Schematic explaination

[J4e8a16n]

Hi,
 I am trying to understand....  The base of the pnp tr.  goes up trough the out and v+ ?

After the diode the capacitor is charged in 2 mS?

The capacitor discharge in the base of the 2n2907?

I calculated that R1*C2*5  is about 2 sec. What does this have to do with the 2 sec rest?

Thanks for your attention,


JP

[flynwill]

R1 * C2 = 3.2 seconds...  That's the time constant, but now you have to work how "how many" time constants.

Just before the circuit switches off the base of Q1 will be at about 1.8 V (assuming 2.4 volt battery voltage), and the voltage at the other side of C2 will be at about 0.5 volts (the saturation voltage of Q2).  When the current flowing through C2 is no longer enough to keep Q1 on (and therefore Q2 on) then it will switch off and collector of Q2 will rise to the battery voltage of 2.4 volts -- a 1.9 volt swing.  So the base of Q1 must also swing that 1.9V postive to end up at 3.7V (1.8 + 1.9).   Now the capacitor starts discharging through R1 again and so the problem is how long does the exponential take to go from 3.7V back to the 1.8V where it current will start to turn Q1 back on again.  Which my calculator says is 0.72 time constants or 2.3 seconds, probably give or take 20%. 

R2 * C2 = 0.4 mS, so you'll need to do a similar analysis to figure out why the discharge is 4 time constants.

[Seekonk]

You were close in the schematic.  Here is the short answer.  The short time period is the 680 ohm discharge, the long is the 10K resistor charge.  The LED has no effect because it looks like a zener and doesn't pass current in the charge.

[dacman]

A1 is a voltage sense circuit with an open drain FET output.  It is basically a switch that is turned on (OUT to GND) when the voltage drops to 2.4 Vdc or less.

When the voltage drops to 2.4 Vdc (or less), R1 is switched to common via A1.  This turns on Q1, which turns on Q2, which turns on the LED.

When Q1 is on, C2 is being charged.  The charge circuit is V+, Q1 (emitter to base), C2, R2, Q2, and common.

When C2 charges enough, it will cause Q1 to turn off, which will turn off Q2, which will turn off the LED.

Note that when Q2 turns off, the difference between two Q2 voltages, both from collector to emitter (VSAT and VCUTOFF), will be coupled across C2, further cutting off Q1.  (Without this Q1 would turn right back on after a slight discharge of C2.)

C2 will now discharge.  The discharge path (traced backwards) is common, A1 FET switch, R1, C2, R2, R4, and V+.  Now for a discharge circuit, we need to get back to common, which from V+ is through the battery.  The battery is aiding the discharge of C2.  Once C2 discharges enough, Q1 will turn back on and the process will repeat.

[J4e8a16n]

I dont know what would be the name of the batt monitor in LT Spice:

       --- Operating Point ---

      2.4 volts         to ground
                  
V(111):   1.22139         0.13923      high de C1      voltage                                   
V(222):    2.4            0.000995998   base Q1      voltage                                   
V(n001):    3            3                  voltage                                   
V(n002):    1.95669         0.892714               voltage                                   
V(p001):    0.742933         0.771645               voltage                                   
V(n003):    1.22139         0.13923      Coll  Q2      voltage                                   
V(n004):    2.4            2.94151               voltage
V(n005):    1.30514         0.771645               voltage                                   

Ic(Q1):    -0.000119619      -0.000461673            device_current                        
Ib(Q1):    -4.74335e-007      -0.211914               device_current                        
Ie(Q1):    0.000120093      0.212376               device_current

Ic(Q2):    0.0223759               0.045122               device_current                        
Ib(Q2):    0.000119619      0.000461673            device_current                        
Ie(Q2):    -0.0224955      -0.0455836                    device_current 

I(C1):    3e-017         3e-017               device_current                        
I(C2):    -8.01452e-019      9.39993e-020            device_current

I(D2):    0.0221987              0.0448359                      device_current                        
I(R1):    4.74335e-007      0.211914               device_current                        
I(R2):    -9.79609e-019      1.22451e-019            device_current                        
I(R3):    -0.000119619      -0.000461673            device_current
I(R4):    0.000177861      0.000286077            device_current                        
I(R5):    0.022198         0.0448359                    device_current                        
I(V1):    -0.0224959      -0.257498               device_current                        
I(V2):    4.74335e-007                        device_current                        

Thanks

JP

[Richard Head]

There should be a pull-down resistor on the base of Q2.

[J4e8a16n]

Oh.  Yes.  The device is shown in  original schematic on the first post up.  The device is a low voltage detector but I dont find it in ltspice.
This is the goal of my LTSpice schematic.

[dacman]

A1 is a voltage sense integrated circuit in a TO-92 package, with a FET output (which has an open drain).  It is basically an open/close switch.  If the input voltage to A1 is > 2.4 Vdc, then the output of A1 is an open, there would be no discharge path for C2, and Q1 would remain cutoff.

For LTSpice, see the below link.
http://www.linear.com/solutions/5735

[J4e8a16n]

Not sure I have it right.

Joinned: the asc file

[dacman]

For the switch, try: Ron=5 Roff=1600Meg Vt=2.5 Vh=-.1
For the pulse, try: 2.7 2.3 0 1u 1u 10 20 2

[J4e8a16n]

Thank you very much.