How can I place stacked vias with the size exact same cut width without metals around?
As the red part only in the image below?

Hello,
I'm trying to characterize a full adder that use transmission gate.
Unfortunately, the power calculation are wrong for the cell are always negative.
Is there any method or commands that can can help in power calculation or add the power consumption by the input pins to the power calculation ?
Another question, Is liberate support the characterization or transmission gate cells as standard cells or I should use liberate AMS for these type of cells ?
Thanks in advance,
Tareq 

When analyzing multiple sessions simultaneously Verisium Manager crashed and reported below error messages:

# A fatal error has been detected by the Java Runtime Environment:
#
#  SIGSEGV (0xb) at pc=0x00007efc52861b74, pid=14182, tid=18380
#
# JRE version: OpenJDK Runtime Environment Temurin-17.0.3+7 (17.0.3+7) (build 17.0.3+7)
# Java VM: OpenJDK 64-Bit Server VM Temurin-17.0.3+7 (17.0.3+7, mixed mode, sharing, tiered, compressed oops, compressed class ptrs, g1 gc, linux-amd64)
# Problematic frame:
# C  [libucis.so+0x238b74]

......

For more details please refer to the attached log file "hs_err_pid21143.log".

Two approaches were tried to solve this problem but neither has worked.
Method.1:

Setting larger heap size of Java process by "-memlimit" options.For example "vmanager -memlimit 8G".

Method.2:

Enlarging stack memory size limit of the Coverage engine by setting "IMC_NATIVE_STACKSIZE" environment variable to a larger value. For example "setenv IMC_NATIVE_STACKSIZE 1024000"

According to "hs_err_pid*.log" it is almost certain that the memory overflow triggered Java's CrashOnOutOfMemoryError and caused Verisium Manager to crash. There are some arguments about memory management of Java like "Xms, Xmx, ThreadStackSize, Xss5048k etc" and maybe this problem can be fixed by setting these arguments during analysis. However, how exactly does Verisium Manager specify these arguments during analysis? I tried to set them by the form of setting environment variables before analysis but it didn't work in analysis and their values didn't change.

Is there something wrong with my operation or is there a better solution?

Thank you very much.

Hi,
I want to parameterized width and length of a NMOS, but it ignore it and I face this error:
*WARNING* Value input must be a number - setting back to previous value.

Does anybody know how I can fix this issue?
Thanks

Hello All,

Following is the einstein's puzzle solved by cadence Incisive92  (solved in less than 3 seconds -> FAST!!!!!!)

Thanks,

Vinay Honnavara

Verification engineer at Keyu Tech

vinayh@keyutech.com

 // Author: Vinay Honnavara

// Einstein formulated this problem : he said that only 2% in the world can solve this problem
// There are 5 different parameters each with 5 different attributes
// The following is the problem

// -> In a street there are five houses, painted five different colors (RED, GREEN, BLUE, YELLOW, WHITE)

// -> In each house lives a person of different nationality (GERMAN, NORWEGIAN, SWEDEN, DANISH, BRITAIN)

// -> These five homeowners each drink a different kind of beverage (TEA, WATER, MILK, COFFEE, BEER),

// -> smoke different brand of cigar (DUNHILL, PRINCE, BLUE MASTER, BLENDS, PALL MALL)

// -> and keep a different pet (BIRD, CATS, DOGS, FISH, HORSES)


///////////////////////////////////////////////////////////////////////////////////////
// *************** Einstein's riddle is: Who owns the fish? ***************************
///////////////////////////////////////////////////////////////////////////////////////

/*
Necessary clues:

