current sense circuit

Shouldn't this be

-->( (3.3-Vbe)*R2/(R2+R3) + 26e-3*log(R3/R4) ) / R1 ans =

There's a point just before it trips where Vbe1=Vbe2

So Ic1 = Ic2 = Ic

3k3*(Ic-I5) - 27k*I5 = 100k*(Ic+I5)

Where I5 is the (negative) current in R5

Solving for Ic gives 97uA

From there, it's easy to get the drop across the 91 ohm sense resistor

This gives around 2.1mA for the trip point.

That was 2.1mA neglecting the fact that Ic2 is flowing through R1.

So it actually predicts ILOAD = 2.0mA

A,

Thanks for stating that Ic1=3DIc2 points to the solution. I'll post the derivation of my equation for the falling trip point later, when I have my notes.

My solution:

(Vn represents voltage across Rn) V5=3D0 implies Ic1/Ic2=3DR3/R4, so V1 =3D V2 + .026 ln (R3/R4) Ic2 =3D (3.3 - V1 - Vbe)/R3 V2 =3D R2*Ic2 =3D R2(3.3-V1-Vbe)/R3 V1 =3D R2(3.3-V1-Vbe)/R3 + .026 ln (R3/R4) solving for V1 V1 =3D R2(3.3-Vbe)/(R2+R3) + (.026R3ln(R3/R4))/(R2+R3) Vbe=3D.6 Iload=3D 1.54mA

Looks good.

My solution:

(Vn represents voltage across Rn) V5=0 implies Ic1/Ic2=R3/R4, so V1 = V2 + .026 ln (R3/R4) Ic2 = (3.3 - V1 - Vbe)/R3 V2 = R2*Ic2 = R2(3.3-V1-Vbe)/R3 V1 = R2(3.3-V1-Vbe)/R3 + .026 ln (R3/R4) solving for V1 V1 = R2(3.3-Vbe)/(R2+R3) + (.026R3ln(R3/R4))/(R2+R3) Vbe=.6 Iload= 1.54mA

OK My (3.3-Vbe) was (3.3-Vbe1) Your (3.3-Vbe) was (3.3-Vbe2) R3/(R2+R3) ~= 1 :-)

"gearhead" wrote in message news: snipped-for-privacy@v19g2000yqn.googlegroups.com... On Feb 21, 10:34 am, "Andrew Holme" wrote:

If you simulate this (elegant) cct at both trip points vbe1==vbe2 and Ic1==Ic2 and the voltage at the top of R3 and R4 is equal.

So that leads me to believe that all you needed to concentrate on was Vout being high or low. Then again I couldn't do the impressive analysis you guys have done.

I don't know if it is possible to recreate Jim's hysteresis graph in LTC

Version 4 SHEET 1 988 680 WIRE 80 -128 -304 -128 WIRE 192 -128 80 -128 WIRE 480 -128 272 -128 WIRE 720 -128 480 -128 WIRE 480 -64 480 -128 WIRE 416 -16 352 -16 WIRE 720 0 720 -128 WIRE 80 64 80 -128 WIRE -304 96 -304 -128 WIRE 480 112 480 32 WIRE 480 112 144 112 WIRE 480 144 480 112 WIRE 80 256 80 160 WIRE 192 256 80 256 WIRE 352 256 352 -16 WIRE 352 256 272 256 WIRE 480 256 480 224 WIRE 480 256 352 256 WIRE 80 304 80 256 WIRE 480 304 480 256 WIRE -304 400 -304 176 WIRE 80 400 80 384 WIRE 80 400 -304 400 WIRE 480 400 480 384 WIRE 720 400 720 80 WIRE 720 400 480 400 WIRE 80 416 80 400 WIRE 480 416 480 400 FLAG 80 416 0 FLAG 480 416 0 SYMBOL pnp 144 160 R180 SYMATTR InstName Q1 SYMATTR Value 2N3906 SYMBOL pnp 416 32 M180 SYMATTR InstName Q2 SYMATTR Value 2N3906 SYMBOL res 64 288 R0 SYMATTR InstName R4 SYMATTR Value 3K3 SYMBOL res 464 288 R0 SYMATTR InstName R3 SYMATTR Value 100K SYMBOL res 464 128 R0 SYMATTR InstName R2 SYMATTR Value 2K SYMBOL res 288 240 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R5 SYMATTR Value 27K SYMBOL res 288 -144 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 91 SYMBOL current 720 0 R0 WINDOW 3 24 28 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value 1m SYMATTR InstName I1 SYMBOL voltage -304 80 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value 3.3 TEXT 536 448 Left 0 !.dc I1 4m 0m 0.1m

..

mA,

uys

I downloaded LTSpice IV. How do I load your script so I can run the simulation?

Hi,

cut and paste the text from the start of the "Version 4" line to the end of the last line into notepad and save it as a filenameofyourchoice.asc file. It will then open in ltspice.