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Hey, I just used one (OK a 3906) and a mosfet wouldn't have worked.On Fri, 28 Oct 2011 07:01:34 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:
On Oct 28, 3:44 am, Jon Kirwan <j...@infinitefactors.org> wrote:
On Thu, 27 Oct 2011 15:57:39 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
First, I'm a transistor idiot.
I wanted to change a 5V 1 MHz clock signal to 12V.
I put this together,
+12V--10kohm----+----Vout
|
|/
in--5V 1MHz-1kohm-+---| 2N3904
| |\
| |
+--1k-+
Gnd
Worked, but it took forever (~100 nS) to turn off.
The base was driven to near 800 mV. When the square wave clock
shut off the base just drifted down slowly for 100ns and then
dropped. The base to ground resistor was added to try and
turn the base off quicker. It barely helped.
If I don't drive it as hard will it turn off faster?
Make the input to base resistor ~5k
to give 0.8V at the base with 5V in?
Thanks
George H.
Oh feel free to suggest a better circuit too!
First thing that crosses my mind is the roughly 4mA of base
drive and the roughly 1mA of collector current. That is hard
saturation and will place a lot of electrons in the collector
drift region (I think.) You might want to run it closer to a
beta of 20 or 30, not 0.25. Also, the other thing I note is
that you are getting 100ns turn off (no surprise) and running
at 1MHz. So 10% turnoff isn't good for you. You want less?
From your numbers, the two 1k resistors (moving to off state)
are in parallel at provide 500 ohms to ground for that charge
to leave. Have you tried a 'speed up' capacitor across your
base drive resistor? I've used them with some success. And
they go back a LONG TIME. I think before Baker clamps.
I'm not that much of a designer. But this is basics. So
that's me. Anyway, here is a shot at it.
Take your above circuit. Get rid of the two 1k resistors and
replace them with something that delivers on the order of a
20 beta. (It is a 2N3904, after all. It can do well.) So
1/20th of 1mA or 20,000 times the 4V or so differential you
will have, less the Vbe of the BJT. That would get near 80k
ohm, but drop it to 56k or thereabouts. At 1mA Ic, you need
about Vbe=0.65V (or less, really.) So 0.65/5*56k is around
7k pull down. But that is R_tot and at 0.65V and 1/20mA
base, this is about R_base=13k for the BJT so the 7k is after
taking that into account, too. This means the pull down is
about the same -- 13k-ish. I'd use 15k. Then put a speed up
cap across the 56k. I think most BJTs have small pF to worry
about. So make it a 12-15pF. Something about that size.
Then see how it goes. You won't be saturating the hell out
of it, anyway. But I've no idea what you are driving,
either.
Just a hobbyist thought.
Jon- Hide quoted text -
- Show quoted text -
I'll give it a quick try. (Reducing the base current and voltage...
KRW suggested the same thing.)
I've reduced the collector resistor to 1k so my collector current is a
bit more now.. but I'll scale things accordingly.
George H.
Use a mosfet! 2N3904s belong in museums next to the 6SN7s.
Yeah, I'll have to order some, I've only got power mosfet's on hand.On Fri, 28 Oct 2011 07:01:34 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
On Oct 28, 3:44 am, Jon Kirwan <j...@infinitefactors.org> wrote:
On Thu, 27 Oct 2011 15:57:39 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
First, I'm a transistor idiot.
I wanted to change a 5V 1 MHz clock signal to 12V.
I put this together,
+12V--10kohm----+----Vout
|
|/
in--5V 1MHz-1kohm-+---| 2N3904
| |\
| |
+--1k-+
Gnd
Worked, but it took forever (~100 nS) to turn off.
The base was driven to near 800 mV. When the square wave clock
shut off the base just drifted down slowly for 100ns and then
dropped. The base to ground resistor was added to try and
turn the base off quicker. It barely helped.
If I don't drive it as hard will it turn off faster?
Make the input to base resistor ~5k
to give 0.8V at the base with 5V in?
Thanks
George H.
Oh feel free to suggest a better circuit too!
First thing that crosses my mind is the roughly 4mA of base
drive and the roughly 1mA of collector current. That is hard
saturation and will place a lot of electrons in the collector
drift region (I think.) You might want to run it closer to a
beta of 20 or 30, not 0.25. Also, the other thing I note is
that you are getting 100ns turn off (no surprise) and running
at 1MHz. So 10% turnoff isn't good for you. You want less?
From your numbers, the two 1k resistors (moving to off state)
are in parallel at provide 500 ohms to ground for that charge
to leave. Have you tried a 'speed up' capacitor across your
base drive resistor? I've used them with some success. And
they go back a LONG TIME. I think before Baker clamps.
