intersection of 1/F and A is located near the second pole of A.
The equation for A is simply the gain bandwidth product which is (Equation 8):
In this equation, fGBW is the frequency where the open loop gain is 0 dB (1 V/V) for unity gain stable op amps. For decompensated parts, if the -20 dB slope was extended to the 0 dB line, the frequency at this point would be the value for fGBW. Using the equation A*F = 1, we now solve for CF (Equation 9):
Because A intersects 1/F at the pole of 1/F, we have the following two relationships (Equations 10a and 10b):
Using these two formulas to solve the AxF equation, the magnitude is found by taking the square root of the sum of the squares (rss) of the real and imaginary parts (s = jω). This results in (Equation 11):
(Click to enlarge image)
The above result simplifies into the equation to use to solve for CF (Equation 12):
(Click to enlarge image)
This formula is difficult to solve for CF. It is best to set up an Excel spreadsheet with all the values and use iteration solve for CF. In almost all applications, the assumption that CIN >> CF will apply.
With this assumption the above formula can be simplified to easily calculate the value for CF as follows (Equation 13):
This completes the theory and calculations for a transimpedance amplifier. The above formula is the one to use for calculating the optimum value for CF. If the value calculated for CF gives too much ringing in the circuit, then some overcompensation can be used to reduce the ringing. Overcompensation will, of course, reduce the bandwidth of the transimpedance amplifier.
Design Examples
National Semiconductor's LMV793 is the op amp used for the design example. This is a medium-speed part with a gain bandwidth product of 88 MHz. It is also a decompensated op amp, which gives the opportunity to investigate the operation of a decompensated op amp used in a transimpedance amplifier.
The input capacitance of the LMV793 is 15 pF. The PIN-HR040 photodiode from OSI Optoelectronics was selected for the detector. This part has bandwidth of about 300 MHz under the conditions being used in the circuit. This bandwidth prevents additional bandwidth limitation due to the photodiode. This diode also has a low capacitance, only 7 pF for the circuit operating conditions. Laser diodes were used for the light source. A laser diode has rise and fall times under 5 nsec, well under the rise and fall times of the op amp. A red laser diode was selected so that the beam of light can be seen, ensuring the safety of the setup.
A high gain will first be selected, setting RF to be 100 kΩ. Figure 7 shows the circuit used for the transimpedance amplifier.
Figure 7: Transimpedance amplifier
(Click to enlarge image)
When the photodiode is exposed to light, the current will flow into the inverting node of the op amp. The direction of the current flow is shown in the equivalent circuit in Figure 2. If the design goal is to have a minimum output voltage output of -20 mV, then the current flow from the photodiode must be 0.2 μA.
The expected current from the photodiode can be calculated from the expected optical power from the light source. For the PIN-HR040, the responsivity to light is given as about 0.38 A/W for the 630 nm wavelength of light from the laser diode being used. So for 0.2 μA of current the optical power on the photo diode is (Equation 14):
Solving for W (light power in Watts), we get (Equation 15):
If the light source has all of its energy on the active area of the photo diode, then this is the only calculation required. The active area of this photo diode is 0.77 mm2 and is a circular pattern.
Fiber optical cable may have an area larger than the photo diode and normally the optical power is measured in W/cm2. Expressing the area of the photo diode in cm2, the result is 7.7x10-3 cm2. For the same 0.2 μA output current from a light source measured in W/cm2, the necessary power would be (Equation 16):
In Figure 7, the bypass capacitors for the power supplies are not shown. Each supply has a 6.8 μF tantalum capacitor for bypassing the lower frequencies. In addition, each supply has a 0.1 μF ceramic capacitor for bypassing the high frequencies. The ceramic bypass capacitor must be located as close as possible to the power pins of the op amp.
For the photodiode CJ = 7 pF. For the op amp CCM = 15 pF. Therefore CIN = 22 pF. Using the formula for CF gives CF = 0.53 pF. Note that not only is CF much smaller than CIN, but with high-gain configurations the value reaches the limits of standard available values. For this experiment, a 0.5 pF capacitor was available, which measured 0.64 pF, so the transimpedance amplifier had slight overcompensation. The bandwidth of the transimpedance amplifier is derived from the RFCF time constant, or about 2.5 MHz. Figure 8 and Figure 9 show the test results of this circuit.
Figure 8: 1 VP-P output
(Click to enlarge image)
Figure 9: 20 mVP-P output
(Click to enlarge image)
From the rise and fall time being around 108 nsec, the estimated bandwidth is 3.2 MHz, which is close to the 2.5 MHz calculated from the RFCF time constant. The overshoot does show a phase margin greater than the 45° phase margin that would come from the optimum value of CF. Waveform averaging was used for the Figure 9 photograph so the actual response of the amplifier for small signals can easily be seen.
