diode in series and parallel working principle?

Thread Starter

TarikElec

Joined Oct 17, 2019
20
Hello,
I would never thought I could not understand a circuit made out of diode but here I am.
1-the first circuit is two diode in series, when i measured the voltage between the them I found it half the voltage of the source which is not logic for me!!! as there is no current flowing through the diode, so how can it be a voltage divider!!!
2-diode in parallel and opposite to each other, for my understand it should show an output of +/-0.7V, but I got a full wave of source signal!!!
can anyone clarify for me these two circuits?
 

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ericgibbs

Joined Jan 29, 2010
9,114
Hi TE,
All diodes have a reverse voltage leakage current, so the two in series, will have leakage current thru them, using a high impedance 'dvm' will show half Vcc.

Post your asc file and will explain the full wave problem for the parallel diodes.
E
 

kubeek

Joined Sep 20, 2005
5,650
second circuit, the voltage source is infinitely powerful. So there will be a sinewave at the output no matter what, check what current is flowing through the diodes, my guess is several amps.
 

Thread Starter

TarikElec

Joined Oct 17, 2019
20
Hi TE,
All diodes have a reverse voltage leakage current, so the two in series, will have leakage current thru them, using a high impedance 'dvm' will show half Vcc.

Post your asc file and will explain the full wave problem for the parallel diodes.
E
Hi Erric, Sorry what do you mean by ´dvm´ and this serial diode is connected to an op amp.
 

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Thread Starter

TarikElec

Joined Oct 17, 2019
20
second circuit, the voltage source is infinitely powerful. So there will be a sinewave at the output no matter what, check what current is flowing through the diodes, my guess is several amps.
Hi, I dont understand how so. if a diode is a short circuit in the positive sine, i should get 0.7V output
 

kubeek

Joined Sep 20, 2005
5,650
Hi, I dont understand how so. if a diode is a short circuit in the positive sine, i should get 0.7V output
First of all this is ideal voltage source, so if the voltage source says there is that voltage across it, then that voltage is there no matter how much current you draw from it. Second thing, the 0.7V figure is just the typical voltage at typical current. At a few microamps of current there will be 0.2-0.3V, at very large current compared to the capability of the diode there will be higher voltage like 1-2V. Look at the IV curve of a diode, the voltage doesn´t just stop at 0.7V with rising current.

The 1N4148 is a 100mA diode. If you put 2V across it, there will be about 2A flowing through the diode. I also checked some 1A rectifier, and the current at 2V was about 45A. Or if you look at it the other way, if you have 45A flowing through the diode there will be 2V voltage drop on the diode.
 

ericgibbs

Joined Jan 29, 2010
9,114
hi T,
A DVM is a Digital Volt Meter, which normally have a very input impedance circuit, so as not to load the point/node being measured.
The LTS 'DVM' equivalent, is an infinitely high impedance as its a software program, so it does not 'load' the circuit.
Do you follow OK.?
E
 

ericgibbs

Joined Jan 29, 2010
9,114
hi T,
Ref the parallel circuit.
Compare these two options, V1 source has Zero output impedance so the current is limited only by the diodes internal resistance.

V2 source has a 100R source impedance, observe how it limits the diode currents and modifies the waveforms.

E
 

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Thread Starter

TarikElec

Joined Oct 17, 2019
20
First of all this is ideal voltage source, so if the voltage source says there is that voltage across it, then that voltage is there no matter how much current you draw from it. Second thing, the 0.7V figure is just the typical voltage at typical current. At a few microamps of current there will be 0.2-0.3V, at very large current compared to the capability of the diode there will be higher voltage like 1-2V. Look at the IV curve of a diode, the voltage doesn´t just stop at 0.7V with rising current.

The 1N4148 is a 100mA diode. If you put 2V across it, there will be about 2A flowing through the diode. I also checked some 1A rectifier, and the current at 2V was about 45A. Or if you look at it the other way, if you have 45A flowing through the diode there will be 2V voltage drop on the diode.
You are right, but in the curve of 1N4148 , I cannot see the current for 2V!
 

Thread Starter

TarikElec

Joined Oct 17, 2019
20
hi T,
A DVM is a Digital Volt Meter, which normally have a very input impedance circuit, so as not to load the point/node being measured.
The LTS 'DVM' equivalent, is an infinitely high impedance as its a software program, so it does not 'load' the circuit.
Do you follow OK.?
E
Yes here I understood. I have put a 1Billion ohm and it shows half the voltage. so, this voltage divider are used mainly in Op Amp as reference voltage for inverting and non inverting inputs?
 

ericgibbs

Joined Jan 29, 2010
9,114
hi,
Normally when using a resistive voltage divider with an OPA, the resistor junction [ Vout node] is connected to the OPA's Non Inverting input as this is a high impedance input, so it does not load the divider.
The Output of the OPA with a divider on the NI input, would be a low resistance, say about 50R, so it would be suitable for using as Reference voltage in some other part of the circuit.
OK.?
E
 

kubeek

Joined Sep 20, 2005
5,650
You are right, but in the curve of 1N4148 , I cannot see the current for 2V!
Well the diode is rated for 100mA. The simulation shows what the value would be, but this tiny diode will not survive that current for very long (maybe less than a millisecond), so practically there is no point putting it on the graph.
 

