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I'm trying to build a circuit with the following properties:

  1. When a switch (SW1) is pressed continuously, it should supply a particular load with 5V (modeled as R4 and LED D2) after about 20sec
  2. When this switch is released, to preserve battery it should not draw any power

After looking at schematics on the internet I've tried the following circuit, unfortunately without any luck.

I've tried simulating with LTSpice, it seems no matter how I change the values the NPN tranistor doesn't want to turn on.

My circuit is probably seriously flawed, could anyone push me in the right direction?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I just answered a similarly odd request here: electronics.stackexchange.com/questions/304232/… \$\endgroup\$ – jonk May 11 '17 at 0:04
  • \$\begingroup\$ Just replace Q with "logic level Nch MOSFET" rated Imax = >5x current needed for low RdsOn. Switch must be rated for Load. If using small signal switch (<2A) then use CMOS logic gate to drive MOSFET for delayed ON \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 11 '17 at 1:01
  • \$\begingroup\$ Thanks for the comment, MOSFETS would have been easier, unfortunately I don't have mosfets available for this project, so I'm hoping to do this with npn's. \$\endgroup\$ – user1712263 May 11 '17 at 1:19
  • \$\begingroup\$ It turns on for me, but for that I either added ic=0 to the capacitor, or made the source pulse 0 5 0 1m. You can also try adding startup or uic to the simulation commands. It's true, though, that the turn on is very slow and unreliable, but that can be changed through R2, for example (as in Finbarr's answer). \$\endgroup\$ – a concerned citizen May 11 '17 at 6:22
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The circuit as given will stabilise with Q1's base around 600mV, which scarcely turns it on. That explains why you get no significant power to the load.

Reducing R1, R2 and R3 to a tenth of the given values solves that problem sufficiently to light the LED, though it would also shorten the time constant. This can be compensated for by increasing C1 by the same factor, to 1000µF.

If a further increase in load power is needed, connecting a second transistor in Darlington configuration might work. You'll need to adjust the timing components for the higher base threshold.

In general, timing long delays using purely discrete, analogue components is fraught with difficulty, especially if you then want a sharp turn-on of a relatively heavy load. Consider using an IC specifically designed for the purpose, such as the 555.

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I wouldn't say it was seriously flawed, it's just that your resistor values are a bit high to get Q1 sufficiently turned on to light the LED.

If you remove C1 from the circuit, you can do a bit of simple analysis to show that you'll get just under 1.6V across R2 and about 50uA of base current into Q1. With a gain of 100 that gives you just 5mA through the LED.

Try lowering some resistor values to get the right base current to turn Q1 on enough for the load you're trying to drive, then determine C1 to get the right delay. Better still, use a logic level MOSFET in its place, then you can stop worrying about the current it needs to switch it on.

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  • \$\begingroup\$ Thanks for your suggestion. I've played around with a lot of values for R1, R2, and R3. Seems like only by changing R1 and R3 to 1 Ohm gave me a higher voltage at the load; 3.6V, which is still too low unfortunately. Also lowering R1 that much, practically removing it, defeats the purpose of the circuit because there is no "start up delay". How would u suggest I change the resistors specifically? PS: I don't have any mosfets available for this project, so I'm hoping to do this with npn's. \$\endgroup\$ – user1712263 May 11 '17 at 1:17
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    \$\begingroup\$ If you connect R4 to the 5V source directly, you can get rid of D1, not that you really need it anyway. Figure out how much current you need to put into the base of Q1 to get it to turn on the LED and work backwards to determine the maximum sum of R1 and R3. You could even consider a second NPN as an emitter follower to get more base current to Q1 if you really get stuck. \$\endgroup\$ – Finbarr May 11 '17 at 8:56
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schematic

simulate this circuit – Schematic created using CircuitLab

Another approach is to use CMOS Logic for Analog designs.either SOT23 SMD or DIP14 with hysteresis inputs (noise free).

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