Next let's look at a transimpedance amplifier with a lower gain. Referring to Figure 7, if RF is now reduced to 10 kΩ, then the gain of the amplifier is one-tenth of the first example. This lower RF resistor gives a wider bandwidth. However, the light output of the laser diode must be ten times brighter for the same output level from the transimpedance amplifier. The higher output level of the laser diode did create a thermal gradient in the optical output, Figure 10.
Figure 10: 1 VP-P output
(Click to enlarge image)
The ideal value for CF is now 1.7 pF, so the standard value 1.8 pF will be used.
The location of the pole must be checked for stability concerns. The pole is located at 8.8 MHz. This makes the noise gain (1/F) equal to 10; which is the minimum stable gain of the LMV793. Although this matches the requirements of an ideal two-pole system for stability, the lab results showed extensive ringing. Thus, this design requires overcompensation for acceptable stability.
There are other poles and zeros internal to the op amp that are located close to the second pole, making this setup far from the ideal two pole system, therefore creating the need for overcompensation. The final value for CF is 2.7 pF. Figure 10 and Figure 11 show the response of the transimpedance amplifier with RF = 10 kΩ and CF = 2.7 pF.
Figure 11: 20 mVP-P output
(Click to enlarge image)
The higher-bandwidth configuration has rise and fall times of about 33 nsec. This gives an estimated bandwidth of 10.6 MHz. The pole, when using the 2.7 pF capacitor for CF, is located at 5.9 MHz.
This article has shown that the transimpedance amplifier is the same as a differentiator amplifier for stability calculations. The only difference between the transimpedance amplifier and a differentiator amplifier is the photodiode, which is a current source attached to the input node of the op amp. The photodiode has no affect on the stability calculations for a differentiator amplifier, once the diode capacitance is added to the total input capacitance of the differentiator amplifier.
Two different gain configurations were used in the lab, confirming a good match between the theory and the circuits tested in the lab. The equation for calculating CF does apply for any differentiator amplifier design. Even though the transfer function for a differentiator amplifier and a transimpedance amplifier is different, the feedback factor used for stability calculations is identical.
About the author
David Westerman is a staff applications engineer for National Semiconductor Corp., Santa Clara, CA. He has worked in several different applications groups at National Semiconductor for the last 16 years. For the last three years, he has been in the amplifiers applications group, covering low-power and precision op amps. He holds a bachelor's degree from the University of California (Berkeley) and a master's degree from the Georgia Institute of Technology (Atlanta), both in electrical engineering. In his spare time, he enjoys square dancing, hiking, and photography.
補充說明:
1)數位器的頻寬最好比要測量的訊號中的最高頻率高 3 ~ 5 倍.
2)Tr=2.2RC ==>RF=10k歐姆和CF=2.7皮法 ==> Tr=59.4ns
==> BW=5.892255Mz(已知RC常數值求BW)
3) Tr=33ns BW=0.35/33ns=10.606060MZ (由示波器測出Tr值求BW)
4)Tr=108奈秒的上升和下降時間，BW=0.35/108ns=3.240740Mhz頻寬約為3.2MHz.
EX: RF=100k歐姆和CF=0.64皮法 Tr=2.2RC = 140.8ns,
BW=0.35/140.8ns=2.485795Mhz==>頻寬約為2.4MHz.
5)另一個和頻寬有關的重要主題是上升時間 (Rise time)。輸入訊號的上升時間是指訊號從最大訊號振幅的 10% 轉換到 90% 的時間，而且與頻寬成反向相關，由以下公式呈現。此公式採用單極模型，R-C 限制輸入反應為基礎。
圖5
這表示100 MHz數位器的輸入途徑的上升時間是3.5 ns。我們建議數位器輸入途徑的上升時間為受測訊號上升時間的 1/3 到 1/5，才能在上升時間誤差最低的情況下測量訊號。測得之上升時間的理論值 (Trm) 可以利用數位器的上升時間 (Trd) 和輸入訊號的實際上升時間 (Trs) 計算而得。
圖6
舉例來說，在使用 100 MHz 高速數位器測量上升時間為 12 ns 的訊號時，測得的上升時間約為 12.5 ns=sqr(3.5^2+12^2)-->(由示波器測出Tr值求真實Tr值,即去掉示波器誤差值)。
沒有留言:
張貼留言
注意：只有此網誌的成員可以留言。