On 05/15/2017 01:18 PM, Steve Wilson wrote:> Jan Panteltje <pNa0nStpealmtje@yahoo.com> wrote: > >> On a sunny day (Mon, 15 May 2017 16:24:56 GMT) it happened Steve >> Wilson <no@spam.com> wrote in >> <XnsA7767E5052407idtokenpost@69.16.179.22>: >> >>> bitrex <bitrex@de.lete.earthlink.net> wrote: >>> >>>> On 05/15/2017 10:09 AM, bitrex wrote: >>>>> On 05/15/2017 10:01 AM, John Larkin wrote: >>>>> >>>>>> BW is 230 MHz as a unity-gain buffer. Each half of the dual will >>>>>> source and sink 88 mA. And there are lots of faster THS-series >>>>>> parts. >>>>>> >>>>>>> >>>>>>> But indeed you can do that with a few NPNs of the old days, >>>>>>> >>>>>> >>>>>> Post it. >>>>> >>>>> With discretes you can have: >>>>> >>>>> 1) wide bandwidth >>>>> >>>>> 2) low parts count >>>>> >>>>> 3) low quiescent current draw >>>>> >>>>> Pick two. >>>> >>>> The two transistor circuit Jan posted does have a pretty wide >>>> bandwidth and a low parts count. It has a pretty nice small signal >>>> output impedance, but its large-signal output impedance is bad. It >>>> also draws a relatively large amount of quiescent current. >>> >>> I somehow missed the two transistor circuit that Jan posted. Can you >>> please repost the link? >>> >>> Thanks >> >> From: >> http://panteltje.com/panteltje/quadcopter/hud.html >> the circuit diagram: >> http://panteltje.com/panteltje/quadcopter/hud-0.1_circuit_diagram_IMG >> _5984.JPG bottom left: >> that has a gain of 2, so can drive a 75 Ohm cable with reasonable Zi. >> The DC path requires an other 75 Ohm at the end of the cable. > > OK, thanks. > > That is a Sziklai Pair with a voltage divider in the feedback path: > > https://en.wikipedia.org/wiki/Sziklai_pair > > It would normally have the same thermal drift as a simple base-emitter > junction, but the divide by two may increase it. > > The output impedance may be increased due to the low loop gain. The > bandwidth is determined by the transistors you use.Its output impedance for large negative-going signals is _dogshit_, regardless of what the small-signal analysis feedback-loop-gain etc. calculations indicate. It can source current okay. It definitely can't sink it. It doesn't really matter for a composite video signal, though, because it's not really that wide a bandwidth signal. The only potentially difficult fast full-level transition I think of would be if you have to go from the "back porch" immediately to "full luminance."
Fast buffer idea
Started by ●May 14, 2017
Reply by ●May 15, 20172017-05-15
Reply by ●May 15, 20172017-05-15
On Mon, 15 May 2017 16:02:50 GMT, Jan Panteltje <pNa0nStpealmtje@yahoo.com> wrote:>On a sunny day (Mon, 15 May 2017 08:56:58 -0700) it happened John Larkin ><jjlarkin@highlandtechnology.com> wrote in ><6sjjhc1mm6atuu3ovmnnmtpke1iu2t4fgh@4ax.com>: > >>On Mon, 15 May 2017 10:18:16 -0400, bitrex >><bitrex@de.lete.earthlink.net> wrote: >> >>>On 05/15/2017 10:09 AM, bitrex wrote: >>>> On 05/15/2017 10:01 AM, John Larkin wrote: >>>> >>>>> BW is 230 MHz as a unity-gain buffer. Each half of the dual will >>>>> source and sink 88 mA. And there are lots of faster THS-series parts. >>>>> >>>>>> >>>>>> But indeed you can do that with a few NPNs of the old days, >>>>>> >>>>> >>>>> Post it. >>>> >>>> With discretes you can have: >>>> >>>> 1) wide bandwidth >>>> >>>> 2) low parts count >>>> >>>> 3) low quiescent current draw >>>> >>>> Pick two. >>> >>>The two transistor circuit Jan posted does have a pretty wide bandwidth >>>and a low parts count. It has a pretty nice small signal output >>>impedance, but its large-signal output impedance is bad. It also draws a >>>relatively large amount of quiescent current. >> >>Iq is undefined, and this is a poster child for thermal runaway. It >>needs more parts! > >???? >TWO transistor ????I was referring to this one: http://panteltje.com/pub/maybe_this_amp_IMG_6183.JPG -- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