1. The British man lives in a red house.
2. The Swedish man keeps dogs as pets.
3. The Danish man drinks tea.
4. The Green house is next to, and on the left of the White house.
5. The owner of the Green house drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the Yellow house smokes Dunhill.
8. The man living in the center house drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blends lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhill.
12. The man who smokes Blue Master drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The Blends smoker lives next to the one who drinks water.
*/




typedef enum bit [2:0]  {red, green, blue, yellow, white} house_color_type;
typedef enum bit [2:0]  {german, norwegian, brit, dane, swede} nationality_type;
typedef enum bit [2:0]  {coffee, milk, water, beer, tea} beverage_type;
typedef enum bit [2:0]  {dunhill, prince, blue_master, blends, pall_mall} cigar_type;
typedef enum bit [2:0]  {birds, cats, fish, dogs, horses} pet_type;




class Einstein_problem;

    rand house_color_type house_color[5];
    rand nationality_type nationality[5];
    rand beverage_type beverage[5];
    rand cigar_type cigar[5];
    rand pet_type pet[5];
        rand int arr[5];
    
    constraint einstein_riddle_solver {
    
        
    
        foreach (house_color[i])
            foreach (house_color[j])
               if (i != j)
                house_color[i] != house_color[j];
        foreach (nationality[i])
            foreach (nationality[j])
               if (i != j)
                nationality[i] != nationality[j];
        foreach (beverage[i])
            foreach (beverage[j])
               if (i != j)
                beverage[i] != beverage[j];
        foreach (cigar[i])
            foreach (cigar[j])
               if (i != j)
                cigar[i] != cigar[j];
        foreach (pet[i])
            foreach (pet[j])
               if (i != j)
                pet[i] != pet[j];
    
    
        //1) The British man lives in a red house.
        foreach(nationality[i])
                (nationality[i] == brit) -> (house_color[i] == red);
                
        
        //2) The Swedish man keeps dogs as pets.
        foreach(nationality[i])
                (nationality[i] == swede) -> (pet[i] == dogs);
                
                
        //3) The Danish man drinks tea.        
        foreach(nationality[i])
                (nationality[i] == dane) -> (beverage[i] == tea);
        
        
        //4) The Green house is next to, and on the left of the White house.
        foreach(house_color[i])        
                 if (i<4)
                    (house_color[i] == green) -> (house_color[i+1] == white);
        
        
        //5) The owner of the Green house drinks coffee.
        foreach(house_color[i])
                (house_color[i] == green) -> (beverage[i] == coffee);
                
        
        //6) The person who smokes Pall Mall rears birds.
        foreach(cigar[i])
                (cigar[i] == pall_mall) -> (pet[i] == birds);
        
        
        //7) The owner of the Yellow house smokes Dunhill.
        foreach(house_color[i])
                (house_color[i] == yellow) -> (cigar[i] == dunhill);
        
        
        //8) The man living in the center house drinks milk.
        foreach(house_color[i])
                if (i==2) // i==2 implies the center house (0,1,2,3,4) 2 is the center
                    beverage[i] == milk;
        
        
        
        //9) The Norwegian lives in the first house.
        foreach(nationality[i])        
                if (i==0) // i==0 is the first house
                    nationality[i] == norwegian;
        
        
        
        //10) The man who smokes Blends lives next to the one who keeps cats.
        foreach(cigar[i])        
                if (i==0) // if the man who smokes blends lives in the first house then the person with cats will be in the second
                    (cigar[i] == blends) -> (pet[i+1] == cats);
        
        foreach(cigar[i])        
                if (i>0 && i<4) // if the man is not at the ends he can be on either side
                    (cigar[i] == blends) -> (pet[i-1] == cats) || (pet[i+1] == cats);
        
        foreach(cigar[i])        
                if (i==4) // if the man is at the last
                    (cigar[i] == blends) -> (pet[i-1] == cats);
        
        foreach(cigar[i])        
                if (i==4)
                    (pet[i] == cats) -> (cigar[i-1] == blends);
        
