On Oct 27, 11:41=A0pm, "k...@att.bizzzzzzzzzzzz" <k...@att.bizzzzzzzzzzzz> wrote:> On Thu, 27 Oct 2011 19:07:56 -0700 (PDT), George Herold > > > > > > <gher...@teachspin.com> wrote: > >On Oct 27, 8:40 pm, "k...@att.bizzzzzzzzzzzz" > ><k...@att.bizzzzzzzzzzzz> 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! > > >> Tie a Schottky from the base (anode) to the collector (cathode). This =will> >> keep the transistor out of saturation and hopefully buy you the speed =you> >> need. Another alternative would be to use an N-FET there instead of th=e NPN.- Hide quoted text -> > >> - Show quoted text - > > >Great, I was wondering if a Fet would work better. =A0The only charge I > >have to worry about is whatever's on the gate capacitance? > > And drain. =A0The 10K pullup resistor has to charge whatever capacitance =is on> the drain (+ load). =A0You can reduce this, obviously at the cost of powe=r. =A0> > Also, you have a 1K base resistor, which is going to give you about 3mA o=f> base current ((5-.7)/1K). =A0Your collector current is only ~1mA (10V/12K=). This> is going to saturate the hell out of the transistor. =A0You might reduce =the> base drive an order of magnitude (or two). > > The problem is the base charge. =A0All that charge in the base needs to f=lush> out through the emitter, taking Beta times as much current from the colle=ctor> with it. =A0The Schottky keeps the transistor out of deep saturation > (technically saturation is Vbc positive) by bleeding current from the bas=e to> the collector. =A0 > > >I'll try the Bakers clamp/ Schottky first since there's already an npn > >'lashed' in place. > > Good plan. =A0But also try increasing the base resistor to at least 10K, =maybe> 50K. > > >using TTL circuitry seems like a nice retro touch. > > That's not a bad idea either, though I'm not sure there are 1G types with=15V> or 30V outputs. > > BTW, if the positive transition needs to be cleaner than the negative, tu=rn> the whole thing upside down and change the sex of the transistor.OK, I can see that, Thanks. George H. - Hide quoted text -> > - Show quoted text -George H.
Common emitter as level switcher
Started by ●October 27, 2011
Reply by ●October 28, 20112011-10-28
Reply by ●October 28, 20112011-10-28
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. (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.
Reply by ●October 28, 20112011-10-28
On Oct 28, 9:44=A0am, George Herold <gher...@teachspin.com> wrote:> On Oct 27, 10:45=A0pm, John Larkin > > > > > > <jjlar...@highNOTlandTHIStechnologyPART.com> 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! > > > 10K is kinda big for the collector pullup. 10K * 10 pF =3D 100 ns, so > > even the scope probe will slow it down. Try 1K. > > Great that's much nicer! =A0The base to collector zener got rid of the--oops that's a base to collector Schottky diode----> 100ns delay, but it was till taking another 100nS to get up to 12 > Volts. =A0(looking with a ~16pF 'scope probe) > > A series inductor > > > might be fun, too. Try L =3D 0.5 * R^2 * C or thereabouts. > > Ouuu.. that's getting a little too complicated. (..tune out a bit of > the C with the L.) =A0I just wanted to add a some voltage to the clock. > And now it's a growing rat's nest on the side of the prototype.http://ima=geshack.us/photo/my-images/828/dscf0043s.jpg/> > > > > A small cap across the base drive 1K is a classic speedup. > > Yeah, I got that. =A0The Bakers clamp 'name' from Jamie was golden. > There was a reference in the first editon of AoE (which lives at > home.) and the 'speed up' cap was right next to the Bakers clamp. =A0At > work today I coudn't find this in the second edition. > > And make > > > the Thevenin voltage drive into the base more like 0,1.2 than the > > existing 0,2.5, namely reduce the lower base resistor, 330 ohms maybe. > > > The 3904 will saturate and turn off slowly. A small mosfet, like a > > 2N7000, with a 1K pullup would be faster. > > Thanks George H. > > > > > > > John- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
Reply by ●October 28, 20112011-10-28
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.Try 100pF, then. Jon
Reply by ●October 29, 20112011-10-29
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. John
Reply by ●October 29, 20112011-10-29
On Sat, 29 Oct 2011 10:04:56 -0700, John Larkin <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:>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.At 5-10x the cost, you just lost 1000 Joerg points.
Reply by ●October 29, 20112011-10-29
"krw@att.bizzzzzzzzzzzz" wrote:> > On Sat, 29 Oct 2011 10:04:56 -0700, John Larkin > <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote: > > >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. > > At 5-10x the cost, you just lost 1000 Joerg points.That's only .01% of a cent. -- You can't have a sense of humor, if you have no sense.
Reply by ●October 29, 20112011-10-29
On Sat, 29 Oct 2011 13:19:12 -0400, "Michael A. Terrell" <mike.terrell@earthlink.net> wrote:> >"krw@att.bizzzzzzzzzzzz" wrote: >> >> On Sat, 29 Oct 2011 10:04:56 -0700, John Larkin >> <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote: >> >> >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. >> >> At 5-10x the cost, you just lost 1000 Joerg points. > > > That's only .01% of a cent.Ok, make that 1000K Joerg points.
Reply by ●October 29, 20112011-10-29
On Sat, 29 Oct 2011 12:10:09 -0500, "krw@att.bizzzzzzzzzzzz" <krw@att.bizzzzzzzzzzzz> wrote:>On Sat, 29 Oct 2011 10:04:56 -0700, John Larkin ><jjlarkin@highNOTlandTHIStechnologyPART.com> wrote: > >>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. > >At 5-10x the cost, you just lost 1000 Joerg points.We pay $ 0.0208 for 2N7002s. And it doesn't need a schottly clamp diode plus two or three passive components in the base drive path. John
Reply by ●October 29, 20112011-10-29
On Sat, 29 Oct 2011 11:32:46 -0700, John Larkin <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:>On Sat, 29 Oct 2011 12:10:09 -0500, "krw@att.bizzzzzzzzzzzz" ><krw@att.bizzzzzzzzzzzz> wrote: > >>On Sat, 29 Oct 2011 10:04:56 -0700, John Larkin >><jjlarkin@highNOTlandTHIStechnologyPART.com> wrote: >> >>>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. >> >>At 5-10x the cost, you just lost 1000 Joerg points. > >We pay $ 0.0208 for 2N7002s. And it doesn't need a schottly clamp >diode plus two or three passive components in the base drive path.Since you bring it up, I never pay more than 0.6 cents for a 2N2222 or 2N3904/6 BJT. And I'm a damned hobbyist. But the main thing is that George already said, "I'll try the Bakers clamp/ Schottky first since there's already an npn 'lashed' in place." And who knows? He might get by with a speed up, even. Jon