Reply by ●May 15, 20172017-05-15
On 05/15/2017 01:28 PM, John Larkin wrote:>>> Snoop the input current of that buffer. At 20 MHz, the overall current >>> gain is less than 1. At 100 MHz, the current gain is about 0.1. You'd >>> be better off with a piece of wire. >> >> I'm not sure what "input current" you're looking at. I'm seeing an >> average current of around 600uA into the MOSFET gates at 100MHz. > > I put a 1 mohm resistor in series with the signal generator and probed > the current, with a sine wave signal input. Try it and see what you > get.How are the gate capacitances of the input MOSFETs supposed to be charged up and down with that fuckin' thing sitting in the way?! Fuck the circuit to hell and then complain it's bad?! Shit! :O
Reply by ●May 15, 20172017-05-15
bitrex <bitrex@de.lete.earthlink.net> wrote:> On 05/15/2017 01:18 PM, Steve Wilson wrote: >> That is a Sziklai Pair with a voltage divider in the feedback path:>> https://en.wikipedia.org/wiki/Sziklai_pair>> The output impedance may be increased due to the low loop gain. The >> bandwidth is determined by the transistors you use.> Its output impedance for large negative-going signals is _dogshit_, > regardless of what the small-signal analysis feedback-loop-gain etc. > calculations indicate. It can source current okay. It definitely can't > sink it.The negative swing depends on the pulldown resistor, same as any single- ended emitter follower. That should be obvious. These issues would be immediately apparent in a proper LTspice analysis. Unfortunately, with some users still devoted to ASCII-art or hand-drawn schematics, it is difficult or impossible to do a reasonable preliminary analysis. The author may have made an incredible blunder in the schematic, or it is impossible to determine the component values in a hand-drawn schematic. Often, you really can't tell if it is worth your time to investigate someone's idea. In most cases, with an ASCII-art or hand-drawn schematic, it is not worth the time to spend any effort on it. The author really doesn't know if it works or not. It could be difficult to detect errors in the posted drawing, since there is no error detection mechanism. But with a working LTspice model, it could be worth investigating. If it runs, it means it follows the author's intent. It depends on who authored it. Unfortunately, many beginners post LTspice circuits that are full of blunders. You can immediately tell their thinking is scrambled beyond hope. This is good information. I simply record their names and ignore anything they post so I don't waste any more of my time.
Reply by ●May 15, 20172017-05-15
John Larkin <jjlarkin@highland_snip_technology.com> wrote:>>TWO transistor ????> I was referring to this one:> http://panteltje.com/pub/maybe_this_amp_IMG_6183.JPGOld idea. I used that in a product in 1979, which won the admiration of engineers at HP, and some orders. The drive capability depends on the pullup resistors in the bases. Quescent current depends on the output emitter resistors, which are not shown in the link. These issues could be observed in a simple LTspice analysis.
