Error the circuit is not grounded simulation requires at least one ground

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A simulation will not run if it has no ground.

When you probe voltages in a simulation, all voltages are measured with respect to ground. Therefore before it can show you any meaningful voltage values, the simulator needs to know were that ground is in your circuit.

So, when you create an active simulation schematic (i.e. a schematic that you wish to run in simulation) then there must be at least one point or node in the circuit that is at ground (0 or 0V).

To establish a ground node, you must place one of the two available ground symbols from the Wiring Tools pallete onto one of the wires in your circuit diagram.

Please see:

<https://easyeda.com/file_view_All-simulation-schematics-MUST-have-a-ground-01_y20YWvRPp.htm>

and:

<https://easyeda.com/file_view_All-simulation-schematics-MUST-have-a-ground-02_4WUSQNLJ5.htm>

in:

<https://easyeda.com/project_view_Spice-tutorials_AysKEWevN.htm>

What is not quite so clear is that every point in a simulation schematic must have a DC path to ground (0). So, for example, a circuit that has two capacitors in series (a favourite of elementary electrical engineering classes and a very tricky problem to solve using a simulator!) must have a DC path to ground from the common point between the two caps.

Similarly, a transformer coupled circuit where the primary side is driven from mains live and neutral and one side of the secondary circuit is connected to earth (which is probably where you would place the ground symbol in your circuit diagram), must have a DC path back to this ground from one or both ens of the transfromer primary pins.

That does not mean that you have to connect such nodes directly to ground. The necessary DC path to ground can be provided by a resistor. Depending on your application, the resistance can be anywhere between a milliOhm right the way up to 1G Ohms (1e9 Ohms).

An example of where very high value resistors would be added is in transformer coupled datacomms networks such a 100BaseTx or 1000BaseT Ethernet. Here, the connections between the two transceivers are floating and so have to have DC paths to ground added to keep the simulator happy.

Using resistances that are high in comparison to the network cable characteristic impedance and hence termination resistance is important in order not to introduce an impedance mismatch and so disrupt the signal integrity. In practice it is simplest therefore to add two resistors: one from each of the two signal wires to ground. This preserves the symmetry of the signalling on the wires and doubles the effective resistance between them.

Note that using resistors beyond these extremes is perfectly possible but can sometimes give simulation convergence problems if you have resistances or impedances at both ends of the extremes in the same circuit. Basically the simulator may run out of dynamic range trying to handle both the very small and very large currents that may be flowing in the two extreme parts of the circuit.

Note also that in ngspice (the simulation engine used by EasyEDA) you cannot have a resistor of zero: it will throw an error. This is basically because any voltage difference across a zero resistance (such as may occur normally in a circuit or even just as a consequence of numerical «noise» in the simulation calculations) will generate an infinite current, which will obviously crash out of the top end of the dynamic range.

See:

<https://easyeda.com/project_view_Transformers-and-coupled-inductors_LWewOI0ic.htm>

for some examples of this use of the ground return path resistor.

Don’t forget that all voltages probed in a schematic are with respect to ground. This is particularly important to remember when you want to probe signals that are floating, such as the transformer coupled examples discussed above. This is when B or E Sources can be used to probe two floating voltages and subtract them to simply generate the difference between them. Alternatively, to probe the difference between two voltages, V(a) and V(b), you can add a spice directive to your schematic to do the subtraction for you. Like this:

>probe V(a)-V(b)

or

>probe V(a,b)

A DC path to ground is often provided by the rest of the circuit but here are some cases to watch out for:

i) All current sources (independent I and dependent B and F sources) are ideal: they have an infinite DC resistance (and AC impedance) and so do not have a DC path through them. Connecting a current source across a capacitor with one side grounded will throw an error even if the current source is set to zero (the capacitor voltage would ramp to infinity if a non-zero current were set and that would throw a different error!). So a resistor must be connected across the current source, to provide the DC path to ground to the other side of the current source / capacitor. A similar problem arises if two current sources are connected in series even if the two currents are identical (if they’re not then the common node flies off to infinity again);

ii) Capacitors (unless using the more advanced options in ngspice) are ideal: they have no parasistic (leakage) DC resistance in parallel with them. This is why both ends of a capacitor must have a DC path to ground either through the external circuit or explicitly by adding a resistor;

iii) switches do not have an infinite open circuit resistance and so they always have their own internal DC path between their terminals. However, this can sometimes cause confusion if you are expecting the switch to actually have an infinite OFF resistance!

See this example for more information:

<https://easyeda.com/file_view_Effects-of-finite-switch-resistances_Ba4mEWewq.htm>

iv) Voltage sources (including independent V and dependent B and E voltage sources) have the opposite problem. They are ideal so they have zero resistance. The same is true for simple inductors. If you connect voltage sources directly in parallel then ngspice will throw an error even if the voltage sources are set to the same value (if they’re not then an infinite current would flow and that would then throw a different error). The same problem arises if you connect a voltage source directly in parallel with a simple, ideal, inductor because the voltage source tries to drive an infinite current through the inductor.

