dBm and impedance question - Page 1
Forum section: electronics / beginners

author: Chris Wilson
author_rank: Super Contributor
timestamp: November 28, 2014, 02:36:17 pm
content: I have a DDS module giving a -12dBm output at 272kHz into a 50 ohm load. If I feed the output of the DDS into a HEF4001 NOR gate CMOS what impedance will the DDS be likely to see and what voltage level on the input of the CMOS is likely to be seen? I calculate -12dBm into 50 ohms as being about0.06V Thanks.http://www.nxp.com/documents/data_sheet/HEF4001B.pdfshows the specs of the device.

author: w2aew
author_rank: Super Contributor
timestamp: November 28, 2014, 02:49:15 pm
content: Quote from: Chris Wilson on November 28, 2014, 02:36:17 pm-12dBm is about 56mVrms, or about 160mVpp. At 272kHz, the input impedance of the HEF4001B is going to be very high - effectively an open circuit. If the output impedance of the DDS is 50 ohms, then you can expect the amplitude to double when loaded with a high impedance. So, you should expect about 320mVpp. Looks like you'll likely need a gain stage between the DDS and the HEF4001B.

author: G0HZU
author_rank: Super Contributor
timestamp: November 28, 2014, 08:52:56 pm
content: I guess you could use a comparator to square it up or if you use the UB (unbuffered) version of the 4001 you can use one or two of the unused NOR gates as amplifiers if you use resistive feedback to bias the first stage into the linear region. From memory, you can tie/bridge both logic inputs of the NOR gate together and then feed a (270K???) resistor from the output of the gate to the bridged input to bias it as an amplifier. Then feed the output of this gate to one input of another gate with the other input tied to ground (to make a basic logic inverter/squarer in the second NOR gate). This should give a fairly good square(ish) signal.Note: you can also tie one input to ground instead of tying both inputs together on the first stage. It should still work like this and might even square up better. See the crude schematic below. I'd expect the first stage to produce about 0.6V (low) to 4V (high) logic swing with a fairly rounded shape and the second stage will square it up. It's not a high performance circuit but it is probably the cheapest if you have a 4001UB version of this chip to handI haven't done this for years though. The first time I did this was when making a very basic shortwave receiver when I was a student and I used the 4001UB NOR gates as a crude RF amplifier.

author: Chris Wilson
author_rank: Super Contributor
timestamp: December 02, 2014, 02:37:56 pm
content: Thanks for the replies Alan and Jeremy. I am still assimilating Jeremy's reply, it is no doubt technically brilliant, but I need to know a lot more about CMOS stuff to fully appreciate the quality of its contentI feel I will be back with more questions!

author: G0HZU
author_rank: Super Contributor
timestamp: December 04, 2014, 10:28:00 pm
content: I wouldn't say my circuit is technically brilliant...It is probably the cheapest and simplest solution for a 272kHz signal if you already have an unbuffered 4001 chip but it is only going to work well below about 1MHz. Probably <500kHz. So if you wanted something to work above 1MHz then this circuit would be no good. However, you could use a similar technique with an unbuffered gate from a faster logic family or go for a discrete squaring circuit or a fast comparator.You can see a similar analogue amplifier circuit in the datasheet below. It's the very last circuit in the datasheet. For small signals this will be a linear RF amplifier circuit. In your case you will be driving it towards a square wave with -12dBm drive at 272kHz. Note this circuit uses the unbuffered version of the chip i.e. the 4001UB.http://www.harrisonelectronics.co.uk/datasheets/hef4001.pdf