Schematic explaination - Page 1
Forum section: electronics / beginners

author: J4e8a16n
author_rank: Regular Contributor
timestamp: May 15, 2015, 08:10:49 pm
content: 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

author: flynwill
author_rank: Regular Contributor
timestamp: May 15, 2015, 11:21:26 pm
content: 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.

author: Seekonk
author_rank: Super Contributor
timestamp: May 16, 2015, 12:00:48 am
content: 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.

author: dacman
author_rank: Frequent Contributor
timestamp: May 16, 2015, 01:31:06 am
content: 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.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: May 18, 2015, 11:06:12 am
content: I dont know what would be the name of the batt monitor in LT Spice:--- Operating Point ---2.4 volts to groundV(111): 1.22139 0.13923 high de C1 voltageV(222): 2.4 0.000995998 base Q1 voltageV(n001): 3 3 voltageV(n002): 1.95669 0.892714 voltageV(p001): 0.742933 0.771645 voltageV(n003): 1.22139 0.13923 Coll Q2 voltageV(n004): 2.4 2.94151 voltageV(n005): 1.30514 0.771645 voltageIc(Q1): -0.000119619 -0.000461673 device_currentIb(Q1): -4.74335e-007 -0.211914 device_currentIe(Q1): 0.000120093 0.212376 device_currentIc(Q2): 0.0223759 0.045122 device_currentIb(Q2): 0.000119619 0.000461673 device_currentIe(Q2): -0.0224955 -0.0455836 device_currentI(C1): 3e-017 3e-017 device_currentI(C2): -8.01452e-019 9.39993e-020 device_currentI(D2): 0.0221987 0.0448359 device_currentI(R1): 4.74335e-007 0.211914 device_currentI(R2): -9.79609e-019 1.22451e-019 device_currentI(R3): -0.000119619 -0.000461673 device_currentI(R4): 0.000177861 0.000286077 device_currentI(R5): 0.022198 0.0448359 device_currentI(V1): -0.0224959 -0.257498 device_currentI(V2): 4.74335e-007 device_currentThanksJP

author: Richard Head
author_rank: Frequent Contributor
timestamp: May 18, 2015, 01:09:27 pm
content: There should be a pull-down resistor on the base of Q2.

author: J4e8a16n
author_rank: Regular Contributor
timestamp: May 18, 2015, 03:14:16 pm
content: 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.

author: dacman
author_rank: Frequent Contributor
timestamp: May 19, 2015, 12:32:12 am
content: 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

author: J4e8a16n
author_rank: Regular Contributor
timestamp: May 19, 2015, 03:20:37 pm
content: Not sure I have it right.Joinned: the asc file

author: dacman
author_rank: Frequent Contributor
timestamp: May 20, 2015, 01:07:26 am
content: For the switch, try: Ron=5 Roff=1600Meg Vt=2.5 Vh=-.1For the pulse, try: 2.7 2.3 0 1u 1u 10 20 2

author: J4e8a16n
author_rank: Regular Contributor
timestamp: May 20, 2015, 10:10:05 am
content: Thank you very much.