Question About Output Voltage for Distribution Amplifier - Page 1
Forum section: electronics / beginners

author: Lovro7
author_rank: Contributor
timestamp: January 29, 2025, 08:17:40 pm
content: Hello,For a project in high school, I am building a distribution amplifier. I used the BG7TBL (only the distribution amplifier part) as a reference. I have a 10 MHz GPS OCXO reference, whose output is 5V TTL.According to the datasheet of the mentioned amplifier, the output power is 13 dBm. This means that the voltage across the load is 1V RMS. If I am not mistaken, the voltage is then 3 × 1V = 3V RMS or 8.5Vpp.For the frequency counter reference, I need a 5V square wave with a 2.5V offset.I was considering using two MOSFETs in a totem-pole configuration for each output. The output would be coupled through a capacitor to remove the DC component and have an impedance of 49Ω at 10 MHz. Additionally, I would place an inductor in series to correct the phase shift.What should the output voltage look like?Should I create an output with an internal impedance of 50Ω and a voltage of 8.5Vpp?Thank you for your help.Best regards,Lovro

author: TimFox
author_rank: Super Contributor
timestamp: January 29, 2025, 08:49:45 pm
content: 1 V rms = 2.8 V pk-pk.

author: CaptDon
author_rank: Super Contributor
timestamp: January 29, 2025, 08:53:55 pm
content: Lovro, where on earth did your math come from? 1.41 X RMS = PK 2.82 X RMS = PK to PK

author: Lovro7
author_rank: Contributor
timestamp: January 29, 2025, 09:05:24 pm
content: Sorry, I misspoke. The voltage across the load is 1V RMS or 1.42Vpp. The load impedance is 50Ω. The cable impedance is 50Ω. The source impedance (distribution amplifier) is 50Ω.So, if I consider the amplifier as an ideal voltage source with a series resistor, the voltage at the source will be 3 × V_load = 3V RMS. From this, I calculated Vpp = Vrms × 2√2.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 29, 2025, 09:07:10 pm
content: Quote from: Lovro7 on January 29, 2025, 08:17:40 pmCorrect that 13dBm is 1 v RMS into 50 Ohms. That is 2.828 v p-p.The usual way of making a DA would be an amplifier with zero (or very close to it) output impedance and 2 v RMS output swing. Then each of the parallel outputs goes through a series 50 Ohm resistor. That both gives the output impedance seen by that load as 50 Ohm and it reduces the level across the load to 1 v RMS.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 29, 2025, 09:08:16 pm
content: Quote from: Lovro7 on January 29, 2025, 09:05:24 pmWhere do you keep getting "3" from?

author: Lovro7
author_rank: Contributor
timestamp: January 29, 2025, 09:14:32 pm
content: Quote from: wasedadoc on January 29, 2025, 09:08:16 pmI used 3 because, in my first post, there were three 50Ω resistors connected in series (the source impedance, cable impedance, and load impedance). Since all resistances are equal, the voltage across each is also equal. That’s why I can multiply by the number of resistors.I know that the correct approach would be to calculate the current and multiply it by the impedance between GND and the point of interest. However, I wasn’t thinking in the way you described in your previous message.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 29, 2025, 09:26:05 pm
content: The cable impedance does not cause voltage loss in the manner you think. For a decent cable of length within a room you can ignore cable loss.

author: Lovro7
author_rank: Contributor
timestamp: January 29, 2025, 09:29:20 pm
content: Quote from: wasedadoc on January 29, 2025, 09:26:05 pmThank you, now I understand. So I can think of the cable as another 50Ω load. The resistance between the cable’s terminals is negligible in most cases.

author: wasedadoc
author_rank: Super Contributor
timestamp: January 29, 2025, 09:35:06 pm
content: Quote from: Lovro7 on January 29, 2025, 09:29:20 pmNo, the cable is not another 50 Ohm load. If there is a 50 Ohm load at one end of it, the total impedance looking into the other end will be close enough to 50 Ohms. The zero impedance output of the amplifier sees the 50 Ohm resistor there plus the 50 Ohm load at the other end of the cable.

