Problem with a simple circuit, why?

[Scimitar]

This is a simple timer for a courtesy light. It delays the turn off of the lamp by around 25 seconds. It works fine once in a while. I had it built on a breadboard for experimenting with values and I have gone through a good half a dozen transistors so far. I have looked at it and done a quick analysis of currents etc and can't see what is going on. It takes out the first PNP 2N3906 and sometimes its complementary too. I put it down to accidental shorts on the breadboard, but it seems that connecting the switch input to ground is doing the dirty. However, the Current is being strangled by the 22K surely?

I have built it up on stripboard now and intend to check it over tomorrow before testing it, so advice gratefully received. What am I missing please? Load is a 5W incandescent bulb and supply is 13.8V ish. Door switches are active low.

[Red to black]

It looks to me that because you have no current limiting resistor in E-C path in the PNP and/or likewise in the B-E path in the NPN you are pulling excessive current through both junctions above as the PNP is being turned on, the full supply is across both the turned on PNP with only the B-E junction of the NPN limiting any current through those paths.

PS. be interesting to see other people's Ideas here.

[Chrispy57]

Could it be that when the first transistor "switches" its collector is (virtually) pulled up to the +12V rail and so also the base of the NPN transistor, which would be driven very hard under those circumstances without any base current or base voltage limiting resistors? Both transistors are described as low current and low power devices.

Cheers
Chris

[MeerKat]

Theoretically the current through the collector-emitter of the 3906 and base-emitter of the 3904 even with a high gain 3906 should not cause problems but could be glitches are the problem?. I would insert a 1.2k resistor in the collector lead of the 3906. I would also put one in series with the diode as well to reduce the surge when the switch closes.

[kalee20]

There's nothing limiting the current through the 2N3906 and 2N3904, except the hfe of the '3906. With 22kΩ base resistor and hfe of 200, you could get 100mA flowing and 1.2W dissipation!

Connect a 4.7kΩ resistor between collector 3906 and base 3904, and your problems will be over!

[trsomian]

It would be much better with a suitable resistor between the collector of the left pnp and the base of the npn.

[paulsherwin]

Agreed, the circuit is overrunning the transistors. In addition to the 4.7k resistor in #5, I would change the transistors to something with a bit more grunt - BC327/337 or 2N2907/2222 should be OK.

[MotorBikeLes]

I am not going to attempt to analyse the circuit, but would make one observation. Both the first and last device are pnp types. My experience is that "like for like" they are far more failure prone than npn types. Could you "turn everything upside down", changing npn to pnp and vice virsa?
les.

[Keith956]

A quick back of envelope calc as follows:

Load is 5W so current in TIP32C is about 5/(12-4) = 0.625A (datasheet VCEsat is 4V at Ic=1A)

hFE of TIP32C is min 25 at Ic=1A, so Ib = 0.04A

So the 2K2 resistor is too high, to supply the 40mA it needs to be (12-Vbe-VCEsat)/0.04 ie R ~= 220 ohms

The 40mA base current os supplied by the 2N3904 which is within its spec of Icmax=200mA. The min hFE is 60 at 50mA so the base current of the 2N3904 Ib = .04/60 = 0.67mA

Assuming 2N3906 VCEsat is about 0.5V and Vbe of the 2N3904 is 0.8V, then the resistor between C of 2N3906 and B of 2N3904 needs to be (12-0.5-0.8)/0.67e-3 ~= 15K

Might want to double check the calcs in case I've made an error!

[The Philpott]

It's likely that i have a spare 'off the shelf' delay circuit of this type Scimitar, if it's simpler for you to opt for that you can have it for postage cost. It simply needs popping out of the vauxhall luminaire and the spade terminals labelled for your application.

Dave

[Herald1360]

As above for problem comments.

Why not use a Schmitt triger circuit to drive the TIP32C?

See attached for suggestion- works fine in simulation!

