Application Note #49 RF Amplifier Output Voltage, Current, Power, and Impedance Relationship

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1 Application Note #49 RF Amplifier Output Voltage, Current, Power, and Impedance Relationship By: Jason Smith; Manager Applications Engineer and Pat Malloy; Sr. Applications Engineer How much output voltage, current and power can RF amplifiers provide? This question is often asked by novice test engineers as well as seasoned RF professionals. Depending on the application, there is often an underlying desire to maximize one of the three parameters; power, voltage or current. hile one would think that a simple application of Ohm s law is called for, this would only apply given ideal conditions, such as when an RF amplifier with a typical 50 Ω output resistance is driving a 50 Ω load. In this rare case where the load impedance perfectly matches the amplifier output impedance, the power delivered to the load is simply the rated power of the amplifier. There is absolutely no reflected power and thus, there is no need to limit or control the gain of the amplifier to protect it from excessive reflected power. Unfortunately, such ideal conditions rarely apply in actual real world applications. Real amplifiers are required to drive varying load impedances. The mismatch between these real loads and the amplifier s output impedance result in a percentage of the forward power being reflected back to the amplifier. In some cases, excessive reflected power can damage an amplifier and precautions that may affect forward power are required. Given these realities, how does one go about determining output voltage, current and power? Again Ohm s law comes to the rescue, but with the caveat that the actual power delivered to the load (net forward power after the application of any VSR protection less reflected power) must be determined before applying Ohm s law. This Application note will highlight some of the major RF amplifier characteristics that impact forward power as well as net power allowing the use of Ohm s law, even when conditions are far from ideal. Back to Basics: Ohm s Law Ohm s law states that the amount of current flowing between two points in an electrical circuit is directly proportional to the voltage impressed across the two points and inversely proportion to the resistance between the points. Thus, the equation I=E/R is the basic form of Ohm s law where the current I is in units of amperes (A), the Electro-motive Force (EMF) or difference of electrical potential E is in volts (V), and R is the circuit resistance given in ohms (Ω). Applying the standard equation relating electrical power to voltage and current (P=V A), cross multiplying and rearranging each of the variables results in the equations shown in the Ohm s law pie chart (see Fig 1) showing the various combinations of the four variables, I, V, Ω and. Let s use Ohm s pie chart to determine the output voltage, current, and power of a 50 Ω amplifier operating under ideal conditions. V A A A A V Volts Ohms Ω V V A Amps atts V A V V A Figure 1: Ohm s Law pie chart 1 of

6 .5 75 atts 50Ω Amplifier 75A400 Current Volts 1.3 Amps Figure 3 plots the voltage and current over the entire range of load impedance. The center point represents the voltage and current produced when the load impedance matches the amplifiers 50 Ω output impedance. Loads greater than 50 Ω are plotted to the right of the center point and loads less than 50 Ω appear to the left. The end points demonstrate the two possibilities of a worst case mismatch; an open where the output voltage is at a maximum with zero current, and a short where the current is maximum with zero voltage. The above graphs are based on the minimum rated output of the amplifier across its entire operating frequency range. There most likely will be spots within the frequency range where the output power will exceed the specified minimum rated output power. To avoid unexpected results, always request a copy of specific production test data before placing an amplifier in service. Example : High Power Solid-State amplifiers by necessity employ active VSR protection. Take, for example the 0000C. 80MHz 00MHz bandwidth 00 watts minimum RF output delivered into a 50 Ω output impedance Active protection kicks in to reduce the gain when reverse power is measured at 500 watts; this is a VSR of 6:1 when using the amplifier at rated power. This fold-back protection limits the reflected power to 500 watts maximum The center point of the graph occurs at the point where the load impedance is matched to the output impedance. Maximum power is delivered to the load only at this point Voltage (Volts) Figure 3: Current vs. Voltage 75A C Power Vs. Load Impedance Forward power Power (atts) 0 Ideal 00 watt amplifer without protection Net power or Power Delivered to Load Load Impedance (Ω) Figure 4: Power vs. Load Impedance 0000C 6 of

7 The 0000C is an example of one of AR s high power amplifiers that folds-back when reverse power reaches 50% of rated power. Even though the amplifier does fold-back, a considerable amount of power is still being delivered to the load. In many cases, other manufacturers of high power amplifiers would not be able to handle such conditions and forward power would either be shut-down completely or reduced drastically. In power critical applications, an impedance matching transformer similar to the one used in the AR 800A3A could be used to match the amplifier to the load. However, since matching transformers tend to be narrow band, this approach may prove impractical if the 0000C were to be operated over its entire frequency band. In this case, a series of narrow-band transformers could be switched in to the application as the frequency dictated or simpler yet, the user could opt for a higher power amplifier C Current (Amps) Ohms 3.6 Volts 4.5 Amps 3 The center point of the graph occures at the point where the load impedance is matched to the output impedance. Maxmium power is delivered to the load only at this point Voltage (Volts) Figure 5: Current vs. Voltage 0000C The above graph demonstrates that even though fold-back occurs at a VSR of approximately 6:1, significant output voltage and current are still delivered to the load. Example 3: Much has been said so far regarding the importance of impedance matching. The 800A3A is an example of a unique amplifier that provides the user with selectable output impedance to match a wide variety of applications. khz 3MHz bandwidth 800 atts minimum output power rating An internal user selectable impedance transformer provides 1.5, 5, 50, 0, 150, 00, or 400 Ω to facilitate a closer match to the load impedance Active protection kicks in when VSR exceeds 6:1 to reduce the gain This fold-back protection limits the reflected power to 400 watts maximum 7 of

8 00 800A3A Net Power Vs. Load Impedance for Different Impedance Transformer Settings Net Power (atts) 1.5 Ω 5 Ω 50 Ω 150 Ω 0 Ω 00 Ω 400 Ω Load Impedance (Ω) Figure 6: Power vs. Load Impedance 800A3A The internal impedance transformer of the 800A3A allows this amplifier to have output impedance that matches that characterized by a variety of applications. External transformers are available to extend the usefulness of the 800A3A to include even more applications. Current (Ampsrms) Ω 50 Ω 0 Ω 15 0 Ω 00Ω 400Ω 1. 5 Ω 800A3A Output for Different Impedance Transformer Settings 0 Volt s 8 Amps Vol t s 5.7 Amps 00 Volt s 4 Amps 8.8 Volt s.8 Amps The diamonds represent voltage and current available for each impedance setting when the load impedance and amplifier output impedance match Volt s.3 Amps 400 Volt s Amps Volt s 1.4 Amps Voltage (Volts rms ) Figure 7: Current vs. Voltage 800A3A Figure 7 clearly highlights the benefits of an amplifier with an internal impedance matching transformer that facilitates a better match with varying loads. The range of output voltage and current is considerably greater than what is provided by a standard 50 Ω amplifier. 8 of

10 Figure 9 plots the available output voltage and current from the 350AH1. The gray area is provided to indicate a more typical output profile. Individual amplifier characteristics will vary and are somewhat influenced by operating frequency and system losses. Summary The age old question of How much output voltage, current, and power can I expect from my amplifier? can in rare cases be answered by merely applying Ohm s law assuming the net power or power delivered to the load is simply the rated power output of the amplifier. In most cases, practical issues such as VSR and forward power concerns must be considered before applying Ohm s law. hile this application note has provided guidance in this matter, AR firmly believes that the best approach is to apply actual test data when calculating output parameters. If you are the least bit uncomfortable with this exercise, feel free to contact one of our Application Engineers. e would be more than happy to guide you through the process. of

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