author: CaptDon
author_rank: Super Contributor
timestamp: January 30, 2025, 02:53:22 pm
content: When a signal generator states 1 volt RMS @ 50 ohms it means the generator will produce 1 volt RMS WHEN CONNECTED to a 50 ohm load! Generally speaking, if the load is disconnected the output voltage will double since the generator source impedance is 50 ohms. Measure your cable conductor to conductor, does it measure 50 ohms, of course not, it measures open circuit. Why then do you assume it will suck up energy as if it were a 50 ohm resistor? The voltage at the load will be equal to the source voltage with the cable being basically lossless from a practical standpoint. If you disconnect the load the signal source voltage would apparently double IF the load is 50 ohms AND the generator or amplifier source impedance is truly 50 ohms. Making any sense now? Seems you have missed some basic circuit concept information.

author: SteveThackery
author_rank: Super Contributor
timestamp: January 30, 2025, 06:24:28 pm
content: Quote from: wasedadoc on January 29, 2025, 09:35:06 pmNewbie question here. What does it mean when a cable is called "50 ohm cable"? I believe it is saying the cable has a 50 ohm "characteristic impedance" ? But I don't know what that means in practical terms. As said earlier, if you put a multimeter between the conductors it doesn't read 50 ohms, it reads open circuit. So what does "50 ohm" cable actually mean? Are there circumstances where it does look like a 50 ohm load?

author: TimFox
author_rank: Super Contributor
timestamp: January 30, 2025, 08:12:42 pm
content: Quote from: SteveThackery on January 30, 2025, 06:24:28 pmSimple answer: the characteristic impedance of a coaxial cable is often 50 ohms. It is happy when the ratio of voltage across the cable divided by current through the cable equals that impedance.If you take a long length (compared to thepulse width) of 50cable and terminate the farend with a 50resistor, then drive it with a narrow positive pulse, there will be no reflection from the load back into the cable, and the pulse will be absorbed by the load.If the termination is > 50, some of the pulse is absorbed, but some is reflected back as a positive pulse.If the termination is < 50, a negative (inverted) pulse is reflected back.With a good 50termination, the signal across the load will be an accurate duplicate of the original pulse.With sine waves, a good termination efficiently transfers power to the load. A bad termination results in a “standing wave” along the cable. Hence, the term VSWR (voltage standing wave ratio) to evaluate the impedance match. That is a good topic to look up in the transmission line literature.For an ideal (no loss) cable,Zo= (L/C)1/2, whereLandCare the inductance and capacitance per unit length of the cable. (Length units cancel: m, ft, or km all work.)

author: Lovro7
author_rank: Contributor
timestamp: January 30, 2025, 08:35:09 pm
content: Thank you for your response; I understand now.So, if I make the connection, the voltage of the ideal source in the distribution amplifier must be 2 Vrms.I also need to remove the DC component. I would achieve this with a capacitor whose impedance would be 49.89 Ω.Should I also add an inductor to correct the phase shift?How could I achieve fast enough switching?I was thinking of designing a circuit with MOSFETs in a totem-pole configuration.

author: TimFox
author_rank: Super Contributor
timestamp: January 30, 2025, 09:50:37 pm
content: To remove a DC component, you want a capacitor in series with the load whose reactance is much lower than the load resistance at the lowest frequency of interest.You can terminate the cable with 50to ground and follow that with a reasonable capacitor driving a resistance much higher than 50.The result, having no DC component, will go positive and negative with respect to 0 V: is that what you want?What is the DC component of the distribution amplifier's output?

author: srb1954
author_rank: Super Contributor
timestamp: January 30, 2025, 09:53:31 pm
content: Quote from: Lovro7 on January 30, 2025, 08:35:09 pmNo. You need a capacitor whose reactance at the operating frequency is much lower than 50.To have a series capacitor with 50reactance at the operating frequency would introduce a high-pass filter effect with a 3dB loss of the fundamental component of the square wave signal. The higher harmonics are less attenuated and the resultant signal would appear to be very peaky and have a sharp rising edge but a downward sloping top.QuoteNot necessary if you have a large enough series capacitor as this will introduce minimal phase shift.Trying to cancel out any phase shift by resonating the capacitor with a series inductor would likely cause more unpredictable phase variations. The phase of a resonant circuit changes very rapidly as the input frequency passes through the LC resonant frequency so any slight deviations in the resonant frequency from the operating frequency will cause wild phase deviations. This deviation of the resonant frequency is inevitable if component tolerances and drift are taken into account.

