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Below is a circuit where on the left there is a floating single-ended source which directly couples as an input to a DC differential amplifier:

enter image description here

So the idea is to be able to simulate a real scenario for a single-ended floating source to a differential-ended inputs of the above amplifier. The source has single-ended outputs but has balanced output impedances as Rs1 and Rs2.

If the source were a bipolar output it would also have a ground which wouldn't be a problem.

But in this case imagine the source is like a battery, which has only two leads. It doesn't have a third lead to connect to common ground of the circuit.

As you see above in DC analysis Vin is increased from 0V to 100mV.

I'm having the following plots:

enter image description here

For the red source output differential voltage (Vin+ - Vin-), LTspice outputs not really a linear plot. There is some sort of nonlinearities.

And if we check Vin+ and Vin- wrt ground (blue and green plots at the bottom), we even see a much nonlinear crazy plot.

But the output Vout is very linear.

My questions:

1) Are those crazy plots for Vin- and Vin- common-mode voltages that the program creates since the source has no ground? But Vout is very free of problem. Is that because the common-mode noise rejected perfectly in simulation?

2) Will this way of coupling(without any use of ground for the source) work in real? I mean wiring exactly like in my schematics.

3-) How to implement this properly without converting the source to a bipolar source and get similar results as in real? I mean in simulation I still want to see the source as two terminal source without any ground but the simulation will work. How to do that?

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    \$\begingroup\$ You should try a transient analysis. These programs are far from perfect and everything that is remotely similar to floating often causes problems if not carefully handled. You could try shunting both with a gigohm to gnd \$\endgroup\$
    – PlasmaHH
    Jan 29, 2018 at 19:16
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    \$\begingroup\$ 1GOhm thing causes Vin+ and Vin- to around 358V in simulation. \$\endgroup\$
    – pnatk
    Jan 29, 2018 at 19:19
  • \$\begingroup\$ In real is the source wired this way like in my schematics i.e with no ground? No ground connection needed for the source? Because I saw in diff ended input inAmps for single ended inputs they use a bias resistor for the current to flow back to the source. \$\endgroup\$
    – pnatk
    Jan 29, 2018 at 19:22
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    \$\begingroup\$ @panicattack Now. Take a look at the bases (without considering the addition of your differential source -- remove that differential source from your mind, completely.) What is the voltage difference between them, left entirely unconnected? Remember, you don't have a differential source applied. These are floating bases. What's the difference? \$\endgroup\$
    – jonk
    Jan 29, 2018 at 19:48
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    \$\begingroup\$ In real (and in simulation), you must provide some path for a common mode input bias current if you want the input transistors to be forward active. \$\endgroup\$
    – The Photon
    Jan 29, 2018 at 19:54

2 Answers 2

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In your circuit, a floating source won't work, because the current flowing out of one of the "Vin" source's terminals must flow back into the other terminal on the Vin source. But for this type of amplifier, you must supply a positive current to both terminals. These currents must flow through the bases of the transistor to ground and be always present. The amplifier actually measures the changes to these base currents.

When perfectly balanced, each emitter resistor is seeing 0.5 ma (splitting your 1 ma negative current source). Each 12K resistor will have 6 volts on it, so your output will be centered around +6V and your base voltage has to be lower than 6 volts our your transistor's VCB is back biased. It obviously must also be positive with respect to the negative supply. A pull up resistor, either to ground or to the positive rail, is required for bias current, and the input can only "float" between +6V and the negative rail.

You also have a current source in series with your voltage source. You have to remember that in simulation world, current sources are infinitely compliant power supplies, and a current source is capable of providing its current at any voltage, no matter how negative or positive. For this reason, a current source in series with a voltage source doesn't make sense unless you set up compliance voltage limits in the model; otherwise the voltage source wouldn't do anything in theory. Best to draw in your constant current circuit.

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  • \$\begingroup\$ Thanks I also see sometimes in these situations especially I see these in inAmps they only use one resistor from the LOW terminal to ground. I dont know why. In this situation would that work as well? I mean sometimes they pull down from only one input. Would that create unbalance problem? \$\endgroup\$
    – pnatk
    Jan 29, 2018 at 21:02
  • \$\begingroup\$ Good point. As long as you have a DC bias and a return path to ground you are ok. Since your worst case is 1 milliamp through a 100 ohm emitter resistor, there is a potential of only 100 mv between the two emitters (and the two bases). So you could reference one side to ground and the other would swing positive and negative with respect to ground. This is great since, ground is more positive than your negative supply, so the current is always positive. \$\endgroup\$
    – John Birckhead
    Jan 29, 2018 at 21:29
  • \$\begingroup\$ @John Birckhead, Would you please clarify what you meant when saying, "You also have a current source in series with your voltage source"? What is the current source and what the voltage source? \$\endgroup\$
    – Circuit fantasist
    Sep 30, 2020 at 10:43
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    \$\begingroup\$ I1 and Vee. THe voltage across I1 will be whatever voltage is necessary to provide the constant current. \$\endgroup\$
    – John Birckhead
    Oct 1, 2020 at 13:25
  • \$\begingroup\$ @John Birckhead, I have asked you since, from my experience, I know that it is misleading for beginners to talk about two separate sources - the current source I1 and the voltage source Vee. Actually, there is a nonlinear (dynamic) current-stabilizing "resistor" I1 connected in series with a voltage source Vee. The combination of these two elements forms a true current source. I have explained it by adding an edit at the end of my answer. \$\endgroup\$
    – Circuit fantasist
    Oct 2, 2020 at 17:13
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Two resistors with relatively high resistance connected between the bases and ground will do the job. They will close the path of the input current IIN. See the picture below:

Differential amplifier with floating input source

Fig. 1. Differential amplifier with floating input source (credit: ResearchGate)

Finally, I would like to comment on some John Birckhead's thoughts about the emitter current "source" at the end of his answer:

You also have a current source in series with your voltage source.

and

... a current source in series with a voltage source doesn't make sense...

From my experience, I have known that it is misleading for beginners to talk about two separate sources - the current source I1 and the voltage source Vee. Actually, there is a nonlinear (dynamic) current-stabilizing "resistor" I1 connected in series with a voltage source Vee. The combination of these two elements forms a true current source. So the "tail" of the differential pair is just a "dynamic resistor".

We can say the same about the humble differential pair above with emitter resistor, where the combination of the emitter resistor RE and the negative voltage source V- forms an imperfect current source.

(In the considerations above, when saying "current source", I mean an element producing power... not only keeping up a constant current.)

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