0
\$\begingroup\$

Is there any circuit to perform such task more efficiently?

I am trying to control a small motor (like a smartphone vibration motor) with specific outputs from the touch IC, so that it can be turned on and off at particular outputs from it. That is, it can turn on at a particular current / voltage input and turn off at another.

\$\endgroup\$
  • 1
    \$\begingroup\$ Look into SCR or thyristor. \$\endgroup\$ – MarkU Jul 30 '18 at 16:05
  • 1
    \$\begingroup\$ Can you define your desired behaviour more precisely? What should happen if the current flowing into e.g. the collector also stops? What are the voltages and currents we're dealing with? There's an infinite (but probably countable) amount of potential solutions to this. In fact, a FET would do this. But it's still probably not what you want. Maybe you mean a thyristor? BUt that requires continuous switched current. Maybe you're looking for a flipflop? Or maybe for a more complex logic device?Who knows!Your question doesn't tell us what you exactly need to happen,or what problem you're solving. \$\endgroup\$ – Marcus Müller Jul 30 '18 at 16:05
  • \$\begingroup\$ Thanks marcus, actually I want to turn on a small motor(like smartphone vibratior type) with a switch system taking its specific input from the tuch-ic(the current output from it) of a tuch screen and turn it off with a specific value. \$\endgroup\$ – Ss1. 0 Jul 30 '18 at 16:33
  • \$\begingroup\$ Please Edit your question to include that info! A potential answerer shouldn't have to read comments to figure this out. \$\endgroup\$ – Marcus Müller Jul 30 '18 at 17:01
0
\$\begingroup\$

What you're talking about is generally done with an SCR. You can do this with an NPN and a PNP transistor:

schematic

simulate this circuit – Schematic created using CircuitLab

Assume that Q1 and Q2 are both off, and the input is zero. Now apply a voltage to R1 (greater than about 0.6 volts). Q1 will begin to conduct, which will provide base current to Q2, and Q2's collector current will provide base current to Q1, keeping it turned on, regardless of the voltage at R1.

EDITED TO ADD - As G36 has commented, it may be necessary to add a few resistors for real-life operation. Equally important, this circuit cannot be used for push-on, push-off operation. In order to turn it off, the current through the load must be interrupted while the input is low. In principle, this can be done by adding another transistor.

schematic

simulate this circuit

In this version, with the load ON, if the ON input is low and the OFF input high, Q3 will pull Q2's emitter low and turn it off. When OFF goes low again, the circuit should remain off. Note that in this version, the circuit turns on when the ON input goes high, and remains on until the OFF input goes high and then low again.

Per G36's comment, R4 is probably critical, diverting leakage through Q2 away from Q1's base.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Have you ever built this circuit in real-life as you have drawn it? \$\endgroup\$ – G36 Jul 30 '18 at 16:49
  • \$\begingroup\$ @G36 it seems like you have an objection, but it would be good if you could be more specific, please. \$\endgroup\$ – Marcus Müller Jul 30 '18 at 17:02
  • 3
    \$\begingroup\$ @MarcusMüller there is a big chance that this circuit will not work as expected when you try to build it with discrete transistors due to high current gain and a leakage current and positive feedback. Additional resistors between the base and emitter are needed in real life circuit. \$\endgroup\$ – G36 Jul 30 '18 at 17:58
  • \$\begingroup\$ @WhatRoughBeast can the circuit be used for the edited scenario! \$\endgroup\$ – Ss1. 0 Jul 30 '18 at 18:35
  • \$\begingroup\$ @G36 - Yeah, I made one - decades ago. As I recall it worked adequately, but I only made it to demonstrate the operation, not for "real" use. \$\endgroup\$ – WhatRoughBeast Jul 30 '18 at 20:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.