Reply by ●May 15, 20172017-05-15
On 05/15/2017 02:15 PM, Steve Wilson wrote:> John Larkin <jjlarkin@highland_snip_technology.com> wrote: > >>> TWO transistor ???? > >> I was referring to this one: > >> http://panteltje.com/pub/maybe_this_amp_IMG_6183.JPG > > Old idea. I used that in a product in 1979, which won the admiration of > engineers at HP, and some orders. The drive capability depends on the > pullup resistors in the bases. Quescent current depends on the output > emitter resistors, which are not shown in the link. > > These issues could be observed in a simple LTspice analysis. >"Diamond buffers are kind of cool, emitter feedback makes them linear and they don't have much phase shift, but they kind of suck balls whenever you want to actually move appreciable current with them which unfortunately is exactly what you'd want to do with a buffer." The circuit I posted is exactly the same topology, except with the following "refinements" MOSFET first stage for higher input impedance Feedback taken to MOSFET drains from the output collectors to keep their Vds approximately constant Output quiescent current is primarily set by the (bypassed for AC) input stage source resistors, not the output emitter resistors "Boosting transistors" which slide the bias up in response to signal so it doesn't draw 150mA sitting around doing nothing.
Reply by ●May 15, 20172017-05-15
On 05/15/2017 02:25 PM, bitrex wrote:> Feedback taken to MOSFET drains from the output collectors to keep their > Vds approximately constantOutput emitters, rather
Reply by ●May 15, 20172017-05-15
On Mon, 15 May 2017 13:38:11 -0400, bitrex <bitrex@de.lete.earthlink.net> wrote:>On 05/15/2017 01:28 PM, John Larkin wrote: > >>>> Snoop the input current of that buffer. At 20 MHz, the overall current >>>> gain is less than 1. At 100 MHz, the current gain is about 0.1. You'd >>>> be better off with a piece of wire. >>> >>> I'm not sure what "input current" you're looking at. I'm seeing an >>> average current of around 600uA into the MOSFET gates at 100MHz. >> >> I put a 1 mohm resistor in series with the signal generator and probed >> the current, with a sine wave signal input. Try it and see what you >> get. > >How are the gate capacitances of the input MOSFETs supposed to be >charged up and down with that fuckin' thing sitting in the way?! > >Fuck the circuit to hell and then complain it's bad?! Shit! :Omohm means milliohm, in SI units and in Spice. -- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
Reply by ●May 15, 20172017-05-15
John Larkin <jjlarkin@highland_snip_technology.com> wrote:> On Mon, 15 May 2017 13:38:11 -0400, bitrex > <bitrex@de.lete.earthlink.net> wrote: > >>On 05/15/2017 01:28 PM, John Larkin wrote: >> >>>>> Snoop the input current of that buffer. At 20 MHz, the overall >>>>> current gain is less than 1. At 100 MHz, the current gain is about >>>>> 0.1. You'd be better off with a piece of wire. >>>> >>>> I'm not sure what "input current" you're looking at. I'm seeing an >>>> average current of around 600uA into the MOSFET gates at 100MHz. >>> >>> I put a 1 mohm resistor in series with the signal generator and >>> probed the current, with a sine wave signal input. Try it and see >>> what you get. >> >>How are the gate capacitances of the input MOSFETs supposed to be >>charged up and down with that fuckin' thing sitting in the way?! >> >>Fuck the circuit to hell and then complain it's bad?! Shit! :O > > mohm means milliohm, in SI units and in Spice.Megohm is MEG in spice. I don't know if caps are required in all versions. However, to measure current, all you need is a voltage source in series with zero volts output. Then plot the current through the device. But in LTspice, you can plot the current in any device using the current probe.
Reply by ●May 15, 20172017-05-15
>"John Larkin" wrote in message >news:937ihcpose7eq6ulu94vium3snbbaqf251@4ax.com...>>The problem with a simple active current source to the input followers >>is quiescent current draw - if you optimize the follower current source >>for the gain you need at the highest frequency the buffer draws a huge >>helping of quiescent current sitting around doing nothing. > >>IC implementations solve this in various clever ways using - you guessed >>it - _more_ active current sources.>Analog ICs can use basically unlimited numbers of radical parts, and >have essentially no interconnect parasitics. And they can spend a lot >of money on design.Indeed.... I remember Jim T boasting..err... I mean informing us all, maybe 15 years ago, that he just paid a tax bill of $500,000 -- Kevin Aylward http://www.anasoft.co.uk - SuperSpice http://www.kevinaylward.co.uk/ee/index.html
