this is an actual circuit that is acutally functioning properly except for the current consumption when pulses stop coming in. I placed a real 1 ohm circuit between the junction of pin 2(input) and the base of the 2907 pnp bipolar transistor. To my astonishment, when pulses stop going into the pin 2 (which is misdrawn as pin 3) my meter reads 78mV across the 1 ohm resistor that goes to the base of the PNP transistor. I can't imagine that there's anything wrong with the transistor, since the circuit works so well. The faulty drawing I made is probably causing endless confusion, pins 2 and 3 swiched, but in real life the base to the pnp (input junction) is at pin 2. When an incoming, a 0 volt pulse hits pin 2/transistor junction, it shorts the timing capacitor (.1 uF) to ground, resetting the timing cycle. So the net effect appears to be that the PNP base, with 12 volts applied, somehow draws 78mA from one of the other leads of the transistor, I'm not sure which. I could use more help with this, and sorry about that drawing. Rich

 

I'll try to re-check the connection of the PNP, when I first installed it I was unsure of the pin-out, it seemed a little odd, but the circuit worked great, so I figured I got it right. It's an SOT something something package, that I've never seen before and I had to look at it several times becuase the manufacturer said the base lead was on the same side of the package as the collector.

 

I suspect that you will trace the excess current to the misconnecting of the SOT-23 transistor package. To be definite about the pinout you will need to consult the manufacturer's datasheet for the device for the correct pinout.

hgmjr

 

Is there a voltmeter test I can perform on this PNP 2n2907 (i have extras)that will tell me the pinout? I'm not confident in the drawing on the datasheet.

 

If your voltmeter has a diode test setting you can identify the base but I don't think it will help you to differentiate the collector from the emitter.

Once you know the base then you can put the device in the circuit and then if the current is very low then you are good. If the current is elevated then reverse the connection between the collector and emitter and the current should drop to a low value.

hgmjr

 

Ah ha! I was hoping it was the transistor. An op-amp running in open loop is not recommended, but it should not be the culprit here.
Going way back to my school days, I believe you could do a simple diode test for a simple BJT.
You can forward bias the BE and BC junctions with a DMM. They both should read ~0.7V for silicon. Reverse biasing the BE and BC junctions should read ~3V IIRC.

 

I put my radio shack DMM volt meter on diode check
I put the black (com) lead on the manufacurer's indication for base
I put the red (+) lead on the manufacurere indication for emitter
I got .599 volts
Then:
I put the black (com) lead on the manufacurer's indication for base
I put the red(+) lead on the manufacturer's indication for collector
I got .589 volts
I reversed polarity of both of these tests, and got no reading (OFF)
Is this right?

 

Yep that sounds about right.

Now you have a firm grip on the base. Take your best shot at collector and emitter then give it a shot. You have a 50/50 chance of getting it right the first guess.

If you get it right the current will drop to a very low value. If you guess wrong you just reverse the emitter and collector connection and the current will drop to less than a milliamp.

hgmjr

 

both of the testers I have show .OF when I connect either polarity between emitter collector.

 

That is normal.

That is why I said there was not an easy way to differentiate the collector and the emitter with certainty using the diode test setting on a voltmeter.

You will just have to install the device in the circuit to make a final determination of collector and emitter.

hgmjr

 

I re installed the PNP transistor, and measured the voltages between the leads, when there were no pulses going into what I think is the base of the PNP.
base to collector, +8.45volts
Base to Emitter, +7.95volts
Emitter to collector, .5 volts.
I'm not sure what the state of the discharge transistor in the 7555 is, but it seems that the external timing capacitor should be discharged. That seems to indicate that the 7555's discharge pin (7) is grounded through the 7555. Thus the Emitter and Collector of my PNP transistor are also both grounded, correct? That leaves the base considerably (8V) more positive than both the emitter and collector. Could that explain the 78mA current at what appears to be the base?

 

The suspicious measurement is the base to emitter voltage. This voltage should be around 0.6 to 0.7 volts with the emitter being more positive than the base. This would correspond to a forward biased base emitter. For the base to be more positive than the emitter in a PNP transistor would be the sign of a reversed biased base-emitter junction. When you reverse bias a base emitter junction the junction usually zeners and starts to conduct at around 5 volts.

hgmjr

 

If I had to guess at what is going on, I would say that it is likely that the collector and the emitter have been reverse connected in the circuit.

hgmjr

 

Can you confirm this for me?

You stated the output pin (pin 3) was essentially open and not connected to anything and your diagram actually is pin 2 (trig) vice pin 3.

This places the base of the transistor and the trigger input at the same point.

If my reading is correct, I'm having a difficult time understanding why you even have the transistor in the circuit.

 

Yes pin 3 is actually disconnected at this time.

Pin 2 actually has the base of the pnp transistor connected to it.

It's a missing pulse detector. The PNP transistor shorts the timing capacitor of the TLC555 to ground, before it can complete the timing cycle. That prevents the pin 3 from going low, where it can sink the load connected to +V and pin 3. Works good, except for that one little problem.

 

The suspicious measurement is the base to emitter voltage. This voltage should be around 0.6 to 0.7 volts with the emitter being more positive than the base. This would correspond to a forward biased base emitter. For the base to be more positive than the emitter in a PNP transistor would be the sign of a reversed biased base-emitter junction. When you reverse bias a base emitter junction the junction usually zeners and starts to conduct at around 5 volts.

hgmjr

I got about 5 volts drop, so I'll try to reverse the connections.

 

Rich,

I don't recall seeing a missing pulse detector configured in such a way as you have done. Typically you see them like the one's illustrated at http://home.cogeco.ca/~rpaisley4/LM555.html#22

 

That's an awesome page, there's got to be something that will work. I figured out that all I needed to change was adding a 14k resistor between right before the base of the PNP, and it still works great, but it doesn't increase the power consumption when it changes state!
Thanks again, Rich