author: CaptDon
author_rank: Super Contributor
timestamp: January 30, 2025, 10:51:35 pm
content: To O.P., You seem to be very lacking in basic circuit theory and concepts. I find it strange you are on to the building phase without a solid foundation in fundamentals. An obvious observation is that a 'coupling capacitor' in this case should appear as nearly a short circuit at the frequency in question. At very high frequencies a smaller capacitance will be adequate, at audio frequencies a much larger capacitor is needed. Fouled logic and wrong assumptions lead you to believe the capacitor should have a reactance of 50 ohms. Proper understanding of circuits including calculating inductive and capacitve reactance would have helped along with knowing why the capacitor in this case should have a reactance far lower than 50 ohms. You really need to spend some study time back at the beginning with D.C. and then A.C. circuit theory and concepts.

author: Lovro7
author_rank: Contributor
timestamp: January 31, 2025, 03:32:51 pm
content: Quote from: CaptDon on January 30, 2025, 10:51:35 pmI understand your concern regarding basic circuit theory. Most of my knowledge comes from hands-on experience with projects, where I’ve applied these concepts in practice. One of my more recent projects involved controlling a MIG/MAG welding machine using an ATmega8, where I had to consider both inductive and capacitive effects in the circuit.Regarding capacitive reactance, I agree that it should be very low in this case to allow proper coupling. I had initially considered a different approach, but based on my calculations, I would use a 22µF ceramic capacitor, as I already have some available.Using the formula:Xc = sqrt(2*pi f*C) =0.0007ohmI calculated phase shift using this formula:\theta = arcTan(Xc/R) = 0.000012oThis confirms that the capacitor will have negligible impact on phase and impedance at the given frequency. I appreciate your input and will double-check my assumptions to ensure the best possible design.I have beginner questions because I’ve recently become interested in RF topics. In school, we are currently covering RCL circuits, and I don’t really have any other way to get explanations besides asking on the forum.

author: TimFox
author_rank: Super Contributor
timestamp: January 31, 2025, 03:44:56 pm
content: Wheredid you get that formula forXc?The correct equation isXc= 1 / (2 pif C)which is very different.

author: Lovro7
author_rank: Contributor
timestamp: January 31, 2025, 03:49:41 pm
content: Story, type error.I was editing wrong formula.

author: TimFox
author_rank: Super Contributor
timestamp: January 31, 2025, 04:15:31 pm
content: You say that you are studying RLC circuits in school.Do you have a textbook?Printed books are better than the internet for looking up basic questions.

author: CaptDon
author_rank: Super Contributor
timestamp: January 31, 2025, 05:31:21 pm
content: I was very lucky to have attended a Vo-Tech High School. 10th through 12th grade. 2 weeks in academic classes then 2 weeks in the technical trade classes throughout the school year. I was in the Electronics 'shop class'. We had a phenomenal shop teacher, Ex Air Force electronics technician. I still have the 10th grade workbook. That class is where we learned to use a slide rule to do impedance and reactance calculations. I still own and sometimes use my 4-color Post 'Versalog' slip-stick. We had very in depth theory study and then built and measured practical circuits under power which also taught multimeter and oscilloscope usage techniques. We had the 'Lab Volt' patchboard learning fixtures. We had the radio and television fixtures where almost every component could be unplugged and substituted with a different component for practical troubleshooting. During our 12th year there was a co-op program where students could actually work for pay out in real industry during the 2 week shop period. Employers submitted a student evaluation report to the school at the end of each individual week. Any failure to show up for work or other unsatisfactory reports would revoke the co-op privliges. I am still grateful for my 10th grade technical teacher!!

author: Lovro7
author_rank: Contributor
timestamp: January 31, 2025, 11:04:34 pm
content: Quote from: TimFox on January 31, 2025, 04:15:31 pmI have a textbook but it doesn’t really help in this project.

author: Lovro7
author_rank: Contributor
timestamp: January 31, 2025, 11:08:18 pm
content: Quote from: CaptDon on January 31, 2025, 05:31:21 pmWe had something similar. For four weeks in the 2nd year of high school (ages 15/16), you work in a company. I worked in the maintenance department of a well-known Slovenian company that produces household appliances and kitchen equipment. About 90% of the work was mechanical, so I didn’t gain much electronics knowledge. However, I learned a lot about repairing heavy machinery (presses, cranes, etc.).