You're all out of your fucking minds... (was 2N2222 transist

[John Fields]

Unless I'm seriously deranged, the OP's got the motor in series with the
transistor's base, not its collector where it should be.



^ ^ Vcc =25V
! !
100k 220ohm
! !
MotoR !
! /
--------B---
\
!
GND


OP:

The 2N2222 should have absolutely no problem driving a 100mA load as
long as it's wired right and you drive it hard enough, like this:


+25V
|
+--------+
| |K
[MOTOR] [1N4001]
| |
+--------+
|
C
+5V>---[390R]----B 2N4401
E
|
GND

BUT... you'll be running the 2N2222 pretty close its maximum
dissipation, so I'd substitute a 2N4401 to eliminate that little
problem.

That should also work OK if you drove the 390 ohm resistor with the
output of an AND gate like one of the ones in a 74HC08.

--
John Fields

 

[John Fields]

On Tue, 24 Feb 2004 14:23:44 -0800, Jamie
<jamie_5_not_valid_after_5_Please@charter.net> wrote:

my charts tell me that a 2N4401 is a 600 Ma while a
2N2222 can do 1.0 amp ?
http://www.fairchildsemi.com/ds/2N/2N4401.pdf
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf
you will notice that a PN2222 will handle 1.0 amp cont.
that is just my observations

---
Learn to read, moron.

1. The OP said 2N2222. Not 2N2222A, not PN2222A, 2N2222.

2. I said dissipation, not current.

--
John Fields

 

[Ian Bell]

John Fields wrote:

Unless I'm seriously deranged, the OP's got the motor in series with the
transistor's base, not its collector where it should be.

I think someone else already pointed that out.


snip

That should also work OK if you drove the 390 ohm resistor with the
output of an AND gate like one of the ones in a 74HC08.

Provided the HC08 can source around 10mA.

Ian

 

[Jamie]

my charts tell me that a 2N4401 is a 600 Ma while a
2N2222 can do 1.0 amp ?
http://www.fairchildsemi.com/ds/2N/2N4401.pdf
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf
you will notice that a PN2222 will handle 1.0 amp cont.
that is just my observations

OP:

The 2N2222 should have absolutely no problem driving a 100mA load as
long as it's wired right and you drive it hard enough, like this:


+25V
|
+--------+
| |K
[MOTOR] [1N4001]
| |
+--------+
|
C
+5V>---[390R]----B 2N4401
E
|
GND

BUT... you'll be running the 2N2222 pretty close its maximum
dissipation, so I'd substitute a 2N4401 to eliminate that little
problem.

That should also work OK if you drove the 390 ohm resistor with the
output of an AND gate like one of the ones in a 74HC08.

 

[John Fields]

On Tue, 24 Feb 2004 22:04:29 +0000, Ian Bell <ian@redtommo.com> wrote:

John Fields wrote:

Unless I'm seriously deranged, the OP's got the motor in series with the
transistor's base, not its collector where it should be.

I think someone else already pointed that out.


snip


That should also work OK if you drove the 390 ohm resistor with the
output of an AND gate like one of the ones in a 74HC08.


Provided the HC08 can source around 10mA.

---
With the beta for a 2N4401 at a minimum of 100, it won't really have to,
even though beta's being "forced" to 10.

--
John Fields

 

[Bob]

"John Fields" <jfields@austininstruments.com> wrote in message
news:fj0n30heccdj206enb096cordm3v5b9l6v@4ax.com...

Unless I'm seriously deranged, the OP's got the motor in series with the
[snip]

Do you kiss your oscilloscope with that mouth?

Bob

 

[Ian Bell]

John Fields wrote:

On Tue, 24 Feb 2004 22:04:29 +0000, Ian Bell <ian@redtommo.com> wrote:

John Fields wrote:

Unless I'm seriously deranged, the OP's got the motor in series with the
transistor's base, not its collector where it should be.

I think someone else already pointed that out.


snip


That should also work OK if you drove the 390 ohm resistor with the
output of an AND gate like one of the ones in a 74HC08.


Provided the HC08 can source around 10mA.

---
With the beta for a 2N4401 at a minimum of 100, it won't really have to,
even though beta's being "forced" to 10.

Unfortunately it will because the base is being fed thru a 390 ohm resistor.

Ian

 

[John Fields]

On Wed, 25 Feb 2004 09:39:05 +0000, Ian Bell <ian@redtommo.com> wrote:

John Fields wrote:

On Tue, 24 Feb 2004 22:04:29 +0000, Ian Bell <ian@redtommo.com> wrote:

John Fields wrote:

Unless I'm seriously deranged, the OP's got the motor in series with the
transistor's base, not its collector where it should be.

