current sense circuit

G

gearhead

Guest
Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]
The term on the left expresses the voltage across R2.
The term on the right accounts for the voltage offset across Q1 and
Q2, which have collector currents in the ratio of R3/R4.
Now, using Vbe=.6 gives sense voltage .14 volts, load current 1.54 mA,
which agrees with the graph on the second page. So far so good.

My equation for the rising trip point
(3.3-Vbe)R2/(R2+(R3||(R4+R5)))
gives 2.35 mA for Vbe = .6.
That doesn't agree with Jim's graph.
Seems like I'm missing something, some simple way of looking at the
hysteresis in the circuit maybe. Anybody?
 
"gearhead" <nospam@billburg.com> wrote in message
news:7559898c-4fd3-4b8b-8cad-c6572084ae68@33g2000yqm.googlegroups.com...
Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]
Click to expand...

Shouldn't this be

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

0.0015564


The term on the left expresses the voltage across R2.
The term on the right accounts for the voltage offset across Q1 and
Q2, which have collector currents in the ratio of R3/R4.
Now, using Vbe=.6 gives sense voltage .14 volts, load current 1.54 mA,
which agrees with the graph on the second page. So far so good.

My equation for the rising trip point
(3.3-Vbe)R2/(R2+(R3||(R4+R5)))
gives 2.35 mA for Vbe = .6.
That doesn't agree with Jim's graph.
Seems like I'm missing something, some simple way of looking at the
hysteresis in the circuit maybe. Anybody?
Click to expand...
There's a point just before it trips where Vbe1=Vbe2

So Ic1 = Ic2 = Ic

3.3-Vbe = 2k*Ic + 100k*(Ic+I5)

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.
 
"Andrew Holme" <ah@nospam.co.uk> wrote in message
news:%ZUnl.29137$TK1.7070@newsfe16.ams2...
"gearhead" <nospam@billburg.com> wrote in message
news:7559898c-4fd3-4b8b-8cad-c6572084ae68@33g2000yqm.googlegroups.com...
Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]


Shouldn't this be

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

0.0015564


The term on the left expresses the voltage across R2.
The term on the right accounts for the voltage offset across Q1 and
Q2, which have collector currents in the ratio of R3/R4.
Now, using Vbe=.6 gives sense voltage .14 volts, load current 1.54 mA,
which agrees with the graph on the second page. So far so good.

My equation for the rising trip point
(3.3-Vbe)R2/(R2+(R3||(R4+R5)))
gives 2.35 mA for Vbe = .6.
That doesn't agree with Jim's graph.
Seems like I'm missing something, some simple way of looking at the
hysteresis in the circuit maybe. Anybody?

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

So Ic1 = Ic2 = Ic

3.3-Vbe = 2k*Ic + 100k*(Ic+I5)

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

So it actually predicts ILOAD = 2.0mA
 
On Feb 21, 10:34 am, "Andrew Holme" <a...@nospam.co.uk> wrote:
"Andrew Holme" <a...@nospam.co.uk> wrote in message

news:%ZUnl.29137$TK1.7070@newsfe16.ams2...





"gearhead" <nos...@billburg.com> wrote in message
news:7559898c-4fd3-4b8b-8cad-c6572084ae68@33g2000yqm.googlegroups.com...
Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]

Shouldn't this be

-->(  (3.3-Vbe)*R2/(R2+R3) + 26e-3*log(R3/R4)  ) / R1
ans  
   0.0015564

The term on the left expresses the voltage across R2.
The term on the right accounts for the voltage offset across Q1 and
Q2, which have collector currents in the ratio of R3/R4.
Now, using Vbe=.6 gives sense voltage .14 volts, load current 1.54 mA,
which agrees with the graph on the second page. So far so good.

My equation for the rising trip point
(3.3-Vbe)R2/(R2+(R3||(R4+R5)))
gives 2.35 mA for Vbe = .6.
That doesn't agree with Jim's graph.
Seems like I'm missing something, some simple way of looking at the
hysteresis in the circuit maybe.  Anybody?

