A 2N3906 Question this time

[Newbie]

Hello All

Thanks for all who replied to my 2N3904 question earlier. I
have one about a 3906 this time. The set up is below

--- 5 V
|
| /
|V
Gnd ------|
|\
\
|
V LED
---
|
\
/ 220 ohm
\
|
| Gnd

When I apply voltage to this circuit the LED does not come on and also
the 2N3906 seems to burn out. I'm pretty sure it does because the 0.7
volt diode drop no longer apears across the BC and BE junctions when I
take it out of the circuit and test it. If I put a 220 ohm resistor in
the base the circuit does seem to work.

I thought the base was pretty high impedance and the current going out
through it would be limited internally as most would go through the
collector. It would seem I am mistaken. Could anyone shed any light on
this? Would a 3904 NPN burn out in a similar situation (base connected
to ground with no resistor and 5V on the collector. I have tried this
and it did not seem to burn out - admittedly the LED+220ohm was before
the collector so this probably limited the current.

Is the LED+220ohm resistor after the collector a good way to configure
a PNP transistor to act as a switch (as in the diagram)? Or should I
have it close to the supply before the emitter?

Many Thanks - I would value your opinions
Dale

 

[Thomas C. Sefranek]

"Newbie" <NoSpam@DontEmailHere.com> wrote in message
news:hbt320p2jf47ulnhfgfa7gkft19ds6v7hd@4ax.com...

Hello All

Thanks for all who replied to my 2N3904 question earlier. I
have one about a 3906 this time. The set up is below

--- 5 V
|
| /
|V
Gnd ------|
|\
\
|
V LED
---
|
\
/ 220 ohm
\
|
| Gnd

When I apply voltage to this circuit the LED does not come on and also
the 2N3906 seems to burn out. I'm pretty sure it does because the 0.7
volt diode drop no longer apears across the BC and BE junctions when I
take it out of the circuit and test it. If I put a 220 ohm resistor in
the base the circuit does seem to work.

I thought the base was pretty high impedance

Why? It's a diode! It's impedance is determined by what current you allow
thru it.

and the current going out
through it would be limited internally as most would go through the
collector. It would seem I am mistaken. Could anyone shed any light on
this? Would a 3904 NPN burn out in a similar situation (base connected
to ground with no resistor and 5V on the collector.

No, the 3904 is an NPN and grounding the base amounts to no on bias.
(The transistor is off!)

I have tried this
and it did not seem to burn out - admittedly the LED+220ohm was before
the collector so this probably limited the current.

Is the LED+220ohm resistor after the collector a good way to configure
a PNP transistor to act as a switch (as in the diagram)? Or should I
have it close to the supply before the emitter?

Use a 39K ohm from the base to ground in your circuit above.

Many Thanks - I would value your opinions
Dale

--
*
| __O Thomas C. Sefranek WA1RHP@ARRL.net
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41, 448.625 MHz

http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

 

[petrus bitbyter]

"Newbie" <NoSpam@DontEmailHere.com> schreef in bericht
news:hbt320p2jf47ulnhfgfa7gkft19ds6v7hd@4ax.com...

Hello All

Thanks for all who replied to my 2N3904 question earlier. I
have one about a 3906 this time. The set up is below

--- 5 V
|
| /
|V
Gnd ------|
|\
\
|
V LED
---
|
\
/ 220 ohm
\
|
| Gnd

When I apply voltage to this circuit the LED does not come on and also
the 2N3906 seems to burn out. I'm pretty sure it does because the 0.7
volt diode drop no longer apears across the BC and BE junctions when I
take it out of the circuit and test it.

The emitter-base junction is a conducting diode in this schematic. It used
to have about 0.7V on its terminals. If you push 5V on it th cuurent will go
skihigh and destroy the transistor.

If I put a 220 ohm resistor in the base the circuit does seem to work.

That's still to low. You may not destroy the transistor anymore, but the led
will not light.

I thought the base was pretty high impedance and the current going out
through it would be limited internally as most would go through the
collector. It would seem I am mistaken. Could anyone shed any light on
this?

A conducting diode has a very low impedance. Nevertheless, the base input
of a transistor can have a pretty high impedance due to the feedback caused
by the load on the emitter. In this circuit you have no load at all.

Would a 3904 NPN burn out in a similar situation (base connected
to ground with no resistor and 5V on the collector. I have tried this
and it did not seem to burn out - admittedly the LED+220ohm was before
the collector so this probably limited the current.

A blocking diode has a high impedance. Moreover, a voltage <0.6V will make
the transistor switched off as well. It does not harm but you will have no
light either.

Is the LED+220ohm resistor after the collector a good way to configure
a PNP transistor to act as a switch (as in the diagram)? Or should I
have it close to the supply before the emitter?

I think the circuit you have in your mind is known an emitter-follower. So
you have to move the load (LED+220ohm) to the emitter. Other side of the
load to +5V and collector to ground. This way the load will limit the
emitter-base current and the LED will light. The LED will be switched off
when the base voltage rises above about 4.4V.

Using a NPN transistor the load should be connected to the emitter and the
ground, collector to +5V. A grounded base will keep the LED off, a plus 5V
on it will light it.

Many Thanks - I would value your opinions
Dale

petrus





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[John Popelish]

Newbie wrote:

Hello All

Thanks for all who replied to my 2N3904 question earlier. I
have one about a 3906 this time. The set up is below

--- 5 V
|
| /
|V
Gnd ------|
|\
\
|
V LED
---
|
\
/ 220 ohm
\
|
| Gnd

When I apply voltage to this circuit the LED does not come on and also
the 2N3906 seems to burn out. I'm pretty sure it does because the 0.7
volt diode drop no longer apears across the BC and BE junctions when I
take it out of the circuit and test it. If I put a 220 ohm resistor in
the base the circuit does seem to work.

I thought the base was pretty high impedance and the current going out
through it would be limited internally as most would go through the
collector. It would seem I am mistaken. Could anyone shed any light on
this? Would a 3904 NPN burn out in a similar situation (base connected
to ground with no resistor and 5V on the collector. I have tried this
and it did not seem to burn out - admittedly the LED+220ohm was before
the collector so this probably limited the current.

You have turned the NPN circuit upside down in every way but one. The
base to emitter diode junction is now connected directly across a
supply with nothing to set its current except for the supply
capability. The base gets connected to the same voltage as the
emitter to turn the transistor off, same as it was with the NPN
version. To turn the transistor on, you have to bring the base
emitter junction into forward bias. In the PNP case, this means
getting the base to be about .6 volts more negative than the emitter,
with enough current passing through this junction that beta times that
current is at least equal to the desired collector current. Providing
a pull down current of a milliamp or so should do this. So the base
should be connected to ground only through something like a 4.7k
resistor.

Is the LED+220ohm resistor after the collector a good way to configure
a PNP transistor to act as a switch (as in the diagram)? Or should I
have it close to the supply before the emitter?

Many Thanks - I would value your opinions
Dale

--
John Popelish