I think someone else already pointed that out.


snip


That should also work OK if you drove the 390 ohm resistor with the
output of an AND gate like one of the ones in a 74HC08.


Provided the HC08 can source around 10mA.

---
With the beta for a 2N4401 at a minimum of 100, it won't really have to,
even though beta's being "forced" to 10.


Unfortunately it will because the base is being fed thru a 390 ohm resistor.

---
Yes, you're right. Sloppy of me; thanks for the reality check!-)

Looking at Figure 16 of On Semi's data sheet for the 2N4401 at

http://www.onsemi.com/pub/Collateral/2N4401-D.PDF

Leads us to believe that for 2mA into the base we can expect a Vce(sat)
of about 0.2V for an Ic of 100 mA, so that's a collector dissipation of
about 200mW, which isn't bad and ought to keep the transistor pretty
cool when it's fully on.

Next, on Figure 17, we see that when Ic = 100 mA the Vbe @ Vce=10 curve
intercepts VOLTAGE (VOLTS) at about 0.75 V, so if we can get 2mA out of
the CMOS gate and still have enough voltage left over to put about 0.75
V across the B-E junction that'll push the 2 mA through it and let the
collector sink the 100 mA we need to run the motor.

Looking at Philips' High Speed CMOS Data Book, we find that for a
standard output sourcing 2 mA from a Vcc of 4.5 V we can expect a worst
case output voltage of about 4 V at 25°C.

Looking at our circuit again, and now realizing that we need to put at
least 2 ma into the base in order to saturate the C-E junction to the
point where we'll get that nice 200 mW dissipation out of the collector
with 100 mA going through the motor, we wind up with:

Rb = (Vin-Vbe)/Ib = (4.0V - 0.75V)/0.002A = 1625 ohms.

The closest 5% is 1600 ohms and that should work OK.


Sooo....

+25V
|
+--------+
| |K
[MOTOR] [1N4001]
| |
+--------+
|
C
Vin>---[1k6]-----B 2N4401
E
|
GND


--
John Fields

 

[John Fields]

On Wed, 25 Feb 2004 00:45:50 GMT, "Bob" <nimby1_not_spmmm@earthlink.net>
wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:fj0n30heccdj206enb096cordm3v5b9l6v@4ax.com...
Unless I'm seriously deranged, the OP's got the motor in series with the
[snip]


Do you kiss your oscilloscope with that mouth?

---
My oscilloscope doesn't have a mouth.

--
John Fields

 

[Ian Bell]

John Fields wrote:

snip

Unfortunately it will because the base is being fed thru a 390 ohm
resistor.

---
Yes, you're right. Sloppy of me; thanks for the reality check!-)

Looking at Figure 16 of On Semi's data sheet for the 2N4401 at

http://www.onsemi.com/pub/Collateral/2N4401-D.PDF

Leads us to believe that for 2mA into the base we can expect a Vce(sat)
of about 0.2V for an Ic of 100 mA, so that's a collector dissipation of
about 200mW, which isn't bad and ought to keep the transistor pretty
cool when it's fully on.

200mW dissipation seems like a good design target to me. OTOH Fig.16 is
just a typical figure and beta in any region is well known to vary widely
from device to device. It is quite likely that in many parts a 2mA base
currrent would give only 0.4V vce(sat) which would double the dissipation.
As a rule it is wise to assume a beta of no more than 20 in the saturation
region which means I would design for a 5mA base current.

Next, on Figure 17, we see that when Ic = 100 mA the Vbe @ Vce=10 curve
intercepts VOLTAGE (VOLTS) at about 0.75 V, so if we can get 2mA out of
the CMOS gate and still have enough voltage left over to put about 0.75
V across the B-E junction that'll push the 2 mA through it and let the
collector sink the 100 mA we need to run the motor.

Looking at Philips' High Speed CMOS Data Book, we find that for a
standard output sourcing 2 mA from a Vcc of 4.5 V we can expect a worst
case output voltage of about 4 V at 25°C.

I don't have this data sheet to hand - what happens if we want to source
5mA?

Ian

 

[John Fields]

On Wed, 25 Feb 2004 18:55:32 +0000, Ian Bell <ian@redtommo.com> wrote:

John Fields wrote:

snip


Unfortunately it will because the base is being fed thru a 390 ohm
resistor.

