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This is a real-world application. My circuit 101 knowledge is not up to the task.
I cannot understand the purpose of the portion of the circuit highlighted in yellow. With the presence of diode D3 and D4, I can't see that R_10K ever gets utilized. The only scenario R_10K will have some current passing through it is when V@pointA < 0.7Vdc and V@pointB > -0.7Vdc, but that's an impossibility, you can't reconcile the currents flowing in R14 and R15. The circuit just seems impossible to operate. I must be missing something here.

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In retrospect, judging from all the answers to this question, the place I was stuck upon was the notion that: Once D3 and D4 are conducting, they will short circuit R_10K.

In a real-world situation, a forward-biased diode is not a short circuit.

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  • \$\begingroup\$ Due to the real world nature of this circuit, I am guessing that, even when a diode is conducting, it is not diverting all the current, tiny amount of current will be flowing thru R_10K. So the purpose of those two diodes are just to limit the output voltage from Pg to be within -0.7 to 0.7 V range. \$\endgroup\$
    – eliu
    Apr 27, 2020 at 21:30

4 Answers 4

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R15 and R15 provide enough current to bias D3 and D4 "on" so that each has roughly 0.7V across it. So the 10K pot has about 1.4V across it and 140uA of current through it.

DR14 will have about (15V - 0.7V)/1.5K or about 9.5mA through it. That current, minus the 140uA in the pot flows through the diode D3. Similar reasoning for the negative branch of the circuit.

So the pot can set a voltage of anywhere from ~ +0.7V to -0.7V depending on the wiper position.

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  • \$\begingroup\$ I guess I am stuck on the false notion that "once a diode conducts, it short circuit the rest of the circuit". That notion is for ideal circuit, so here, diode and a resistor can be in parallel and share currents amonst them, right? \$\endgroup\$
    – eliu
    Apr 27, 2020 at 21:32
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    \$\begingroup\$ @eliu that's not even for an ideal circuit. That's a very drastically reduced model of a diode as perfect rectifier. That's simply rarely how you model diodes. An ideal diode has perfect exponential current over voltage, if you ask me – but for other people, the ideal diode would be what you describe (but you, really, can almost never forget about the forward voltage of the diode). \$\endgroup\$ Apr 27, 2020 at 21:39
  • \$\begingroup\$ @MarcusMüller I have viewed diodes the wrong way my whole life. In all my text books, they are either blocking or short circuit. I think I was able to get away with the wrong notion, because diodes were used in that way most of the time. \$\endgroup\$
    – eliu
    Apr 27, 2020 at 21:43
  • \$\begingroup\$ @eliu: Perhaps you have been misled by the incorrect saying "Current always follows the path of least resistance". Beginners often seem to read that as "only follows...". In fact, current follows all possible paths. By Ohm's Law, more current flows in a low resistance path than in a higher resistance path. \$\endgroup\$ Apr 27, 2020 at 21:46
  • \$\begingroup\$ @PeterBennett it is really the "short circuit" part that threw me off. I should have known about the internal resistance of the silicon, just didn't click here until now. \$\endgroup\$
    – eliu
    Apr 27, 2020 at 21:49
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A is 0.7V, B is -0.7V, approximately. 10K pot has about 1.4V over it. Pot can select between -0.7 and 0.7V by turning the pot. 1k5 has slightly less than 10mA flowing.

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If you remove the diodes, there would be about 23 volts across the 10 K pot. However, with the diodes, D3 will be forward biased so will hold point A at about +0.7 volts. Similarly, D4 will hold point B at -0.7 volts.

Therefore, the 10K pot will have about 1.4 volts across it, and will pass 0.14 mA. The voltage on the pot wiper can vary between -0.7 volts and +0.7 volts. This circuit will give a fairly stable range of voltages from the pot wiper.

You could have much higher value resistors as R14 and R15, and still have this work.

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While John, Peter and Justme have answered your 'how' question exactly, you may still be wondering 'why'.

These diodes are being used as a cheap voltage regulator. If the diodes were not present, and if the resistors were adjusted to provide +/- 0.7 V by themselves, then if the rails changed to (say) +/- 10 V, the bias voltage would drop in the same ratio. Even worse, if just one rail moved to 10 V, then the bias would change from being centred around zero, to being offset well away from zero.

The 'diodes have a 0.7 V drop' model ...

The diodes clamp the pot voltages to +/- 0.7 V to allow the rails considerable freedom in how stable they have to be to get a stable adjustable bias voltage.

The 'diodes have an exponential voltage/current relation' model ...

The diodes reduce the sensitivity of the pot voltages to the rail voltages to allow the rails considerable freedom in how stable they have to be to get a stable adjustable bias voltage. Let's say we get a change of rail from 15 V to 10 V, and the current drops from 10 mA to 6 mA. At the average current of 8 mA, the diodes will have a dynamic resistance (that's dV/dI, not V/I) of about 3 ohms (approximately 25 mV / diode I, for silicon diodes), so the 5 V 33% rail change would give an expected voltage variation of about 12 mV or less than 2%.

As you can see, a change of 10 mV in the roughly 700 mV across them, is 'clamped at 0.7 V', for most purposes.

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  • \$\begingroup\$ thank you, since we are here: clamping +-15V all the way down to +-0.7V, although justified, but seems a bit extreme, and a waste of most 15V too. Besides, we are wasting all power driving most currents to the ground. Clamping down to +-5V seems more "comfortable". Could you provide a link to a more intelligent way of doing this, if you have time, thank you =) \$\endgroup\$
    – eliu
    Apr 28, 2020 at 14:33
  • \$\begingroup\$ @eliu Don't worry too much about 'wastefulness' in circuits. You have +/- 15 V because that's what your amplifiers need. You can limit your adjustment to +/- 0.7 V, because that's all you need (presumably?). Silicon diodes are dirt cheap, to go to +/- 5 V would be more expensive, with zeners or regulators, and then you'd need to pot down the adjustment range anyway. The wastefulness is running 10 mA down an adjustment chain, I'm sure 1 mA would be adequate, perhaps even less. \$\endgroup\$
    – Neil_UK
    Apr 28, 2020 at 14:44
  • \$\begingroup\$ That's very reassuring, thank you! \$\endgroup\$
    – eliu
    Apr 28, 2020 at 14:51

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