What is the voltage as shown in this circuit?

[powersys]

The circuit is open (switch is open). If the probes of the voltmeter is placed as shown in the figure below, what is the reading given by the voltmeter? 12V or less? Please advise.

 

[hermin]

since the diode is an active deivce, the voltmeter would register 0V, becuase the diode wont be in forward bias...any objections?

 

[barrybear]

I set your circuit up and measured the voltage at 11.7 volts . Diode was a in 4002. measured with a digital multimeater, an analoge meter would make a diferance due to the change in current through the diode.

Barrybear

 

[mutevaggil]

and i set up your circuit in a simulate program (proteus).

here is the result:

 

[powersys]

Different results obtained from simulation and test. My understanding is if the circuit is open, there will be no current flowing and therefore, there will be no voltage drop across diode and the reading should 12V. But like barrybear, when I carried out test, I found out the reading is less than 12V (e.g. 11.7V).

May be for an ideal voltmeter (no current drawn by voltmeter), the reading should be 12V as shown by the simulation result from mutevaggil. In practice, I think there is a finite current drawn by the voltmeter, and therefore there is a slight voltage drop across the diode. Pls correct me if I'm wrong.

Can someone please advise why? Which is one more correct?

 

[maxwelleb]

neglecting voltmeter: if it is a real diode then i think it should be 12V, because no current flows and the only point where there is 0A current in a real diode is when the diode voltage is 0V and this explains the simulation result above;
with the finite resistance of the voltmeter: the current should be very small, and the voltage would be little below 12V.

but if we take the simple 0.7V-switch-model of the diode, then: neglecting the voltmeter influence; there is no current; then we can replace the the diode with an open switch; the voltage on this can be anything but the current is 0A; after replacing you get a piece of "wire" (at the cathode side) which is isolated from the rest of the circuit; and you ask what is the potential on this wire; i think with circuit theory it can be anything higher than 11.3V (where the open switch model is valid). circuity theory only wants KCL KVL to be fullfilled;
when we only use KCL and KVL and dont think further about field theory, there is no reason the voltage can even be 100V; so the diode voltage drop would be -88V; switch is open; and KCL and KVL are ok then and everything is consistent;
but in a real setup i think you can solve this only with field theory; you have some battery and some potentials in the space and anywhere at some distance is a piece of wire; solve laplace equation and see what is the potential; here then the distance s become important;
11.7V above: i think the resistance of the multimeter is not high enough, becaus 0,3V on real diode means current flows;

 

[powersys]

neglecting voltmeter: if it is a real diode then i think it should be 12V, because no current flows and the only point where there is 0A current in a real diode is when the diode voltage is 0V and this explains the simulation result above;
with the finite resistance of the voltmeter: the current should be very small, and the voltage would be little below 12V.

but if we take the simple 0.7V-switch-model of the diode, then: neglecting the voltmeter influence; there is no current; then we can replace the the diode with an open switch; the voltage on this can be anything but the current is 0A; after replacing you get a piece of "wire" (at the cathode side) which is isolated from the rest of the circuit; and you ask what is the potential on this wire; i think with circuit theory it can be anything higher than 11.3V (where the open switch model is valid). circuity theory only wants KCL KVL to be fullfilled;
when we only use KCL and KVL and dont think further about field theory, there is no reason the voltage can even be 100V; so the diode voltage drop would be -88V; switch is open; and KCL and KVL are ok then and everything is consistent;
but in a real setup i think you can solve this only with field theory; you have some battery and some potentials in the space and anywhere at some distance is a piece of wire; solve laplace equation and see what is the potential; here then the distance s become important;
11.7V above: i think the resistance of the multimeter is not high enough, becaus 0,3V on real diode means current flows;

Let's refer to REAL setup (as shown in the figure in my first post). Assume that the voltmeter is an IDEAL voltmeter, can I say that the reading should be exactly 12V?

Now let's consider a REAL voltmeter. If I add a resistor with very large resistance in series with the voltmeter's probes, I think we might get 12V if the resistance is large enough, right?

 

[Sigma|Six]

Hi Powersys, I agree with Maxelleb's thought.
To make thing simple, you are measuring the potential of the 2 points, volt meter is always in high impedance when measuring voltages so no current occur. No current occur means Diode not possible to bias. Hence forger about the current flow. in actually setup, I am prettry sure you will get 12V if you measure with a standard calibrated DMM.

 

[Nimer]

hi

i think it is 12 V for rela circuit implementation

regards

 

[mutevaggil]

May be for an ideal voltmeter (no current drawn by voltmeter), the reading should be 12V as shown by the simulation result from mutevaggil. In practice, I think there is a finite current drawn by the voltmeter, and therefore there is a slight voltage drop across the diode. Pls correct me if I'm wrong.

i think you re right

in theory, voltage =12V
in practice, voltage<12V

 

[powersys]

Hi Powersys, I agree with Maxelleb's thought.
To make thing simple, you are measuring the potential of the 2 points, volt meter is always in high impedance when measuring voltages so no current occur. No current occur means Diode not possible to bias. Hence forger about the current flow. in actually setup, I am prettry sure you will get 12V if you measure with a standard calibrated DMM.

I tested the circuit and I got 11.7V (or less than 12V). I used Fluke 12B DMM. Probably I will use Agilent 34401A to measure it.

 

[artem]

Voltage drop accross diode wil be determined by current through diode which is function of voltmeter input impedancen and diode drop voltage/current characteristic and of course should be less than 12 volts.

 

[platonas]

Voltage drop accross diode wil be determined by current through diode which is function of voltmeter input impedancen and diode drop voltage/current characteristic and of course should be less than 12 volts.

artem is 100% true. The multimeter has usually an internal resistance of 10M (for digital one). I want to believe that circuit simulation programs would give better results if you connect a resistor of 10MΩ across the multimeter since they consider infinite input resistance for their instruments.

 

[v_c]

If you are measuring about 11.7V, then the current in the diode is roughly 1µA assuming the input impedance of the voltmeter is 10M?. So the quiescent values for the diode current and voltage are (1µA, 0.3V). Now the question is ... are these values consistent with the i-v curve of the diode that you are testing. In other words, are the quiescent values on the i-v curve on the datasheet. This might be hard to see since the current is so low but you might be able to estimate.

With an ideal voltmeter (infinite input impedance), the quiescent i-v values for the diode will be (0A, 0V).

Best regards,
v_c