---
Yes, you're right. Sloppy of me; thanks for the reality check!-)

Looking at Figure 16 of On Semi's data sheet for the 2N4401 at

http://www.onsemi.com/pub/Collateral/2N4401-D.PDF

Leads us to believe that for 2mA into the base we can expect a Vce(sat)
of about 0.2V for an Ic of 100 mA, so that's a collector dissipation of
about 200mW, which isn't bad and ought to keep the transistor pretty
cool when it's fully on.

200mW dissipation seems like a good design target to me. OTOH Fig.16 is
just a typical figure and beta in any region is well known to vary widely
from device to device. It is quite likely that in many parts a 2mA base
currrent would give only 0.4V vce(sat) which would double the dissipation.
As a rule it is wise to assume a beta of no more than 20 in the saturation
region which means I would design for a 5mA base current.



Next, on Figure 17, we see that when Ic = 100 mA the Vbe @ Vce=10 curve
intercepts VOLTAGE (VOLTS) at about 0.75 V, so if we can get 2mA out of
the CMOS gate and still have enough voltage left over to put about 0.75
V across the B-E junction that'll push the 2 mA through it and let the
collector sink the 100 mA we need to run the motor.

Looking at Philips' High Speed CMOS Data Book, we find that for a
standard output sourcing 2 mA from a Vcc of 4.5 V we can expect a worst
case output voltage of about 4 V at 25°C.

I don't have this data sheet to hand - what happens if we want to source
5mA?

---
Vout goes to about 3.8V with a 4.5V supply, so the base resistor would
go to about 600 ohms, and I'd agree with your choice of 5mA for the base
current.

I looked at the wrong curve last time, and with 2 mA out and a 4.5V
supply the output voltage would go to about 4.3 V, making the base
resistor about 1750 ohms, so I would have gone with 1600 anyway.

--
John Fields

 

[Jamie]

What ever, go ahead and twist the post all around.
i have seen many like you over the years
and i am sure i will see many more.. i'll keep with my own
practices thank you. they have served me well for the last 30 years
in this field.
But thank you for your inspiring comments, those always make
my day.





John Fields wrote:

On Tue, 24 Feb 2004 14:23:44 -0800, Jamie
jamie_5_not_valid_after_5_Please@charter.net> wrote:


my charts tell me that a 2N4401 is a 600 Ma while a
2N2222 can do 1.0 amp ?
http://www.fairchildsemi.com/ds/2N/2N4401.pdf
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf
you will notice that a PN2222 will handle 1.0 amp cont.
that is just my observations


---
Learn to read, moron.

1. The OP said 2N2222. Not 2N2222A, not PN2222A, 2N2222.

2. I said dissipation, not current.

 

[Ian Bell]

John Fields wrote:

snip

I don't have this data sheet to hand - what happens if we want to source
5mA?

---
Vout goes to about 3.8V with a 4.5V supply, so the base resistor would
go to about 600 ohms, and I'd agree with your choice of 5mA for the base
current.

I looked at the wrong curve last time, and with 2 mA out and a 4.5V
supply the output voltage would go to about 4.3 V, making the base
resistor about 1750 ohms, so I would have gone with 1600 anyway.

Isn't it about this point in the thread that someone pops up and says we
would be better using a logic FET anyway ;-)

Ian

 

[John Fields]

On Thu, 26 Feb 2004 13:27:45 +0000, Ian Bell <ian@redtommo.com> wrote:

John Fields wrote:

snip


I don't have this data sheet to hand - what happens if we want to source
5mA?

---
Vout goes to about 3.8V with a 4.5V supply, so the base resistor would
go to about 600 ohms, and I'd agree with your choice of 5mA for the base
current.

I looked at the wrong curve last time, and with 2 mA out and a 4.5V
supply the output voltage would go to about 4.3 V, making the base
resistor about 1750 ohms, so I would have gone with 1600 anyway.


Isn't it about this point in the thread that someone pops up and says we
would be better using a logic FET anyway ;-)

---
Don't put beans in your nose?^)

--
John Fields

 

[John Fields]

On Thu, 26 Feb 2004 02:00:29 -0800, Jamie
<jamie_5_not_valid_after_5_Please@charter.net> wrote:

What ever, go ahead and twist the post all around.

---
You fucked up, you got caught, you're a moron. Get over it.

i have seen many like you over the years
and i am sure i will see many more..

---
Keep on fucking up and I'm sure you will in the future as you amit you
have in the past.
---

i'll keep with my own practices thank you. they have served me well
for the last 30 years in this field.

---
I'm sure. When you're cleaning toilets it doesn't make any difference
whether you know the difference betweena watt and and an ampere.
---

But thank you for your inspiring comments, those always make
my day.

---
Get a life, you pathetic wretch.

--
John Fields