This problem often occurs when driving a transformer from a voltage source. Adding a small series resistance fixes this little gotcha (in practice there will always be a finite winding resistance anyway!).

A

1.
A capacitor is open-circuited.

Конденсатор
в схеме разомкнут.

2. A
current source is open-circuited.

Источник
тока в схеме разомкнут.

3. A
subcircuit cannot be created.

Субсхема
не может быть создана.

4. A
subcircuit may not contain instruments.

Подсхема
не может содержать измерительные
приборы.

5. A
voltage source or switch is short-circuited. In Electronics
Workbench, batteries and switches have no resistance. If you are
using batteries and/or switches in parallel, insert a 1-№
resistor in series with each one.

В схеме обнаружено короткое замыкание
источника напряжения или переключателя.

В EWB
источники напряжения и переключатели
не имеют сопротивления. Если источники
и/или переключатели нужно включить в
схеме параллельно, то включите
последовательно с каждым из них
сопротивление в 1 Ом.

6. All
changes to your circuit will be lost!

Все
изменения в схеме будут утеряны!

7. An
inductor is short-circuited.

Короткое
замыкание катушки индуктивности.

8. An
instrument cannot be cut; return it to the shelf.

Прибор
нельзя вырезать из схемы; просто
перетащите его на полку.

9. An
instrument cannot be deleted; return it to the shelf.

Прибор
нельзя удалять из схемы; просто
перетащите его на полку.

C

10.
Cannot simplify; the truth table has unknown value(s).

Минимизация невозможна. В таблице
истинности есть неопределенные
значения.

Поставьте
0,1илиXтам, где в столбцеOUTстоит знак?.

11. Check
your circuit configuration, there may be something wrong with the
construction of your circuit. If it includes a transformer or
controlled source, make sure both sides (primary and secondary)
are grounded. Otherwise, try adjusting the points per cycle and/or
the tolerance in the Analysis Options dialog box.

Проверьте
конфигурацию вашей схемы, в ней что-то
неправильно с точки зрения построения.
Если она включает трансформатор или
управляемый источник, убедитесь, что
обе стороны (первичная и вторичная)
заземлены. Если в схеме все в порядке,
попробуйте изменить количество точек
на период и/или погрешность в диалоговом
окне AnalysisOptions(Параметры
моделирования, см. раздел 5).).

D

12.
Delete selected components?

Удалить
выделенные компоненты?

13.
Dividing by a zero-valued matrix element.

Деление
на элемент матрицы, равный нулю.

E

14.Electronics
Workbench cannot reach a solution. The maximum number of
iterations has been exceeded. Try adjusting the points per cycle
and/or the the tolerance in the Analysis Options dialog box.

EWB не
может получить решение. Превышено
максимальное количество итераций.
Попробуйте изменить количество точек
на период и/или погрешность в диалоговом
окне Analysis Options(Параметры
моделирования, см. раздел 5).

15.
Electronics Workbench cannot reach a solution. The time step
required for convergence is too small. Try adjusting the points
per cycle and/or the the tolerance in the Analysis Options dialog
box.

EWB не
может получить решение. Шаг по времени,
требуемый для сходимости, слишком
мал. Попробуйте изменить количество
точек на период и/или погрешность в
диалоговом окне Analysis Options(Параметры
моделирования, см. раздел 5).

F

16. File <имя
файла> already
exists. Replace existing file?

Файл <имя файла> уже существует.
Заменить существующий файл?

H

17.
Holding current must be less than Turn-on current

Ток
удержания должен быть меньше тока
срабатывания.

I

18.
Instruments cannot be copied.

Приборы
не могут быть скопированы.

19.
Instruments cannot be rotated.

Приборы
нельзя поворачивать.

20.
Internal Error — Electronics Workbench has encountered a
mathematical problem. If the circuit includes an
ammeter/multimeter, set its resistance to 1 m№. If there is a
voltmeter/multimeter, set its resistance to 1 M№.

Внутренняя
Ошибка — EWB столкнулся с математической
проблемой. Если в схеме есть амперметр
(из мультиметра), то установите его
сопротивление на 1 мОм (m№). Если в схеме
есть вольтметр (из мультиметра), то
установите его сопротивление на 1 МОм
(M№).

L

21.
Leakage faults need 2 terminals selected.

Для
задания дефекта утечки нужно выделить
не менее 2 контактов.

N

22.
Not enough memory available to store more data.

Недостаточно
памяти для сохранения данных.

O

23. Open
faults need at least 1 terminal selected.

Для
задания дефекта обрыва нужен хотя бы
один выделенный контакт.

S

24.
Save changes made to Circuit <имя
файла схемы>?

Сохранить
изменения, сделанные в файле схемы
<имя файла>?

25. Short
faults need at least 2 terminals selected.

Для
задания дефекта короткого замыкания
нужно выделить не менее 2 контактов.