        
        //11) The man who keeps horses lives next to the man who smokes Dunhill.
        foreach(pet[i])
                if (i==0) // similar to the last case
                    (pet[i] == horses) -> (cigar[i+1] == dunhill);
        
        foreach(pet[i])        
                if (i>0 & i<4)
                    (pet[i] == horses) -> (cigar[i-1] == dunhill) || (cigar[i+1] == dunhill);
                    
        foreach(pet[i])        
                if (i==4)
                    (pet[i] == horses) -> (cigar[i-1] == dunhill);
                    


        //12) The man who smokes Blue Master drinks beer.
        foreach(cigar[i])
                (cigar[i] == blue_master) -> (beverage[i] == beer);
        
        
        //13) The German smokes Prince.
        foreach(nationality[i])        
                (nationality[i] == german) -> (cigar[i] == prince);
        

        //14) The Norwegian lives next to the blue house.
        foreach(nationality[i])
                if (i==0)
                    (nationality[i] == norwegian) -> (house_color[i+1] == blue);
        
        foreach(nationality[i])        
                if (i>0 & i<4)
                    (nationality[i] == norwegian) -> (house_color[i-1] == blue) || (house_color[i+1] == blue);
        
        foreach(nationality[i])        
                if (i==4)
                    (nationality[i] == norwegian) -> (house_color[i-1] == blue);
        

        //15) The Blends smoker lives next to the one who drinks water.            
        foreach(cigar[i])            
                if (i==0)
                    (cigar[i] == blends) -> (beverage[i+1] == water);
        
        foreach(cigar[i])        
                if (i>0 & i<4)
                    (cigar[i] == blends) -> (beverage[i-1] == water) || (beverage[i+1] == water);
                    
        foreach(cigar[i])        
                if (i==4)
                    (cigar[i] == blends) -> (beverage[i-1] == water);
        
    } // end of the constraint block
    


    // display all the attributes
    task display ;
        foreach (house_color[i])
            begin
                $display("HOUSE : %s",house_color[i].name());
            end
        foreach (nationality[i])
            begin
                $display("NATIONALITY : %s",nationality[i].name());
            end
        foreach (beverage[i])
            begin
                $display("BEVERAGE : %s",beverage[i].name());
            end
        foreach (cigar[i])
            begin
                $display("CIGAR: %s",cigar[i].name());
            end
        foreach (pet[i])
            begin
                $display("PET : %s",pet[i].name());
            end
        foreach (pet[i])
            if (pet[i] == fish)
                $display("THE ANSWER TO THE RIDDLE : The %s has %s ", nationality[i].name(), pet[i].name());
    
    endtask // end display
    
    
endclass




program main ;

    initial
        begin
            Einstein_problem ep;
            ep = new();
            if(!ep.randomize())
                $display("ERROR");
            ep.display();
        end
endprogram // end of main

        

Che(read more)

This blog describes how to efficiently use Virtuoso Visualization and Analysis XL.(read more)

Concurrent Layout Editing enables more than one designer to work in a hierarchy at the same time. Check out this blog to know more. (read more)

Optimizing PCB thermal performance is essential in today’s high-density designs, as it ensures stability, prolongs component life, and prevents potential thermal issues. One of the first steps to achieving this is with strategic component placement. Positioning high-power components—such as regulators, power transistors, or processors—away from heat-sensitive parts can prevent thermal interference, and placing them near the edges of the PCB often helps dissipate heat more effectively. It’s also beneficial to group components by their heat generation, creating dedicated thermal zones that can manage localized heating and reduce impact on other areas of the board.

Using thermal vias is another effective technique. By placing thermal vias under components like BGAs or power ICs, heat can be transferred from the surface to internal layers or ground planes. Increasing the size and number of these vias, or using thicker plating, enhances heat conductivity and helps manage heat more evenly across layers in multilayer boards. Increasing copper thickness on the PCB also has a major impact. Opting for thicker copper layers (e.g., 2 oz or even 3 oz copper) significantly boosts the heat dissipation capabilities of power planes and traces, especially in high-current areas. Large copper planes, such as dedicated ground or power planes, are equally effective in spreading heat efficiently. Adding thermal pads directly beneath heat-generating components improves this heat distribution.