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

So Ic1 = Ic2 = Ic

3.3-Vbe = 2k*Ic + 100k*(Ic+I5)

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
Click to expand...
Thanks for stating that Ic1=Ic2 points to the solution.
I'll post the derivation of my equation for the falling trip point
later, when I have my notes.
 
On Feb 21, 7:21 am, "Andrew Holme" <a...@nospam.co.uk> wrote:
"gearhead" <nos...@billburg.com> wrote in message

news:7559898c-4fd3-4b8b-8cad-c6572084ae68@33g2000yqm.googlegroups.com...

Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]

Shouldn't this be

-->(  (3.3-Vbe)*R2/(R2+R3) + 26e-3*log(R3/R4)  ) / R1
 ans  
    0.0015564
Click to expand...
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

The term on the left expresses the voltage across R2.
The term on the right accounts for the voltage offset across Q1 and
Q2, which have collector currents in the ratio of R3/R4.
Now, using Vbe=.6 gives sense voltage .14 volts, load current 1.54 mA,
which agrees with the graph on the second page. So far so good.

My equation for the rising trip point
(3.3-Vbe)R2/(R2+(R3||(R4+R5)))
gives 2.35 mA for Vbe = .6.
That doesn't agree with Jim's graph.
Seems like I'm missing something, some simple way of looking at the
hysteresis in the circuit maybe.  Anybody?

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

So Ic1 = Ic2 = Ic

3.3-Vbe = 2k*Ic + 100k*(Ic+I5)

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.
Click to expand...
Looks good.
 
"gearhead" <nospam@billburg.com> wrote in message
news:975c9413-1786-40b0-8cb8-ed50bb472d45@r41g2000yqm.googlegroups.com...
On Feb 21, 7:21 am, "Andrew Holme" <a...@nospam.co.uk> wrote:
"gearhead" <nos...@billburg.com> wrote in message

news:7559898c-4fd3-4b8b-8cad-c6572084ae68@33g2000yqm.googlegroups.com...

Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]

Shouldn't this be

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

0.0015564
Click to expand...
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" <nospam@billburg.com> wrote in message
news:845efc16-63b3-4d5d-81ad-20413209d390@v19g2000yqn.googlegroups.com...
On Feb 21, 10:34 am, "Andrew Holme" <a...@nospam.co.uk> wrote:
"Andrew Holme" <a...@nospam.co.uk> wrote in message

news:%ZUnl.29137$TK1.7070@newsfe16.ams2...





"gearhead" <nos...@billburg.com> wrote in message
news:7559898c-4fd3-4b8b-8cad-c6572084ae68@33g2000yqm.googlegroups.com...
Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]

Shouldn't this be

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

0.0015564

The term on the left expresses the voltage across R2.
The term on the right accounts for the voltage offset across Q1 and
Q2, which have collector currents in the ratio of R3/R4.
Now, using Vbe=.6 gives sense voltage .14 volts, load current 1.54 mA,
which agrees with the graph on the second page. So far so good.

My equation for the rising trip point
(3.3-Vbe)R2/(R2+(R3||(R4+R5)))
gives 2.35 mA for Vbe = .6.
That doesn't agree with Jim's graph.
Seems like I'm missing something, some simple way of looking at the
hysteresis in the circuit maybe. Anybody?

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

So Ic1 = Ic2 = Ic

3.3-Vbe = 2k*Ic + 100k*(Ic+I5)

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

Thanks for stating that Ic1=Ic2 points to the solution.
I'll post the derivation of my equation for the falling trip point
later, when I have my notes.
Click to expand...
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
 
"gearhead" <nospam@billburg.com> wrote in message
news:4cba2f08-9508-42b1-bbc0-8dd4e75422f1@t13g2000yqc.googlegroups.com...
On Feb 22, 4:02 am, "IanM" <nobody@no_where.co.uk> wrote:
"gearhead" <nos...@billburg.com> wrote in message

news:845efc16-63b3-4d5d-81ad-20413209d390@v19g2000yqn.googlegroups.com...
On Feb 21, 10:34 am, "Andrew Holme" <a...@nospam.co.uk> wrote:





"Andrew Holme" <a...@nospam.co.uk> wrote in message

news:%ZUnl.29137$TK1.7070@newsfe16.ams2...

