# Hows does this 12v relay works with transistor voltage drop?

I'm working my way through "Make: Electronics" by Charles Platt and came across the following schematic:

As someone who just recently got started with electronics, what I can't wrap my head around is how R1 relay works after Q1's voltage drop. I have built the same circuit only with different power source and relay, both 5V instead of 12V as suggested by the book and what seems to be happening is that the emitter voltage is bellow the relay trigger voltage (12V in the schematics and 5V in my case since I'm using 5V relay and source). From what I have been reading around the internet and communities like this, the emitter voltage in this configuration is supposed to be limited to the base voltage minus ~0.4V or so. Due to the 10K resistor, once the transistor is triggered, the base voltage will be under 12V, thus the emitter voltage will also be under 12V and the relay will not close. That's basically how my thoughts are at the moment and I need help spotting my mistakes.

Edit: I believe this question is being misunderstood. I'm not asking how a self-locking relay works. I'm asking how a 12V relay is being triggered when the voltage on its coil (from Q1 emitter) is bellow 12V (I'm checking it with a multimeter and theres a voltage drop across the transistor).

• Is it okay to have relay between +12V and the collector? Jan 27, 2020 at 0:16
• The circuit you are showing is a self-locking relay circuit. The preceding page show exactly how this works: Jan 27, 2020 at 0:22
• you have the wrong type of switches on the doors and windows ... they should have normally closed contacts (NC) ... the caption should say switches deactivated ... Jan 27, 2020 at 0:39
• About your edit: like I said in bold, the transistor is no longer useful once it is activated once. The whole circuit it uses is bypassed by the relay. The relay doesn't need a trigger, its coil is maintaining the relay on by itself. Jan 27, 2020 at 0:49
• @NorthernSage I'm not sure in all this mess if someone has said so, but it is usually the case that relays are specified to "engage" at 70% their rated voltage. So if you have a 12 V relay, it is usually designed so that it is guaranteed to operate at 8.4 V. (What it does below that value, but near it, is usually not specified.)
– jonk
Jan 27, 2020 at 1:35

It seems the problem can be solved by lowering the value of the resistor between the transistor's base and +12V source. The voltage drop still happens, although it's not so big to the point of not triggering the relay as before. I'll wait for a few days since getting a more detailed explanation on the issue would be much appreciated and mark my own answer as accepted in case nothing better is posted.

• Hmm, nowhere in your initial question is there written that the relay doesn't trigger? I will edit my post. The circuit you are making is a circuit where the relay will never close at all? Jan 27, 2020 at 1:20

This circuit is to make an alarm where the relay will never open until power is reset.

This is called a self-locking relay circuit. Basically, the transistor is only there to provide a one-time only current. This will shift the relay into the closed state, at which point the relay powers itself through it's own contacts. The transistor no longer has any incidence on the circuit since it's completely bypassed.

The page just before your schematic explains clearly how it works:

Source: Make Electronics by Charles Platt

To make it easy to understand:

When the switches are activated, the transistor gets a positive supply to the Base. Which then makes the transistor sink current through. The coil activates on the relay, which then bypasses the transistor and supplies 12V to each side of the transistor.