disobedient NPN BJT

Hey everyone, I have run into a very strange problem, and I am hoping you can enlighten me. I have tried to test the attached circuit. However when I do the BJT conducts even when the base is grounded. The blue values are measured. I am using the BJT as a switch and I know I have already pushed my BJT way past the max power. All other components are high power rated. But as the circuit is, it should not even be conducting, because Vce max is 40V. When I designed this high current circuit, I assumed 6.3V would drop on the diode, collector to emitter saturation voltage would be about a volt, and the rest would drop across the resistor. However it seems that about a volt is being dropped across the resistor and the rest across CE. Is this because my resistance is so small? How can I remedy this? I was thinking of just using a higher power rated BJT but I don't know if that will stop it from conducting when it is not supposed to. I don't really want to change my resistor or diode because I want high current. Thanks for the help.

 

Yes, this circuit should be off, therefore you have a bad component or a mistake in connecting things. However, why would you expect 6.3 volts across a forward biased diode?

"A volt" is being dropped across the resistor...20 ohms...that's .05 amps, not .25 amps.

Keep talking.

ps, "high current" is relative. In the last few days, several people have come here asking about 30 or 40 amps.

 

Your Zener needs to be reverse biased (cathode to + rail). When forward biased as you show, the voltage drop will be like a normal diode of 0.6 volts.

 

Fooled one of us. That's not a zener symbol. Still, it's forward biased, and that is less than a volt, zener or not.

 

Lets just say the voltages I mentioned earlier are true and the current is unknown. The design is on perf board and I cannot easily remove components. There are no shorts anywhere. Lets also say that the pictured diode is a "normal" diode with an unusually large forward bias potential.

 

If the Bjt is conducting, you can calculate Ic with this:

Ic = (Vcc-Vce-Vd)/R1 = (12V-1V-6.3V)/20 = 235mA.

I still don't understand why the diode would be dropping 6.3V. Is it a Zener? Something else?

I assume your Bjt is broken, or you connected it incorrectly if it is conducting when it shouldn't.

TheComet

 

A diode cannot have 6.3 volts across it when forward biased. Cannot.

So it is not a diode. You may have bought a diode but it is no longer such. It is broken.

Unless... how long since you checked the batteries in your digital meter? My meter reads low when the batteries are nearly dead but yours may work differently.

Also TheComet makes good points about the transistor. It too is broken.

 

Wolfgang,

It is the easiest thing in the world to miswire a circuit, we have all done it.

I recommend that you check carefully that the transistor is actually connected as shown and not reversed.

Then repost the circuit with accurate measurements and all component types, including the 'diode' and list which end is marked as cathode on your diode.

I say accurate measurements because the reported voltage drops add up to
greater than the 12 volts supply. Your current calculations have already been questioned.

 

Pull enough current throug it and it will. Only problem is that the current needs to be in tens of amps and the diode will melt very shortly after.

 

Are you maybe talking of 0.63 V instead?

 

I made it work with 2.8W BJTs

 

There is no path for the collector leakage current (Icbo), so it flows through the base-emitter causing the transistor to partially turn on.

 

The base-to-ground resistor is present, and it's only 680 ohms.

I think he has the collector and emitter swapped, as studiot suggested. If so, the EB junction will break down at around 6V, leaving Vec≈Veb+0.7V, where 0.7V is the now-forward-biased BC junction. Approximately 1mA will flow through the 680Ω resistor.

 

What gets me is ...

5V + 6.3 V only leaves 0.7 volts across the 20 ohm resistor or 0.035 A of current, considerably less than the 0.25 posted.

Now ... using the 0.25A posted, you'd have IR of 5V + Vce 5V leaving 2V acrossed a forward biased diode. Now, I've seen diodes drop almost 2V when 10 amps was put through them, but that would change the dynamics of the whole circuit as well.

Things just aren't adding up. The OP left off the Vbe voltage.

Why doesn't the OP post the component values, like their 6.3V diode?