"gearhead" <nos...@billburg.com> wrote in message
news:7559898c-4fd3-4b8b-8cad-c6572084ae68@33g2000yqm.googlegroups.com...
Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]

Shouldn't this be

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

0.0015564

The term on the left expresses the voltage across R2.
The term on the right accounts for the voltage offset across Q1 and
Q2, which have collector currents in the ratio of R3/R4.
Now, using Vbe=.6 gives sense voltage .14 volts, load current 1.54
mA,
which agrees with the graph on the second page. So far so good.

My equation for the rising trip point
(3.3-Vbe)R2/(R2+(R3||(R4+R5)))
gives 2.35 mA for Vbe = .6.
That doesn't agree with Jim's graph.
Seems like I'm missing something, some simple way of looking at the
hysteresis in the circuit maybe. Anybody?

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

So Ic1 = Ic2 = Ic

3.3-Vbe = 2k*Ic + 100k*(Ic+I5)

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
Thanks for stating that Ic1=Ic2 points to the solution.
I'll post the derivation of my equation for the falling trip point
later, when I have my notes.

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- Hide quoted text -

- Show quoted text -

I downloaded LTSpice IV. How do I load your script so I can run the
simulation?
Click to expand...
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.
 
On Feb 22, 4:02 am, "IanM" <nobody@no_where.co.uk> wrote:
"gearhead" <nos...@billburg.com> wrote in message

news:845efc16-63b3-4d5d-81ad-20413209d390@v19g2000yqn.googlegroups.com...
On Feb 21, 10:34 am, "Andrew Holme" <a...@nospam.co.uk> wrote:





"Andrew Holme" <a...@nospam.co.uk> wrote in message

news:%ZUnl.29137$TK1.7070@newsfe16.ams2...

"gearhead" <nos...@billburg.com> wrote in message
news:7559898c-4fd3-4b8b-8cad-c6572084ae68@33g2000yqm.googlegroups.com....
Take a look at this circuit:

http://www.analog-innovations.com/SED/CurrentSense.pdf

I wanted to come up with an analytical expression for the voltages at
the transition points.

First, I considered that with logicout high and load current falling,
the circuit trips when R3 and R4 see equal voltage. R5 has no current
at that moment.
So I redrew the circuit without R5 and wrote this expression for the
voltage across R1:
[(3.3-Vbe)(R2/(R2+R3))] + [(R3/(R2+R3)) .026 ln (R3/R4)]

Shouldn't this be

-->( (3.3-Vbe)*R2/(R2+R3) + 26e-3*log(R3/R4) ) / R1
ans
0.0015564

The term on the left expresses the voltage across R2.
The term on the right accounts for the voltage offset across Q1 and
Q2, which have collector currents in the ratio of R3/R4.
Now, using Vbe=.6 gives sense voltage .14 volts, load current 1.54 mA,
which agrees with the graph on the second page. So far so good.

My equation for the rising trip point
(3.3-Vbe)R2/(R2+(R3||(R4+R5)))
gives 2.35 mA for Vbe = .6.
That doesn't agree with Jim's graph.
Seems like I'm missing something, some simple way of looking at the
hysteresis in the circuit maybe. Anybody?

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

So Ic1 = Ic2 = Ic

3.3-Vbe = 2k*Ic + 100k*(Ic+I5)

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
Thanks for stating that Ic1=Ic2 points to the solution.
I'll post the derivation of my equation for the falling trip point
later, when I have my notes.

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- Hide quoted text -

- Show quoted text -
Click to expand...
I downloaded LTSpice IV. How do I load your script so I can run the
simulation?
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top