[lesmw0sec]

Originally Posted by Red to black

It looks to me that because you have no current limiting resistor in E-C path in the PNP and/or likewise in the B-E path in the NPN you are pulling excessive current through both junctions above as the PNP is being turned on, the full supply is across both the turned on PNP with only the B-E junction of the NPN limiting any current through those paths.

PS. be interesting to see other people's Ideas here.

I would support this observation.

[Scimitar]

Thanks for all the help and advice. It's obvious once it is pointed out of course! As I get older, I swear I get less intelligent. I will run up a few tests a little later and report back.

These have to be built as direct replacements into the original can, so that will be another bit of fun for me!

[RF Burn]

The Base Emitter junction of the 3904 is just a diode, it will try to pass as much current as it can to limit Vbe to 0.65V, The 3904 will be fully saturated when the input is pulled to ground and hence the current will only be limited by the power supply!

Result (1) if the 3904 BE junction fails open then the 3906 will probably survive the experience.
Result (2) if the 3904 BE junction fails short then the 3906 will probably pop it's clogs too due to excessive Collector current or dissipation.
Which are exactly the results that the OP mentions

The circuit needs a resistor between the collector of the 3906 and the base of the 3904, say 22K which would limit the 3904 base current to about 0.5mA, plenty to saturate it, and make the circuit work! As an industrial engineer I would use a second resistor from base of the 3904 to ground, say another 22K just to soak up any leakage current through the 3906 when it is 'off' from causing the lamp to light dimly because of the very high gain of the 3904/TIP32 complementary darlington arrangement.

[stuarth]

RF Burn makes a good point. You need to limit the current through the first 2 transistors to stop them being destroyed, but you also need to think about leakage current.

I assume the circuit is always powered up, so any leakage current will drain the battery. Leakage through the middle transistor multiplied by the gain of the third transistor could be a significant drain on the battery, particularly with the hot weather we are enjoying at the moment. Electronics in the roof light could get very hot, and with a less than perfect battery, you might have difficulty starting the engine when you got back from your hols. Or when this lockdown is lifted.

Stuart

[Red to black]

I have read most responses so far, just glad my early "skim read" of the circuit diagram and the subsequent responses from other posters broadly agreed with my own thoughts and I am not too far wide of the mark

Although I had my "electrician's head" on at the time ie. following the current path, I forgot totally about the Hfe factor (Kalees reply) having an overall effect too!
I also could have been a bit clearer with my wording too, but I knew what I meant

I purposely didn't try to analyse the circuit in any depth for three reasons, i) because I was never a circuit designer in the first place, ii) there are far cleverer people on here that would have ripped me to shreds on any shortcomings on my maths which is rusty at best , and iii) it was just an off the top of my head observation of the circuit topology used.
A fourth could be added here too, namely I was too lazy !!

In response to Les, I suppose the OP needed an 'active low input' and a PNP input device provides that easily. I know that another negation by swapping over devices polarity wise would have provided that condition, however it may have considerably increased current consumption (read battery drain) and from reading subsequent post justifies his choice of circuit elements.

Herald (Chris) does have a good point about Schmitt triggers, it is better to have cleaner (somewhat defined) transition points and a bit of hysteresis thrown in for good measure in this type of application, however as the OP has indicated in a later post then space is at a premium.

[Scimitar]

Well so far it has taken out a 3904 again, so busy trying to find out why.

[Mooly]

Here is your circuit with a 5k added to safeguard the 2N3904. The door closes 10 second in. The final PNP output is low on gain and doesn't saturate (bulb dependent of course).

How about using a FET instead of all this?

[Mooly]

And a single FET. This switches fairly cleanly and delivers the full supply to the load.

[duncanlowe]

Originally Posted by Mooly

And a single FET. This switches fairly cleanly and delivers the full supply to the load.

Using a FET has the advantage of less dissipation in the device (so long as you choose an appropriate RDSon). Bipolar transistors always have some volt drop across the junction that you can't get over.

The other thing to remember about a circuit like this, is the cold resistance of a bulb is much lower than when it's lit, so the initial current through the final device, will be much higher than the normal running current.