LT Spice "Diode's relation with the circuit voltage"

Thread Starter

Joy1999

Joined Jul 26, 2022
9
Hi all,

I am simulating a circuit on LT Spice, and using a diode to restrict the flow of current in the opposite direction. The diode is restricting the current to flow in the negative direction but they are increasing the voltage by a lot. I'm using a 10V voltage source and when it passes the diode it's peak reaches approximately 700V. I am attaching screenshot of the schematic and the waveform here.

Thank you
 

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ericgibbs

Joined Jan 29, 2010
16,432
hi Joy,
Please post your LTS asc file when querying a simulation, it will improve the replies.
Together with any special models.
E
 

ericgibbs

Joined Jan 29, 2010
16,432
hi Joy,
Your current source, means it will drive that current through whatever resistance is in its conduction path.
A reverse biassed diode is not an infinite open circuit, it has a high resistance, so the current will develop a high voltage across it.
Also consider the forward resistance of the diode.

What exactly are you trying to do with the project.?
E
EG 1446.png
 

Thread Starter

Joy1999

Joined Jul 26, 2022
9
So the current source is like a dc motor where regeneration takes place, I want the negative current from the current source to charge the capacitor and use a switch before I1 so that during regeneration the switch between capacitor and current source is close and other time the switch between voltage source and I1 is closed.

Thanks
 

Thread Starter

Joy1999

Joined Jul 26, 2022
9
hi Joy,
Your current source, means it will drive that current through whatever resistance is in its conduction path.
A reverse biassed diode is not an infinite open circuit, it has a high resistance, so the current will develop a high voltage across it.
Also consider the forward resistance of the diode.

What exactly are you trying to do with the project.?
E
View attachment 275901
So the current source is like a dc motor where regeneration takes place, I want the negative current from the current source to charge the capacitor and use a switch before I1 so that during regeneration the switch between capacitor and current source is close and other time the switch between voltage source and I1 is closed.

Thanks
 

crutschow

Joined Mar 14, 2008
30,824
So the current source is like a dc motor where regeneration takes place,
But a motor is more like a voltage source, than a current source.
An ideal current source (as the simulation has) can generate a near infinite voltage, which the motor cannot.
You need a more realistic model for the motor when it is providing regeneration.
 

Thread Starter

Joy1999

Joined Jul 26, 2022
9
hi joy,
You cannot charge it this way.

What is the purpose of this project.?
E
Hi E
My purpose is to increase the efficiency of the LiIon battery by using the regenerative current from the DC Motor, to store in the capacitor and use it to power it or charge the LiIon battery. Basically I want to harvest energy generated from the DC Motor.

Is there any article or link for DC Motor modeling in LTSpice? I already have the PWL file which has current vs time profile for the Motor. So instead of voltage source I used current source as a motor.

Thanks
 

Thread Starter

Joy1999

Joined Jul 26, 2022
9
What is causing this "regeneration current" from the motor and how often does it occur?
Is this a vehicle?
Hi

I am using this to create a prosthetic knee in which when the knee swings it regenerates current. It regenerates around 30% of the total energy it consumes.
 

MrAl

Joined Jun 17, 2014
9,558
Hi

I am using this to create a prosthetic knee in which when the knee swings it regenerates current. It regenerates around 30% of the total energy it consumes.
Hi there,

This sounds very interesting could make a cell phone charger with this idea i bet.

There are sneakers that do this with inserts i think in the soles, but i dont know where to find them.
 

WBahn

Joined Mar 31, 2012
27,496
I am simulating a circuit on LT Spice, and using a diode to restrict the flow of current in the opposite direction. The diode is restricting the current to flow in the negative direction but they are increasing the voltage by a lot. I'm using a 10V voltage source and when it passes the diode it's peak reaches approximately 700V. I am attaching screenshot of the schematic and the waveform here.
FWIW, let's see if we can understand the behavior of the simulation circuit you first posted:

1663383386510.png

Let's assume, for the moment, that the Schottky diode has a sufficiently high reverse breakdown voltage that, in this simulation, it never breaks down. Let's also assume that it's forward voltage is 0.4 V (a bit below the 0.45 V max spec). What would you expect V1 to be at the end of the rising current output of your PWL source (i.e., when it is outputting 2 A)?

