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I am trying to use a Differential Op Amp to receive and transmit over a Differential Signalling Bus.

I am basing this on this excellent op amp tutorial.

The Bus I am using has has 2 lines, Low(v1) and High(v2) both either 2.5V or 0V. When v1 is at 2.5V, v2 should be 0V, when v2 is 0V v1 should be 2.5V.

To receive I am reading the output of the op-amp.

I need the output to be 5V (so I need the difference between v1 and v2 gained/amplified by 2x).

In what I've built, V- is the GND rail and V+ is 5V. R1=R3=33ohms R2=R4=15 ohms.

When I disconnect R4 from the V- rail, V2 seems to go high.

Can I have an NPN transistor switch R4 from V- to transmit?

Does this design really make sense or do I need multiple op-amps (1+ for receive, 1+ for transmitting)?

Should this design work for receiving?

Before building the rest of my receiver (the MCU to connect to Vout), I want to make sure that it'll work, but my multi-meter shows 3V or 4V (it varies each time I check), because the bus is running at 25khz which is probably faster than my auto-sensing multi-meter.

I am using a LM741 (NTE941M) op amp.

enter image description here

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  • \$\begingroup\$ Why not use something like a DS16F95A? \$\endgroup\$ Commented Sep 4, 2013 at 2:18
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    \$\begingroup\$ Aren't all opamps differential?? \$\endgroup\$ Commented Sep 4, 2013 at 5:55

2 Answers 2

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Differential Amplifiers are used to measure the difference between two inputs. Often they also amplify this difference. Like OpAmps (DA's are often build from OpAmps) DA's are linear circuits, and their speed is limited compared to digital circuits.

What you want is not the difference of the two voltages, but a simple answer: is one voltage higher, or the other? You don't mention speed, but you seem to be talking about digital transmission, so you might well require a speed that is beyond what can be achieved with OpAmps and/or DA's.

The basic circuit that you could use is a comparator (for instance LM311). A comparator checks which of its two inputs is higher, and outputs a high or low voltage accordingly. Note that a comparator can not be used to measure how much the inputs differ: that is the realm of DA's.

Differential transmission is so common that there are lot's of chips specifically designed for this purpose, both transmitter, receivers, and combinations of the two (transceivers). Maybe you can use an RS485 transceiver.

PS a 741 opamp should remain in the curiosities cabinet, along with vacuum tubes and selenium rectifiers. If you realy want to use OpAmps get something more recent that can at least work on a single 5V rail.

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  • \$\begingroup\$ Yup, and 555 timers belong next to the 741 opamps too. \$\endgroup\$ Commented Sep 4, 2013 at 12:09
  • \$\begingroup\$ I disagree about 555's: there are far better opamps now that 741's, but which chip should replace a 555? (okay, a C555) \$\endgroup\$ Commented Sep 4, 2013 at 15:26
  • \$\begingroup\$ It depends on the situation, but I recently used a PIC12F629. \$\endgroup\$ Commented Sep 4, 2013 at 17:20
  • \$\begingroup\$ I might do that myself too, but it requires access to programming equipment, and is FAR less tolerant in its power supply voltage. (And a PIC can't match the output current a 555 can deliver.) \$\endgroup\$ Commented Sep 4, 2013 at 17:22
  • \$\begingroup\$ How would I wire up an LM311 for my purpose? \$\endgroup\$
    – Matt
    Commented Sep 4, 2013 at 17:24
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Is this the crux of your question: -

How do I convert differential 2.5V logic to a single-ended 5V logic?

If it is then forget about the 741 (or derivatives) - these are not rail-to-rail output devices and they are pretty slow too. Ideally you need a reasonably fast R2R device for running from a single 5V supply. You also need to bias the mid-point to 2.5V because the average signal output level you require is 2.5V (0V and 5V). Here's a picture of the design you linked to: -

enter image description here

I've circled in red the problem areas. Firstly R4's ground connection. This would be OK if you were using input signals that went above and below ground equally and you have a negative supply voltage but you don't. You need to use an op-amp like an AD8605 - this is capable of doing what you appear to want.

Also, you don't need a gain of 2 - you need a gain of 1 because your input signal is 5V peak to peak and that's what your output requirement is.

Instead of using low values of resistors try 1k to 10k. R1 to R3 = (say) 4k7 and R4 becomes two 10ks; one down to ground as shown in the diagram and one pulling up to 5V - together they split the power rail and feed +Vin with 2.5V.

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  • \$\begingroup\$ What's a common rail-to-rail op amp? \$\endgroup\$
    – Matt
    Commented Sep 4, 2013 at 17:23
  • \$\begingroup\$ What does it mean to be rail-to-rail? \$\endgroup\$
    – Matt
    Commented Sep 4, 2013 at 17:24
  • \$\begingroup\$ Rail to Rail means it's output can swing to within a few millivolts of the power rails. It also means the input range includes the power rails. \$\endgroup\$
    – Andy aka
    Commented Sep 4, 2013 at 19:29

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