W
Watson A.Name - \"Watt Su
Guest
I am *so* sick and tired of the idiots that 'design' a circuit like this
one http://circuitos.tripod.cl/schem/r95.gif
that use such a high resistance in the base that it's literally
_impossible_ for it to work, especially with a 2N3904. So I decided to
try an experiment with both the 2N4401 and the 2N3904, and instead of a
relay, use a 100 ohm resistor and a 10VDC source. The usual 12V relay
has a coil resistance of 120 to 150 ohms, so it needs about 100 mA at
12V to work at its optimum. Obviously it might work with less, but
that's not where it's at.
So I put a 100 ohm 1W resistor as the collector load, and 10VDC, and
then a 20k pot with 1k in series to limit the maximum base current. My
objective is to get most of the supply voltae to drop across the load
resistor with 0.2V or less across the transistor C to E.
I was sadly disappointed by the 2N3904. I had to crank the base current
up to 5.75 mA to get the Vce to .2V. Getting the 2N3904 to take 100 mA
is really pushing it. The beta or current gain drops off drastically at
that current, in this case, the gain is only 17. something. What I'm
getting at is that this is a very poor choice for a relay driver
transistor, and you couldn't use it in the circuit in that URL above,
because there would be only a volt or two across the relay, and the rest
across the transistor.
So I put the much heftier 2N4401 in the circuit and did the measurements
again. I adjusted the pot for .2V Vce, and it took only .93 mA base
current to get the 100 mA Ic. This is a gain of over a hundred at that
current, six times better than the puny 2N3904, and might be almost good
enough to get the circuit to work in that URL. But that circuit really
should have had a base resistor of 4.7k or less instead of 100k; with
the 100k, it will never work with either transistor, there's just not
enough base current. As a matter of fact, the 741 output can't go close
to the negative rail, so it really needs a couple diodes in series with
the output to drop a volt or two. A red LED works well, too. Better
yet, replace the 741 with an opamp that go to the negative rail.
So don't use a puny transistor pushed to its maximum to drive a relay.
Use the right transistor with the adequate amount of base current to get
it to drive the relay adequately. The last thing you want is a half
pulled in relay.
--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@
one http://circuitos.tripod.cl/schem/r95.gif
that use such a high resistance in the base that it's literally
_impossible_ for it to work, especially with a 2N3904. So I decided to
try an experiment with both the 2N4401 and the 2N3904, and instead of a
relay, use a 100 ohm resistor and a 10VDC source. The usual 12V relay
has a coil resistance of 120 to 150 ohms, so it needs about 100 mA at
12V to work at its optimum. Obviously it might work with less, but
that's not where it's at.
So I put a 100 ohm 1W resistor as the collector load, and 10VDC, and
then a 20k pot with 1k in series to limit the maximum base current. My
objective is to get most of the supply voltae to drop across the load
resistor with 0.2V or less across the transistor C to E.
I was sadly disappointed by the 2N3904. I had to crank the base current
up to 5.75 mA to get the Vce to .2V. Getting the 2N3904 to take 100 mA
is really pushing it. The beta or current gain drops off drastically at
that current, in this case, the gain is only 17. something. What I'm
getting at is that this is a very poor choice for a relay driver
transistor, and you couldn't use it in the circuit in that URL above,
because there would be only a volt or two across the relay, and the rest
across the transistor.
So I put the much heftier 2N4401 in the circuit and did the measurements
again. I adjusted the pot for .2V Vce, and it took only .93 mA base
current to get the 100 mA Ic. This is a gain of over a hundred at that
current, six times better than the puny 2N3904, and might be almost good
enough to get the circuit to work in that URL. But that circuit really
should have had a base resistor of 4.7k or less instead of 100k; with
the 100k, it will never work with either transistor, there's just not
enough base current. As a matter of fact, the 741 output can't go close
to the negative rail, so it really needs a couple diodes in series with
the output to drop a volt or two. A red LED works well, too. Better
yet, replace the 741 with an opamp that go to the negative rail.
So don't use a puny transistor pushed to its maximum to drive a relay.
Use the right transistor with the adequate amount of base current to get
it to drive the relay adequately. The last thing you want is a half
pulled in relay.
--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@