Let's assume, for the moment, that at the end of the rising current pulse that all of the current is flowing through the lower branch because the capacitor in the upper branch is basically fully charged. That means that there will be 2 A of current flowing through the diode (which is twice it's max spec, but we'll ignore that) and through the 10 Ω resistor. That will make V1 equal to

V1 = (+10 V) - (0.45 V) - (10 Ω)(2 A) = -10.05 V

Zooming in on your waveform and measuring the pixel distances, it looks like your V1 start out at -10.3 V, so this is pretty close. The resolution is very course at 2.06 V/pixel

Then the current changes direction and starts ramping down from 2 A to -1 A over the course of 1 s, so a rate of -3 A/a. This means that it will change polarity at 2/3 s and that, until that time, the voltage at V1 will be dominated by the lower branch because the diode clamps the voltage. When the current is just about to reach zero, the voltage at V1 should be about +9.55 V. Looking at where the V1 voltage starts taking off put it between 8.2 V and 10.3 V at a time of 0.676 s (the time resolution is about 5.5 ms/pixel). So this is in very good agreement with the simulation results.

If we assume that no current is flowing in the upper branch, then C1 is charged to about 10 V at this point. In fact, since V1 is changing, there will be some current in the upper branch, resulting in some drop across the two 10 Ω resistors. So let's get an estimate of what that might be. V1 is changing at about

dV1/dt ~= (3 A/s)(10 Ω) = 30 V/s

If C2 is charging at that rate, the current flowing in it would be

i_C2 = C dV1/dt = (470 uF)(30 V/s) = 14.1 mA

This is pretty small on the scale of the +2 A and -1 A swings of the current source, plus it makes the voltage drop across the resistors only about 0.28 V. So we can probably ignore it pretty safely.

When the current reaches zero, essentially no current will flow in either branch because the voltage source, V1, is trying to drive current into a capacitor that is already essentially charged to the correct voltage to block any further flow. This is backed up by noting that the time constant associated with C1 in series with all three resistors is 14 ms, so it tracks the changes very closely.

Now what happens as the current starts going from 0 V to -1 A over the course of the next 1/3 s?

The diode becomes reverse biased and so no current will flow in the lower branch, forcing all of the current to flow in the upper branch. The total charge delivered by the source comes in two portions. First, during the ramp down when the current has passed zero and is on its way to -1 A, which takes place over 1/3 s and is linear, so the charge delivered during this period is the area under the triangle:

Q_source = (1/2)*C*T = (1/2)*(1 C/s)*(1/3 s) = 1/6 C

Then, after the current source peaks at -1 A and starts ramping back up toward +2 A again, there is another 1/3 s during which the current is negative. It ends up having the same area, so the total charge delivered is 1/3 C. The time at which the capacitor stops increasing in voltage occurs 1/3 s after when the current source changed, so 1.33 s.

This increases the capacitor voltage by

Q = CV
V = Q/C
V_change = Q_source / C = (1/6 C) / (470 uF) = (1/3 C) / (470 uC/V) = +709 V

This is added to the ~10 V starting point, and you now have 719 V.

At this point the current source is back at zero output, so there is no drop across the two resistors in the top branch.

Going back to the screen shot and measuring the pixels, the peak voltage is 718.5 V at 1.32 seconds, which is in excellent agreement with this analysis (well within the coarseness of the image scaling).

The major takeaway from this is that

1) The device model for your diode does not incorporate any notion of reverse breakdown.
2) The high voltages you are seeing are a reflection of driving a capacitor with a current source, coupled with that capacitor being treated as ideal with no leakage and no breakdown voltage, either.
 
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