simulate this circuit – Schematic created using CircuitLab
I know you can use 555 and other stuff but I'm trying to do this with just these few components. This circuit is the best that I managed to do.
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- 1\$\begingroup\$ Welcome to EE.SE. Post a schematic of your circuit - there's a button on the editor toolbar - so we can understand your thinking process and comment on it. \$\endgroup\$Transistor– Transistor2018-06-16 10:12:00 +00:00Commented Jun 16, 2018 at 10:12
- \$\begingroup\$ What functional improvements do you think your circuit needs? \$\endgroup\$Andy aka– Andy aka2018-06-16 10:36:06 +00:00Commented Jun 16, 2018 at 10:36
- \$\begingroup\$ A pulse is not possible using only passive components and one switch, you need another active switch. \$\endgroup\$Voltage Spike– Voltage Spike ♦2018-06-19 05:02:42 +00:00Commented Jun 19, 2018 at 5:02
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simulate this circuit – Schematic created using CircuitLab
Figure 1. OP's circuit redrawn with the switch in the positive.
I've redrawn your schematic with the switch in the positive leg. This allows us to connect the LED cathode to ground and, since we usually take voltage readings referenced to ground, it makes analysis simpler.
I've replaced SW1 with a time-delay contact so we can use the simulator.
Figure 2. Voltage at the top of D1 and current through D1.
When switch 1 closes the following happens:
- C1 is discharged at the time of switch-on.
- The left side of C1 starts to charge at a rate determined by R1. With a 5 V supply and a 1k resistor a maximum of 5 mA can flow.
- The right side of C1 will start to rise towards 5 V too but the LED will turn on and prevent the voltage getting any higher than the VF (forward voltage) of the LED. You can see this is about 1.7 V on the graph.
- The capacitor will supply current for a short time. You can get a rough idea of the time from the RC time-constant. \$ \tau = RC = 1k \times 220 \ \mathrm {\mu s} \$. The current should have dropped by 63% after this time and you can see that this is about right on the Amperes chart of Figure 2. (Actual value is more like 300 ms.)
- After the capacitor has discharged R2 provides a very small current - about 40 µA - which is enough to forward bias the LED at about 1 V but the current is tiny and you probably won't see any light unless the room is completely dark.
Your question doesn't explain what you would like to happen. Hopefully this will help you understand what is happening.
- \$\begingroup\$ So what you're saying is that the voltage across diode cant get higher than 1.7V and the current is limited with R1, so until the cap fills up there will be current flowing and the LED will be on until the cap fills up and the current stops. I think I get that? Also that very small current is keeping the diode slightly on but I added a resistor parallel to it and now I'm measuring 45mV on the diode so its off now. I dont get it why its now off. Im really confused with these circuits... \$\endgroup\$Dangz1– Dangz12018-06-16 11:49:10 +00:00Commented Jun 16, 2018 at 11:49
- \$\begingroup\$ It's going to be difficult to calculate the current through 'a resistor'. What's it's resistance value? \$\endgroup\$Transistor– Transistor2018-06-16 12:13:56 +00:00Commented Jun 16, 2018 at 12:13