OBW0549

Joined Mar 2, 2015
3,101
1-the first circuit is two diode in series, when i measured the voltage between the them I found it half the voltage of the source which is not logic for me!!! as there is no current flowing through the diode, so how can it be a voltage divider!!!

2-diode in parallel and opposite to each other, for my understand it should show an output of +/-0.7V, but I got a full wave of source signal!!!
There are some important things to keep in mind when using Spice simulators such as LTSpice. Spice and the actual behavior of real-world, physical, components are not the same:
  • Voltage sources in Spice are perfect: they will apply their specified voltage to whatever they are connected to, no matter how much current they need to provide in order to do so.
  • Likewise for Spice current sources: they will force their specified current through whatever they are connected to, no matter how much voltage they need to generate in order to do so.
  • For passive components like R's, C's and L's, component values are precise and are exactly what you specify; for example, a 3.3 kΩ resistor will be exactly 3.3000000000000... kΩ.
  • As a consequence of the above, two or more identically-specified components will have exactly the same value.
  • The above also applies to semiconductor components like diodes, BJTs and MOSFETs. For example, two 2N3904 transistors in a circuit will have exactly the same characteristics and, unlike in real-life circuits, will therefore be perfectly matched.
  • Spice doesn't care whether the voltages or currents in a circuit exceed a part's maximum ratings; for example, it will happily shove 2 amps of current through a poor little 1N4148 diode (if the voltages are right) even if such a huge current would make the diode explode.
  • In Spice, there are no parasitic capacitances between circuit nodes. In the real world, every single thing in the universe has mutual capacitance with every other thing; in a circuit, these capacitances between circuit nodes may be only a few picofarads (or even a fraction of a picofarad), but in some circumstances they can affect circuit operation.
  • Likewise with inductance and resistance. Wires in Spice have neither; in real life, they have both. The amount of inductance and/or resistance in a wire connecting two points may not be much, but it can make a difference.
For these reasons (and others), Spice has to be used with some caution. "Designing in the simulator" can be problematic if differences between Spice and the real world are not taken into account.

(EDIT: added remarks about parasitics)
 
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ronsimpson

Joined Oct 7, 2019
341
I wish SPICE components exploded when you abuse them. That would greatly help engineers. It is very common to see someone using a part 100x it rated current or voltage.

In another SPICE I used years ago there was a way to say all the resistors & caps are +/-5% and have the computer look for all the possible effects of parts imperfections. (may take hours to work) Some time when R1= 5% high and R10 is 5% low things go wrong. There are some types of oscillators that will not start up in SPICE because "perfect" parts do not work like real parts.

There is a saying: "What happens when an unstoppable force meets an immovable object?" In SPICE it is a voltage source connected to a short. or A current source not connected.
 

OBW0549

Joined Mar 2, 2015
3,101
There are some types of oscillators that will not start up in SPICE because "perfect" parts do not work like real parts.
In the ancient version I use (ICAPS4/RX from Intusoft), oscillators almost never start up by themselves, usually because the Spice components are noiseless; they require a "kick" of some sort to get them going.
 

ronsimpson

Joined Oct 7, 2019
341
This circuit and multivibrators do not start when the two transistors are a perfect match. If you change the gain of one of the transistors or change one of the resistor even 1% the unbalanced circuit will start up better.
1575386681889.png
I found the example of why a two transistor oscillator will not start up using perfectly matched parts.
Two identical twins sitting on a perfectly balanced board with their feet not reaching the ground. (not fun)
Put a rock in one child's pocked and soon his feet will reach the ground and away we go.
1575387003012.png
 

Thread Starter

TarikElec

Joined Oct 17, 2019
20
hi T,
A DVM is a Digital Volt Meter, which normally have a very input impedance circuit, so as not to load the point/node being measured.
The LTS 'DVM' equivalent, is an infinitely high impedance as its a software program, so it does not 'load' the circuit.
Do you follow OK.?
E
Yes I got you
 

Bordodynov

Joined May 20, 2015
2,455
I wish SPICE components exploded when you abuse them. That would greatly help engineers.
Adding this feature will complicate the program algorithm and increase the time of scheme simulation. And this is only to help a very limited number of users of the program. A competent engineer will study the currents and powers of the elements himself. Some elements withstand tens and hundreds of times the nominal currents for a short time. And it complicates the decision of a question - will let a smoke or will explode a radio element or not.
 
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