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I'm continuing development from this question.

The schematic given by Jasen worked as a charm but little did I know that was with the connector plugged out, as soon as I plugged the connector in I found out that it no longer works. After a bunch of tries and failures I determined that while I'm essentially connecting a pair of pins, another one must be disconnected(?).

The connector consists of 4 pins in total which I've labeled P1 to P4 below, which correspond as shown to the previous question:

P1: Unlock
P2: Common
P3: -
P4: Lock

Basically to unlock pins P1 and P2 need to be connected and P3 must be disconnected, or open. To lock pins P2 and P4 need to be connected and P1 needs to be open, where by open I mean there must be no connection between the plug and the socket.

This slightly complicates and adds on to the schematic from the previous question, here is what I came up with:

enter image description here

I see how it can probably be slightly confusing, labeled P1 to P4 are coming from the car and P1' to P4' go in the socket on the door, I've omitted P2' and P4' because they are directly connected to P2 and P4 (spliced connection).

I'm using an LM2902 as a logic not gate, since I already have it and I'm using only 1 channel of it so I decided I'll just use 2 more instead of buying a not gate. Basically the paths from P1 to P1' and P3 to P3' should be broken respectively as described in the scenarios above.

I also found out that the voltage between P2 and P3 (12V/0V) tells me the state of the lock but that's not relevant to the question - I'm just mentioning in case someone wonders.

Now there is definitely something wrong because only the lock part works and the BC556 that's connected to P1 and P2 immediately gets hot when it shouldn't especially as there is no signal on the unlock line to begin with. Cables around it got brown - that hot. After replacing it, that is. First one had decomposed with it's collector leg broken apart. I'm probably going to go ahead and use relays because I've spent way too much time on this problem, but I really wish to know what's wrong with my current plan because I just got excited about transistors..

PS: I apologize there are no labels on the components.
PS2: All 3.7k resistors were replaced with 3.3k resistors except the one on the state divider, but it probably doesn't really matter.

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  • \$\begingroup\$ why are you interrupting the P3 wire and not the P4 wire? \$\endgroup\$
    – Jasen
    Mar 13 '20 at 3:33
  • \$\begingroup\$ @Jasen because that is how i found it works, I just noticed there is a mistake in the question, fixing now \$\endgroup\$
    – php_nub_qq
    Mar 13 '20 at 6:27
  • \$\begingroup\$ @Jasen if i connect P1 and P2 while P3, notice not P4, is connected to the socket I get a short circuit. My cable justexploded. \$\endgroup\$
    – php_nub_qq
    Mar 13 '20 at 7:10
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BC337 and BC557 are rated for 800mA and 200mA respectively... how much current does the load (the door locks) want to draw? Might be able to measure it with a handheld meter.

Assuming this is an automotive door lock, one actuator will likely draw 1A or more to function. This means >4A is needed (as a short pulse) to lock and unlock all the doors simultaneously. This might be why these transistors burn up - they can't handle that much current. Not only that, only a short pulse is probably needed and both are never shut off.

Try a transistor like a TIP1xx, 2N3055, 2SC6144, etc. Something beefier which won't die the first time it's used or within 10 seconds. Mount it to a heat sink, at least while experimenting, to prevent them from getting too hot. Then, consider altering the circuitry to provide a pulse to lock and unlock, not a steady current. Do things one step at a time - get the transistors working to actually lock and unlock without failing first, then figure out how to "pulse" them appropriately.

Another thing to consider, since the door actuators are likely just solenoids, is "do they have built-in freewheeling diodes?" If they do not, then the moment current is removed, a large negative voltage spike might appear across them. If the door actuators are instead motorized mechanisms, then this might be slightly less of an issue, but still something to be concerned about. Transistors have a voltage limit, and if exceeded, they will short-out and die. Research freewheeling diodes and try to determine if you need these. Should be able to tell from a multimeter if this is happening by the voltage reading negative right after current is interrupted.

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  • \$\begingroup\$ Thank you for your time. As mentioned I already tested the "left" part of the circuit and it works fine. There is some kind of hydraulic mechanism on the locks which makes me think it's centrally operated and those are just signals, they do not source the current for the actual locking/unlocking. I measured the current with a clamp and it was 0.1A during locking but it's hall based and I'm not 100% sure about the accuracy. What I'm sure about though is that I messed up something in the circuit because this transistor should not get hot at all, there should not be any current flowing as is \$\endgroup\$
    – php_nub_qq
    Mar 12 '20 at 19:03
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LM2902 inputs are current sources (upto 250nA "input bias current"). say 200nA current from the LM2902 weakly activates the input transitor, this get amplified about 200 times to 40uA in the NPN. 40uA through the base of the PNP gets amplified another 200 times to about 8mA through the PNP

8mA at 12V is about 100mW so it heats up.

Connect a resistor from lock to ground and from unlock to ground, 10K or lower should be fine.

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  • \$\begingroup\$ Wait what, how are the input sources? I thought they sink current to ground, how else would it work as a comparator? \$\endgroup\$
    – php_nub_qq
    Mar 13 '20 at 9:00
  • \$\begingroup\$ It's in the data sheet where it says "input bias current" if you look at the schematic the LM2902 inputs is a PNP transistor bases, so they will produce current rather than accept it. ti.com/lit/ds/symlink/lm2902.pdf \$\endgroup\$
    – Jasen
    Mar 13 '20 at 9:02
  • \$\begingroup\$ Looks like this is where the confusion lies. But how would a circuit like this work if it did not sink current? \$\endgroup\$
    – php_nub_qq
    Mar 13 '20 at 12:41
  • \$\begingroup\$ Or this circuit which I'm using to determine whether the engine is running \$\endgroup\$
    – php_nub_qq
    Mar 13 '20 at 13:05
  • \$\begingroup\$ first circuit makes no assumptions, second circuit has an out of range input voltage, so anything could happen. \$\endgroup\$
    – Jasen
    Mar 13 '20 at 22:49

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