Thermal relief pads help regulate heat flow for through-hole components by controlling heat transfer, which reduces thermal stress during soldering and prevents excessive heat spread to nearby sensitive areas. Performing thermal analysis with software tools like Celsius can be invaluable, as it allows you to simulate and model heat distribution, spot potential thermal issues, and refine your design before finalizing it.

Using heat sinks and thermal pads provides a direct way to draw heat from high-power components. Heat sinks can be attached with thermal adhesives, screws, or clamps, while thermal interface materials (TIMs), such as thermal pads or conductive adhesives, further reduce thermal resistance, enhancing heat-transfer efficiency. Optimizing the PCB layer stackup is also a key factor. Dedicated ground and power layers improve heat conduction across the PCB, enabling heat transfer between layers, particularly in high-density and multilayer PCBs.

In designs with high power requirements, active cooling options like fans, blowers, or heat pipes can be essential, helping to direct airflow across the PCB and further improving heat dissipation. Adding ventilation slots around hot zones and considering passive cooling paths enhance natural airflow, making the design more thermally efficient. By combining several of these techniques, you can create a PCB that handles heat effectively, resulting in a robust, long-lasting, and reliable product.

Let us know if you’ve had any challenges with thermal management in your designs—I’d be glad to discuss further!

In the era of Artificial Intelligence, front-end designers need a magical key to empower them with technology that enables fully optimized RTL for implementation handoff and provides RTL designers with capabilities to accurately assist in the implementation convergence process.

The magic lies with Cadence Joules RTL Design Studio, an expert system that leverages generative AI for RTL design exploration, triages possible causes of violations, and additional insights that empower designers to understand how to address issues in their RTL, leading to smarter and more efficient chip design.

This unlocks the immense debugging and design analysis capabilities from a single, unified cockpit, enabling fully optimized RTL design prior to implementation handoff for the front-end designers and addresses all aspects of physical design by adding visibility into power, performance, area, and congestion (PPAC)

One critical component is the clock tree, which distributes the clock signal to all sequential elements, such as flip-flops and latches. Designers need the right techniques in the beginning stage to optimize the clock tree structure, ensuring that their designs meet the required timing specifications, reduce power consumption, maintain signal integrity, and increase reliability.

 This incredible feature is part of the Joules RTL Design Studio.

How do you efficiently explore the clock tree structure to optimize the results using Joules RTL Design Studio?

Joules Studio allows viewing a simplified version of the clock structure. This feature is primarily designed to help display clock frequency scaling through clock dividers. You can customize colors, symbols, and design elements using an input file. Additionally, you can cross-probe the custom clock tree structure to other widgets and the main schematic view in Joules Studio.

Moreover, with the clock tree preference features of the ideal clock tree wizard in Joules Studio GUI, you can highlight clock path, generate clocks and master clock, set case analysis, fold and unfold instances, undo and redo, set sense and disable timing, color preference, etc.

You can binge on these features through the channel videos posted on the support portal, which covers the Joules RTL Design Studio GUI Clock Tree Structure and Features of Ideal Clock Tree Wizard.

You can refer to the videos on Cadence Online Support (Cadence login required).

​Video Links:
Viewing Custom Clock Tree Structure in Joules RTL Design Studio (Video) 

Exploring Clock Tree Preference Widget of Ideal Clock Tree Wizard in Joules RTL Design Studio (Video) 

Want to learn more?

Explore the one-stop solution Joules RTL Design Studio Product Page on Cadence Online Support (Cadence login required).

Related Resources 

Related Training Bytes:

Understanding Prototype Design Flow in Joules RTL Design Studio (Video)

Running Prototype Implementation Flow in Joules RTL Design Studio (Video)

Understanding Analyze Timing By Hierarchy In Joules RTL Design Studio (Video)

Related Courses:

• Joules Power Calculator

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