[lunar7 (OP)]

I did this circuit question, but would like other opinions, in the hope it matches mine.

[alancalverd]

Depends on the diode characteristic.

[evan_au]

We try not to do people's homework for them.
- Post your working here and we will comment on it.

You could start with: Is the diode forward biased or not?

[alancalverd]

Ge, Si or PiN? Or a Zener?

[Bored chemist]

Or a Zener?

The symbol isn't  a Zener.

[alancalverd]

It's certainly forward-biassed, assuming the battery symbol is conventional, but the diode symbol is nonstandard and generic - could even be a LED or a photodiode  as far as the question is concerned

[Petrochemicals]

A good place to start

https://www.bbc.co.uk/bitesize/guides/zqxb4qt/revision/6#:~:text=When%20resistors%20are%20connected%20in%20parallel%2C%20the%20supply%20current%20is,same%20potential%20difference%20across%20them.

https://electronicsclub.info/ohmslaw.htm

[Bored chemist]

A good place to start

https://www.bbc.co.uk/bitesize/guides/zqxb4qt/revision/6#:~:text=When%20resistors%20are%20connected%20in%20parallel%2C%20the%20supply%20current%20is,same%20potential%20difference%20across%20them.

https://electronicsclub.info/ohmslaw.htm

Which of those "good places to start" tells you what the diode's characteristics are?
For example, which one tells you if it's germanium or silicon?
As Alan pointed out, it's impossible to answer the question without that data.

(Obviously, if it was a light emitting diode it would have little arrows pointing away from it, so it's not one of those. A photodiode would add even more complexity since the result would then depend on whether the circuit is in the dark but, if it was, the diode would have little arrows pointing away from it and it doesn't.
Without the arrows, we can rule out those cases and without the "wings" on the cathode, we know it's not a Schottky or Zener diode. We can't expect Alan to get everything right...)

[vhfpmr]

Firstly, the usual rule for exam questions is that if you aren’t given information, it’s because you aren’t expected to use it. Secondly, considering the trivial nature of the exercise, and the non-triviality of solving it analytically with a diode characteristic, I’d consider that to be a hint, too. Thirdly, my experience of a career in electronics is that you rarely if ever get all the information you want, but the boss won’t be very impressed if you just throw your hands in the air and tell him it’s all hopeless, and there’s no point in doing anything. Generally, you’re expected to do what’s reasonable and defensible with the information you have, and to state what assumptions you have made, if any.

No manufacturer’s data sheet I’ve ever seen specifies the diode characteristic numerically as a polynomial, they will normally specify voltage at one or two currents and temperatures, and include a graph or two. Since the voltage across a diode is similar between types, and remains relatively independent of current, it is usual to assume the voltage is a fairly constant standard for most purposes. Indeed in many applications it would be considered bad design if the diode characteristic were critical.

Assuming a constant voltage drop across the diode leaves a simple exercise that can be solved by the application of Ohm’s and Kirchoff’s laws. Alternatively, a typical diode characteristic in graphical form can be used to solve the exercise by drawing the load line on the characteristic to find the operating point at the intersection of the two lines.

[lunar7 (OP)]

No it is not homework.
The diode is in forward bias.
In order for the diode to 'kick-in' it's needs about 0.6 V. So, the pd across the 4ohm resistor is 11.4 V. Then it should be easy.

[vhfpmr]

It is easy. Based on 0.6V for the diode, which is about typical for silicon, the current is 12/6+(12-0.6)/4 = 4.85A.

[lunar7 (OP)]

That's it.

[Bored chemist]

No manufacturer’s data sheet I’ve ever seen specifies the diode characteristic numerically as a polynomial,

Good.
Because it's much better to model it as an exponential.

Now, can you tell me where I can get a boring run-of -the-mill diode which will drop just 0.6 volts at 3 amps?

[Petrochemicals]

Firstly, the usual rule for exam questions is that if you aren’t given information, it’s because you aren’t expected to use it.

It's my experience if you are not given complete information in exams, you can use an assumption, providing you reason your assumption. Just take it as 0.1v drop.

[alancalverd]

(Obviously, if it was a light emitting diode it would have little arrows pointing away from it, so it's not one of those. A photodiode would add even more complexity since the result would then depend on whether the circuit is in the dark but, if it was, the diode would have little arrows pointing away from it and it doesn't.
Without the arrows, we can rule out those cases and without the "wings" on the cathode, we know it's not a Schottky or Zener diode. We can't expect Alan to get everything right...)

By no means obvious. If you want to  know the voltage drop across the 4 ohm resistor, it doesn't matter whether it is a photodiode, LED, Schottky, Zener, tunnel, avalanche, laser,  PiN, germanium, silicon, vacuum, copper oxide  or selenium  (remember them?) diode. What matters is its forward voltage in the 3 amp region. There's no need to add arrows and tails to a generic diode-ish symbol in this schematic, because you could use any of the above to ensure the current only goes one way through the resistor, but you must state the diode type to do the calculation. It might  be a stacked copper oxide rectifier with a very bizarre forward performance or even a corroded terminal in a car.

Looks like a fine example of an unanswerable question set by an ignorant examiner.

[Bored chemist]

because you could use any of the above to ensure the current only goes one way through the resistor,

I think you will find that the battery does that.

[lunar7 (OP)]

Firstly, the usual rule for exam questions is that if you aren’t given information, it’s because you aren’t expected to use it.

It's my experience if you are not given complete information in exams, you can use an assumption, providing you reason your assumption. Just take it as 0.1v drop.

Not sure what is implied here, but the 0.6 V I quoted is from a course specification, so if it were an exam question then a student would be expected to know the starting pd for a diode.

[evan_au]

the 0.6 V I quoted is from a course specification

Silicon diodes typically have a forward voltage drop of around 0.7V (but 0.6V would be seen at low currents).
- The older Germanium diodes have a lower forward voltage drop, as little as 0.3V, due to the smaller semiconductor band-gap (but they also have higher leakage and lower breakdown voltages than silicon).
- Real diodes have internal resistance, which increases the voltage at high currents (and takes it away from a pure exponential)
- As vhfpmr said, given the manufacturer's data sheet, you can solve it graphically, as shown here:
https://en.wikipedia.org/wiki/Diode_modelling#Graphical_solution

can you tell me where I can get a boring run-of -the-mill diode which will drop just 0.6 volts at 3 amps?

In modern high-voltage power electronics, they don't tend to use boring run-of-the-mill PN diodes, as the forward voltage drop increases power dissipation...

Active rectification is a technique for improving the efficiency of rectification by replacing diodes with actively controlled switches such as transistors, usually power MOSFETs or power BJTs. Whereas normal semiconductor diodes have a roughly fixed voltage drop of around 0.5-1 volts, active rectifiers behave as resistances, and can have arbitrarily low voltage drop.

See: https://en.wikipedia.org/wiki/Active_rectification
One of the devices graphed there has a voltage drop of 0.1V at 10 Amps

[Bored chemist]

In modern high-voltage power electronics, they don't tend to use boring run-of-the-mill PN diodes, as the forward voltage drop increases power dissipation..

I know.
But do you really think those active rectifiers are on the syllabus?

[evan_au]

The most spectacular form of diode must be the mercury-arc rectifier.
- These are like a giant, glowing octopus (or light globe), 1 or 2 meters high, furiously radiating ultraviolet light.
- They were used for many years to convert the city high-voltage 3-phase AC voltage supply into DC for the electric trains.
See: https://en.wikipedia.org/wiki/Mercury-arc_valve

PS: A mercury-arc diode would appear open-circuit if fed from a 12V DC battery...