I'm not that much of a designer. But this is basics. So
that's me. Anyway, here is a shot at it.
Take your above circuit. Get rid of the two 1k resistors and
replace them with something that delivers on the order of a
20 beta. (It is a 2N3904, after all. It can do well.) So
1/20th of 1mA or 20,000 times the 4V or so differential you
will have, less the Vbe of the BJT. That would get near 80k
ohm, but drop it to 56k or thereabouts. At 1mA Ic, you need
about Vbe=0.65V (or less, really.) So 0.65/5*56k is around
7k pull down. But that is R_tot and at 0.65V and 1/20mA
base, this is about R_base=13k for the BJT so the 7k is after
taking that into account, too. This means the pull down is
about the same -- 13k-ish. I'd use 15k. Then put a speed up
cap across the 56k. I think most BJTs have small pF to worry
about. So make it a 12-15pF. Something about that size.
Then see how it goes. You won't be saturating the hell out
of it, anyway. But I've no idea what you are driving,
either.
Just a hobbyist thought.
Jon- Hide quoted text -
- Show quoted text -
I'll give it a quick try. (Reducing the base current and voltage...
KRW suggested the same thing.)
I've reduced the collector resistor to 1k so my collector current is a
bit more now.. but I'll scale things accordingly.
George H.
Use a mosfet! 2N3904s belong in museums next to the 6SN7s.
John- Hide quoted text -
- Show quoted text -
OK, That's interesting. How does that help get the the charge out ofIn addition to the other suggestions (lowering Rcollector
and reducing base drive), you might consider
adding an emitter resistor, 0.05xRcollector, in parallel with a
capacitor, so the input can pull charge out of
the base. It's less expensive than a Baker clamp.
If the stage driving this transistor is low impedance in the low level, it'sOn Oct 30, 5:01 pm, whit3rd <whit...@gmail.com> wrote:
In addition to the other suggestions (lowering Rcollector
and reducing base drive), you might consider
adding an emitter resistor, 0.05xRcollector, in parallel with a
capacitor, so the input can pull charge out of
the base. It's less expensive than a Baker clamp.
OK, That's interesting. How does that help get the the charge out of
the base-collector junction? (From the limited reading I was doing
over the weekend that seems to be the cause of the slow time response
due to saturation.) But I'll give it a try...
OK the emitter resistor just keeps everything a bit above ground, andOn Mon, 31 Oct 2011 07:36:48 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
On Oct 30, 5:01 pm, whit3rd <whit...@gmail.com> wrote:
In addition to the other suggestions (lowering Rcollector
and reducing base drive), you might consider
adding an emitter resistor, 0.05xRcollector, in parallel with a
capacitor, so the input can pull charge out of
the base. It's less expensive than a Baker clamp.
OK, That's interesting. How does that help get the the charge out of
the base-collector junction? (From the limited reading I was doing
over the weekend that seems to be the cause of the slow time response
due to saturation.) But I'll give it a try...
If the stage driving this transistor is low impedance in the low level, it's
another path to sweep charge out of the base. You don't want the charge swept
out the emitter because that requires Beta(+1)*charge. However, this won't
often be as good as not allowing the saturation charge to build up in the base
(Schottky clamp) in the first place.
That's the way I see it. It increases the output impedance (and low level) ofOn Oct 31, 12:46 pm, "k...@att.bizzzzzzzzzzzz"
k...@att.bizzzzzzzzzzzz> wrote:
On Mon, 31 Oct 2011 07:36:48 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:
On Oct 30, 5:01 pm, whit3rd <whit...@gmail.com> wrote:
In addition to the other suggestions (lowering Rcollector
and reducing base drive), you might consider
adding an emitter resistor, 0.05xRcollector, in parallel with a
capacitor, so the input can pull charge out of
the base. It's less expensive than a Baker clamp.
OK, That's interesting. How does that help get the the charge out of
the base-collector junction? (From the limited reading I was doing
over the weekend that seems to be the cause of the slow time response
due to saturation.) But I'll give it a try...
If the stage driving this transistor is low impedance in the low level, it's
another path to sweep charge out of the base. You don't want the charge swept
out the emitter because that requires Beta(+1)*charge. However, this won't
often be as good as not allowing the saturation charge to build up in the base
(Schottky clamp) in the first place.
OK the emitter resistor just keeps everything a bit above ground, and
that adds a bit of 'push' when the base is grounded?
Its purpose is to keep the base-collector junction from forward biasing. Yes,The Schottky does this too.. well at least keeps the collector further off ground..
(yeah, you know all this.)
I don't see how it will help, in this case. There isn't the charge that canI found a bag of 2N7000's in a drawer
today. That and a 1k drain resistor worked great.
Should I put a bit of R in gate lead?