Jul 12, 2012 #1 Jushoraj Member level 1 Joined May 22, 2012 Messages 34 Helped 5 Reputation 10 Reaction score 5 Trophy points 1,288 Location Bangalore Activity points 1,548 Assume a pad and a block is placed 1mm apart from each other. If they are connected together by metal M1 carrying a current of 100mA, What will be the width of M1. I need the total solution. Can any one help me to solve this out?
Assume a pad and a block is placed 1mm apart from each other. If they are connected together by metal M1 carrying a current of 100mA, What will be the width of M1. I need the total solution. Can any one help me to solve this out?
Jul 12, 2012 #2 F findsriharsha Member level 3 Joined Sep 27, 2009 Messages 65 Helped 7 Reputation 14 Reaction score 8 Trophy points 1,288 Location California US Activity points 1,644 Re: Analog layout: Solve You can use the EM check tool in cadence and find out what should be the width required to carry 100mA through M1 of length L..
Re: Analog layout: Solve You can use the EM check tool in cadence and find out what should be the width required to carry 100mA through M1 of length L..
Jul 12, 2012 #3 erikl Super Moderator Staff member Joined Sep 9, 2008 Messages 8,108 Helped 2,695 Reputation 5,370 Reaction score 2,308 Trophy points 1,393 Location Germany Activity points 44,123 Re: Analog layout: Solve Jushoraj said: I need the total solution. Can any one help me to solve this out? Click to expand... W = L*RA * I / Vdrop W = width [µm] L = length [µm] RA = area-resistance [Ω/◻] I = current [mA] Vdrop = acceptable voltage drop [mV] Your(?) example: L=1000µm ; RA(M1) ≈ 0.1[Ω/◻] ; I=100mA ; Vdrop = 100mV W = ... = 1000 * 0.1 = 100µm
Re: Analog layout: Solve Jushoraj said: I need the total solution. Can any one help me to solve this out? Click to expand... W = L*RA * I / Vdrop W = width [µm] L = length [µm] RA = area-resistance [Ω/◻] I = current [mA] Vdrop = acceptable voltage drop [mV] Your(?) example: L=1000µm ; RA(M1) ≈ 0.1[Ω/◻] ; I=100mA ; Vdrop = 100mV W = ... = 1000 * 0.1 = 100µm
Jul 13, 2012 #4 Jushoraj Member level 1 Joined May 22, 2012 Messages 34 Helped 5 Reputation 10 Reaction score 5 Trophy points 1,288 Location Bangalore Activity points 1,548 Re: Analog layout: Solve erikl said: W = L*RA * I / Vdrop W = width [µm] L = length [µm] RA = area-resistance [Ω/◻] I = current [mA] Vdrop = acceptable voltage drop [mV] Your(?) example: L=1000µm ; RA(M1) ≈ 0.1[Ω/◻] ; I=100mA ; Vdrop = 100mV W = ... = 1000 * 0.1 = 100µm Click to expand... Really thank you for your reply......
Re: Analog layout: Solve erikl said: W = L*RA * I / Vdrop W = width [µm] L = length [µm] RA = area-resistance [Ω/◻] I = current [mA] Vdrop = acceptable voltage drop [mV] Your(?) example: L=1000µm ; RA(M1) ≈ 0.1[Ω/◻] ; I=100mA ; Vdrop = 100mV W = ... = 1000 * 0.1 = 100µm Click to expand... Really thank you for your reply......