Single wire in line delay circuit, not switching on hard enough - Page 1
Forum section: electronics / projects

author: Bendba
author_rank: Regular Contributor
timestamp: January 11, 2018, 05:37:53 am
content: Hi,I just got one of those cheap useless box from eBay and I wanted to add as simple as possible of a circuit to add a slight delay to the reaction.This is what I came up with:It works pretty good but as I expected, the circuit doesn't let enough current through. The little motor starts after about a second and a half but isn't strong enough to flick the switch.Is there any chance I can improve without having to take a different approach?ThanksBenHere is the full circuit:My add on could be on the return side of the motor, it wouldn't matter.

author: rstofer
author_rank: Super Contributor
timestamp: January 11, 2018, 03:58:57 pm
content: Have you measured the voltage drop across the Darlington? It is probably dropping a major fraction of the battery voltage.

author: Bendba
author_rank: Regular Contributor
timestamp: January 11, 2018, 08:23:43 pm
content: Quote from: rstofer on January 11, 2018, 03:58:57 pmHi,it is indeed dropping about 0.98V, so the transistor don't quite saturate, otherwise the drop would be slightly lower. I'm trying to make them saturate but if I reduce the 1.2k ohm resistor, it starts turning on before the capacitor charges.The only thing I can think of is using germanium transistors.Buy the way, I believe the configuration is called zakli pair (one NPN and one PNP), gives you the advantage of a lower drop out voltage then a Darlington pair (two NPN's or two PNP's)

author: rstofer
author_rank: Super Contributor
timestamp: January 11, 2018, 10:51:31 pm
content: That voltage divider keeps the NPN base too low. I tried 100 Ohms and got the output very close to 2.1V and that's about all there will be from this type of circuit. I have attached the LTspice circuit and results circuitpng and results.pngYou simply can't get the base of the NPN high enough to get the PNP into saturation. The same problem exists when using N-channel MOSFETs for high side switching.I tried low side switching with no noticeable improvement.See Circuitt1.png and Results1.png

author: martys
author_rank: Regular Contributor
timestamp: January 11, 2018, 11:00:10 pm
content: If you can find a used/discarded PC motherboard, you can find and salvage a no-cost wealth of N-Chan 20-30V, 30-AMP power MOSFETs to create a lo-side switch. Connect the motor negative lead to the delay circuit(Source of the N-Chan MOSFET), but charge the capacitor between MOSFET gate an source from the + connection of the power supply.You use the same kinda circuit, but a single large resistance resistor and a much smaller capacitor value can be used to create a turn-on delay without any real power loss after the motor turns on.Depending on delay time and the power of the motor, a small heatsink may be required.

author: rstofer
author_rank: Super Contributor
timestamp: January 12, 2018, 12:36:13 am
content: The reason I haven't jumped at the MOSFET approach is that I'm not aware of a device that can be turned on with only 3V (at most) Vgs. There are Logic Level MOSFETs but I haven't researched them. Maybe this:https://www.adafruit.com/product/355You might be able to get 3A with 3Vgs - see Figure 3 The thing is, Figure 12 doesn't go as low as 3Vgs. It stops at 4V.https://cdn-shop.adafruit.com/datasheets/irlb8721pbf.pdfAll of which means that it might be necessary to use some kind of boost converter to derive the gate voltage. Or maybe 3A is enough - it probably is considering the application.

author: SeanB
author_rank: Super Contributor
timestamp: January 12, 2018, 05:04:12 am
content: Try replacing the darlington with a SCR. A TIC106d will be sensitive enough to trigger with sane RC values, and still will work at 3V, and will only lose around a half volt once triggered.

author: Bendba
author_rank: Regular Contributor
timestamp: January 13, 2018, 11:25:24 am
content: Hi,I did try the SCR transistor equivalent but couldn't get the voltage drop below 0.85V in a working circuit.I tried making it out of germanium transistors and got the voltage drop down to 0.45V (edit: down to 0.28V now) but it seems they have to much leakage and do not want to stay off at power up. Is there a trick?I wanted to try to stick with a circuit that inserts in line to keep the quiescent current to zero.I might just have to take another approach or change the battery to 4.5V instead of 3VBy the way, here is the circuit diagram:If I replace the transistors with silicon transistors, it works, even though the voltage drop is too high.