John Larkin a �crit :> 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. >Hey, I just used one (OK a 3906) and a mosfet wouldn't have worked. Maybe the 6SN7 would have worked, because I needed that grid current too at positive bias... -- Thanks, Fred.
Common emitter as level switcher
Started by ●October 27, 2011
Reply by ●October 29, 20112011-10-29
Reply by ●October 30, 20112011-10-30
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.
Reply by ●October 31, 20112011-10-31
On Oct 29, 1:04=A0pm, John Larkin <jjlar...@highNOTlandTHIStechnologyPART.com> wrote:> On Fri, 28 Oct 2011 07:01:34 -0700 (PDT), George Herold > > > > > > <gher...@teachspin.com> wrote: > >On Oct 28, 3:44=A0am, 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, > > >> > =A0 =A0 =A0 =A0 +12V--10kohm----+----Vout > >> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | > >> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |/ > >> > in--5V 1MHz-1kohm-+---| =A0 2N3904 > >> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | =A0 |\> > >> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 | =A0 =A0 | > >> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 +--1k-+ > >> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 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. =A0The base to ground resistor was added to try and > >> >turn the base off quicker. =A0It 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. =A0That is hard > >> saturation and will place a lot of electrons in the collector > >> drift region (I think.) =A0You might want to run it closer to a > >> beta of 20 or 30, not 0.25. =A0Also, the other thing I note is > >> that you are getting 100ns turn off (no surprise) and running > >> at 1MHz. =A0So 10% turnoff isn't good for you. =A0You 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. =A0Have you tried a 'speed up' capacitor across your > >> base drive resistor? =A0I've used them with some success. =A0And > >> they go back a LONG TIME. =A0I think before Baker clamps. > > >> I'm not that much of a designer. =A0But this is basics. =A0So > >> that's me. =A0Anyway, here is a shot at it. > > >> Take your above circuit. =A0Get rid of the two 1k resistors and > >> replace them with something that delivers on the order of a > >> 20 beta. =A0(It is a 2N3904, after all. =A0It can do well.) =A0So > >> 1/20th of 1mA or 20,000 times the 4V or so differential you > >> will have, less the Vbe of the BJT. =A0That would get near 80k > >> ohm, but drop it to 56k or thereabouts. =A0At 1mA Ic, you need > >> about Vbe=3D0.65V (or less, really.) =A0So 0.65/5*56k is around > >> 7k pull down. =A0But that is R_tot and at 0.65V and 1/20mA > >> base, this is about R_base=3D13k for the BJT so the 7k is after > >> taking that into account, too. =A0This means the pull down is > >> about the same -- 13k-ish. =A0I'd use 15k. =A0Then put a speed up > >> cap across the 56k. =A0I think most BJTs have small pF to worry > >> about. =A0So make it a 12-15pF. =A0Something about that size. > > >> Then see how it goes. =A0You won't be saturating the hell out > >> of it, anyway. =A0But 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. =A0(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 -Yeah, I'll have to order some, I've only got power mosfet's on hand. But honestly, at the moment the npn with diode is just fine. George H.
Reply by ●October 31, 20112011-10-31
On Oct 30, 5:01=A0pm, 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. =A0It'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... George H.
Reply by ●October 31, 20112011-10-31
On Mon, 31 Oct 2011 07:36:48 -0700 (PDT), George Herold <gherold@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.
Reply by ●October 31, 20112011-10-31
On Oct 31, 12:46=A0pm, "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. =A0How does that help get the the charge out of > >the base-collector junction? =A0(From the limited reading I was doing > >over the weekend that seems to be the cause of the slow time response > >due to saturation.) =A0But I'll give it a try... > > If the stage driving this transistor is low impedance in the low level, i=t's> another path to sweep charge out of the base. =A0You don't want the charg=e swept> out the emitter because that requires Beta(+1)*charge. =A0However, this w=on'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? The Schottky does this too.. well at least keeps the collector further off ground.. (yeah, you know all this.) I 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? George H.
Reply by ●November 1, 20112011-11-01
On Mon, 31 Oct 2011 18:41:19 -0700 (PDT), patricia herold <pmdherold@gmail.com> wrote:>On 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?That's the way I see it. It increases the output impedance (and low level) of the driver. It's not something I've ever done.>The Schottky does this too.. well at least keeps the collector further off ground.. >(yeah, you know all this.)Its purpose is to keep the base-collector junction from forward biasing. Yes, it also adds some to Vce(sat).>I 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?I don't see how it will help, in this case. There isn't the charge that can be swept out. I suppose it helps some to take Vgs negative.
