Variant on Gerry Sweeny's 10MHz Distribution Amp - Input Attenuation Help Needed - Page 1
Forum section: electronics / beginners

author: David D
author_rank: Contributor
timestamp: February 04, 2024, 08:59:31 pm
content: In my desire to obtain a frequency standard to reference my gear to, I purchased a Leo Bodnar GPSDO (single output version). I also came across Gerry Sweeney's video on converting an Extron video distribution amp and thought that looked like a fun and simple project to distribute the 10MHz GPSDO signal to my counter, scope, and signal generators, all which have 10MHz reference inputs. I found and purchased an Extron DA2 RGBHV which looked like a good candidate for the project.This particular Extron is a 1 x 2 RGBHV distribution amp with switchable AC/DC coupling. I am only using the RGB portion and have abandoned the HV part of the circuitry. I replaced the 75 ohm resistors with 50 ohm. I also took the last output on each of the two op amps and added the unused BNC outputs.Right now, I have a single input that feeds the the op amp with a 50 ohm terminating resistor from the input to ground. My question is, I would like to add a second input with a 20dB pad. The Leo Bodnar GPSDO output signal is a little hotter than I need, even at the lowest output strength (8 mA). Can I do this using the circuit as I have it drawn below? My thought is to remove the 50 ohm terminating resistor and replace it with the 20 dB PI attenuator circuit as shown.

author: fourfathom
author_rank: Super Contributor
timestamp: February 04, 2024, 09:20:26 pm
content: Sure, that will work well enough. The input resistance of the non-attenuated input port will be about 43 Ohms, but that shouldn't bother anything as long as you use reasonably short cables. The rise-time on the Bodnar output looks like about 10-15 ns (sorry, I don't have a fast scope here), at 10 MHz, maximum drive, 50 Ohm load, so this will be pretty forgiving. You could always adjust the 249 Ohm and right-hand 51 Ohm values to get a better compromise match, but this isn't critical.*** better answer below ***

author: MarkT
author_rank: Frequent Contributor
timestamp: February 04, 2024, 09:25:41 pm
content: A 20dB attenuator would be more like 56R/470R/56R, given the load is an opamp input, not another 50 ohm load.

author: fourfathom
author_rank: Super Contributor
timestamp: February 04, 2024, 09:37:14 pm
content: Quote from: MarkT on February 04, 2024, 09:25:41 pmYes, these values work very well (assuming high-Z amplifier inputs which should be the case). I should have thought harder before I replied above.

author: David D
author_rank: Contributor
timestamp: February 04, 2024, 09:42:38 pm
content: I apologize, but I didn't realize my schematic resolution was so poor. The resistors are actually61ohms and 249 ohms. I based these values on a 50 ohm circuit with a 20dB drop.I found the data sheet for the op amp used in this, and it is a Texas Instruments OPA3691. According to the data sheet, it has a non-inverting input impedance of 100K ohms / 2pF. This appears to be a relatively high input impedance to me. Does this change anything with respect to the attenuating resistor values?

author: fourfathom
author_rank: Super Contributor
timestamp: February 04, 2024, 10:06:49 pm
content: Quote from: David D on February 04, 2024, 09:42:38 pmYour values (61 / 249 / 61) are calculated for a 50 Ohm load. With the high-Z amplifier load, you will have an attenuation of 14 dB and an input impedance of 50.9 Ohms at either port.MerkT's values (56 / 470 / 56) give an unloaded attenuation of 19.4 dB and a port impedance of 50.6 Ohms.

author: David D
author_rank: Contributor
timestamp: February 04, 2024, 11:10:06 pm
content: How does one calculate for different input and output impedances? I just tried a calculator for a non-symmetrical 20dB Pi attenuator using 50 ohms for the input and 100K ohms for the output. The input shunt resistance R1 comes out as 49.23 ohms, the series resistance R2 is 11.07 kilohms, and the output shunt R3 comes in at -12477515.41 megohms. What am I doing wrong?

author: fourfathom
author_rank: Super Contributor
timestamp: February 04, 2024, 11:44:42 pm
content: Quote from: David D on February 04, 2024, 11:10:06 pmObviously that calculator isn't being very useful! All I did was check the attenuation and port resistance given the resistor values, using simple math. To arrive at the exact values I would to use a little algebra. I did a quick Google search and couldn't find an on-line calculator for this case. Here is a link with equations for an asymmetric-impedance Pi attenuator, but I haven't checked them:https://www.electronics-tutorials.ws/attenuators/pi-pad-attenuator.html. While it's nice to get the math exact, in your case I personally wouldn't try too hard -- the impedance and attenuation values aren't particularly critical.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: February 05, 2024, 04:22:21 pm
content: The input voltage on those op amps is typical CMOS, about 3.3 volts, as stated in the spec sheet. CMOS components are voltage rather than current devices. We do not have the complete schematic, there are other resistors around the op amps including input matching,I think the amp was made to take CMOS level signals and distribute them.Leo Bodnar output is CMOS level.I do not think you need any attenuation.I think this is just a CMOS Buffer, so CMOS volts in one input and CMOS volts out, multiple outputs.If you wish to add attenuation I would just put a higher value resistor after your 50 ohm input resistor.The input signal is not just going to one op amp but all of them.Even Leo is limited by ohms law, The input to the op amps will be very low milliamps.It was probably unnecessary to change the 75 ohm resistors, This would only matter if precise measurements were required, this is not the case here.

author: fourfathom
author_rank: Super Contributor
timestamp: February 05, 2024, 04:56:45 pm
content: Quote from: Wallace Gasiewicz on February 05, 2024, 04:22:21 pmI'm using one of these surplus video distribution amps in the stock 75 Ohm configuration. It works just fine distributing my 10 MHz lab reference (Bodnar, unattenuated, probably max output setting), to various signal generators, spectrum analyzers, transmitters, etc, using 50 Ohm coax and connectors. The longest cable run is probably about fifteen feet.And I agree that there's no reason that the output of a Bodnar GPSDO should need to be attenuated before feeding it to the distribution amp.

author: David D
author_rank: Contributor
timestamp: February 05, 2024, 06:39:24 pm
content: Quote from: fourfathom on February 04, 2024, 11:44:42 pmYou're right about the calculator not being very useful!Quote from: Wallace Gasiewicz on February 05, 2024, 04:22:21 pmI have searched the web for a schematic for this Extron amp and have found zilch, so my information is limited to what I can see and ring out on the board. Unfortunately, the board is a layered board, so not all of the traces are visible. I did just go back and check my measurements again and realized that I had measured the output of the Leo Bodnar unit without a 50 ohm termination. With the 50 ohm termination it reads about 1.3 volts P-P, so it looks ok now.Quote from: fourfathom on February 05, 2024, 04:56:45 pmThis has been mostly a learning exercise for me to see how to modify this Extron the "proper" way, even if it is "good enough" without any of the mods (75 to 50 ohm resistors, etc.). The truth is that with the input and output resistors changed to 50 ohms the Bodnar output looks the same or better on my scope after passing through the Extron, so it's probably fine.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: February 05, 2024, 07:59:52 pm
content: You could use one of these CD4050 Buffershttps://www.ti.com/lit/ds/symlink/cd4050b.pdf?ts=1707161223579(Or any number of differently configured CMOS logic chips i.e. OR gates, AND gates,)to get the same output.But then you would have to buy some sort of box and a bunch of SMA connectors and a board .....When I built my GSPDO, I used a TTL Logic Buffer chip ( 5 V ) for all the various outputs, same idea.