Dec 26, 2007 #1 S samuel Full Member level 3 Joined Nov 15, 2004 Messages 171 Helped 1 Reputation 2 Reaction score 0 Trophy points 1,296 Location world Activity points 1,143 (1) The initial value of a capacitor is 5V; (2) Then a constant current discharge this capacitor, the discharge waveform is as Fig.1 (3) This discharging current consists two part, I=Ibak+Isig. (4) Now, we can sample this capacitor respectivly at t1 and t2. My question is how to get the voltage Isig generated at the capacitor excludes the voltage Ibak generated at the capacitor? thanks. samuel.
(1) The initial value of a capacitor is 5V; (2) Then a constant current discharge this capacitor, the discharge waveform is as Fig.1 (3) This discharging current consists two part, I=Ibak+Isig. (4) Now, we can sample this capacitor respectivly at t1 and t2. My question is how to get the voltage Isig generated at the capacitor excludes the voltage Ibak generated at the capacitor? thanks. samuel.
Dec 27, 2007 #2 leo_o2 Advanced Member level 4 Joined Sep 3, 2004 Messages 1,322 Helped 278 Reputation 558 Reaction score 241 Trophy points 1,343 Location China Activity points 5,761 copy the current Ibak, charging a cap with same value as sampled cap from t1 to t2. You can get a voltage Vx on the cap. V(t1)-V(t2)+Vx is what you want.
copy the current Ibak, charging a cap with same value as sampled cap from t1 to t2. You can get a voltage Vx on the cap. V(t1)-V(t2)+Vx is what you want.
Dec 30, 2007 #3 G gpwu Member level 2 Joined Dec 7, 2007 Messages 49 Helped 6 Reputation 12 Reaction score 2 Trophy points 1,288 Activity points 1,486 1. put Isig=0, measure V(t2-t1)=dv1 2. add normal Isig, measure V(t2-t1)=dv2 Then the dv2-dv1 should be Isig's contribution
1. put Isig=0, measure V(t2-t1)=dv1 2. add normal Isig, measure V(t2-t1)=dv2 Then the dv2-dv1 should be Isig's contribution