26. Stop
circuit simulation?

Остановить
моделирова­ние?

27.
Subcircuit may not contain a copy of itself.

Субсхема
не может содержать свою копию.

T

28.
Temporary file is full.

Рабочий
файл заполнен.

29.
The circuit uses a different version of model <имя
модели> than in library
<имя
библиотеки>.

Схема
использует версию модели <имя
модели> отличающуюся от версии,
имеющейся в библиотеке <имя
библиотеки>.

30.
The circuit uses model <имя
модели> from
an unknown library <имя
библиотеки>.

Схема
использует модель <имя модели>
из неизвестной библиотеки <имя
библиотеки>.

31.
The circuit uses an unknown model <имя
модели>
from the library <имя
библиотеки>.

Схема
использует неизвестную модель <имя
модели> из библиотеки <имя
библиотеки>

32. The
dimension of the matrix is zero.

Размерность
матрицы равна нулю.

33. The
generated circuit will not fit in the workspace.

Полученная
схема не уместится на монтажном столе.

34. The
new model needs a name.

Для
новой модели требуется задать имя.

35.
The subcircuit <имя>
already exists. Replace existing subcircuit?

Субсхема
<имя> уже существует. Заменить
существующую субсхему?

36. The
temporary data file has reached its maximum size.

Рабочий
файл данных достиг своего максимального
размера.

37. The
temporary file used for storing simulation results has reached its
maximum size.

Рабочий
файл, используемый для хранения
результатов моделирования, достиг
своего максимального размера.

38. This
subcircuit requires %d pins, but the maximum number of pins is %d.
If you choose to proceed, some of the pins will be lost.

Для
этой субсхемы нужно хх контактов, а
максимально допустимо — уу контактов.
Если будете продолжать, то некоторые
контакты будут потеряны.

39. To
use the Bode plotter you must have an AC source in the circuit.

Чтобы
применить построитель диаграмм Боде,
нужно иметь в схеме источник переменного
тока.

U

40.
Unknown library

Неизвестная
библиотека

41.
UnknownModel

Неизвестная
модель

W

42.
Warning: the terminal that you want to delete is attached to a
wire in some copies of this subcircuit. If you proceed, the
wire(s) will be removed.

Предупреждение:
контакты, которые вы хотите удалить,
присоединены к проводам в какой-то
копии этой субсхемы. Если вы продолжите,
провода будут удалены.

43.
Warning: This change will apply to the model parameters of ALL
digital devices.

Предупреждение:
эта замена будет применена ко всем
цифровым устройствам с этой моделью.

Y

44. Your
circuit exceeds the component limit.

В
схеме превышены ограничения на
количество элементов.

45. Your
circuit requires a reference point for simulation. Attach a ground
component to the circuit.

Схема
требует ссылочной точки для моделирования.
Подсоедините элемент земляв
схему.

I will try to answer your question in a strictly math sense..
You will understand why the poor chap called the simulator is unable to find a solution here.

Your Circuit is as follows

Lets call the upper terminal A and the lower one B as shown in figure.

Now we know for your voltage source
$ V_A — V_B = 1V $

We also know from the resistor side that

$ I = (V_A — V_B) / R $
which is
$ I= 1 $

Now I can give you infinite number of $ V_A $ and $ V_B $
which still satisfy $ V_A — V_B = 1 $

For example
$ V_A=5 $ & $ V_B=4 $
still gives $ V_A-V_B=1$
and this is a valid solution.

Can you see why the simulator cannot solve it ?
Bcz there is no unique solution to $ V_A $ or $ V_B $ here.

All the voltage differences and currents are still defined..
But absolute voltages are not..

For absolute voltage to be defined ( and for ur simulator to throw a solution ).
You need a reference.
That reference is usually chosen as ground.

When you define either $ V_A $ or $ V_B $
then a unique solution exists.

General practice being..
make $ V_B=0$

We see that then the Solution for $ V_A=1 $
You Can force $ V_B $ to any other voltage.

Think of a even simpler situation:

Lets say that there is 20 storied building ( lets say that each floor is 10ft in height).
and lets say you are standing on the 12th floor of the building.

If someone asks you at what height you are standing..

What would your answer be ?

120ft ? Are you sure ?

What if the building is on Mt.Everest ( which itself is some 29000ft from sea level ) ?

Although you can say that from the 0th floor to you.. the difference is 120ft.
Although you can say that from the 1st floor to you.. the difference is 110ft.
You cant define your absolute height unless you know from where you are measuring.

If you are building is on Mt.Everest and your reference is Sea level.
then the height at which you are standing is 29000ft + 120ft.

If however your reference is the 0th floor, the height is 120ft.

I hope you understand the difficulty the simulator is facing.

Simulate the below two Circuits and you will understand what I am speaking of..

1. Zeroth floor being the reference :
   

2. Sea level being the reference :

All the best !!