How to calculate Vce in between cascode BJT's  - Page 1
Forum section: electronics / projects

author: MathWizard
author_rank: Super Contributor
timestamp: January 19, 2022, 08:03:41 am
content: In a circuit like below where the bottom of the current mirror meets the diff-pair, and you have Q2 over Q4, 1 Vce over another Vce, I can find the current , and guess the BJT's are in active mode with reverse based CB, but what calculation gives the number for Vce? What have I missed or don't remember , maybe it's in the Eber's Moll type models? How do you calculate the base voltage of Q5 thenI could set the output voltage/I and work back, and then that set's the Vce ? But otherwise how do u do it ?? I don't know the BJT semi-conductor details much yet.I was skimming through cascode amp's, current mirrors and other multi transistor amps, and I never saw the answer in there either. Calling a BJT a voltage controlled current source, I know the BJT is happy with a range of Vce voltages (for each operating region)Being a diff amp, I could see it setting Q5's current to about the same IB as the Q2's IB. But in a pure cascode, what set's that voltage in between Q2/4 ??

author: Zero999
author_rank: Super Contributor
timestamp: January 19, 2022, 10:36:36 am
content: It's not a cascode, but a differential pair and current mirror.Normally you have negative feedback, so the output voltage stabilizes at 0V. The inputs are connected to 0V, so the output will be unpredictable and dependent on the mismatch between the transistors. Try connecting Q3's base to the output, to close the loop.The voltages on Q3 and Q4's collectors are roughly a diode drop below the positive supply voltage, which is to be expected.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 19, 2022, 11:45:51 am
content: Two collectors facing, is an undefined condition: two currents fighting it out. The voltage gain is extremely high (limited by Early effect). The voltage either must be stabilized by feedback, or by dumping into a load impedance; or letting one or the other saturate (perhaps for a comparator instead?), but saturation is undesirable as it severely reduces hFE, the worst case being for Q4, thus drawing tail current (in part or whole) through the input pin.Here, the load impedance is Q5 Vbe. So the collector voltages are mostly irrelevant, a diode drop as said above, and what's being controlled (well, when the diff inputs aren't shorted together, obviously) is a subtraction of currents and thus the difference in currents (the diff amp output as such) goes into Q5 Ib. And as long as its hFE is quite large, this again gives very high gain, and so needs feedback stabilization yadda yadda, and you get the usual op-amp situation that's familiar.Tim

author: magic
author_rank: Super Contributor
timestamp: January 19, 2022, 11:50:31 am
content: Yep, node voltage tends to be determined by the lowest impedance present on the node, which is Q5 base here.This assumes good pairwise matching of transistors and Ic(Q4) > Ic(Q2), as it should be considering the current mirror and its base currents.If Ic(Q2) > Ic(Q4) then their collectors will be pinned to the positive railAnd, technically, Ic(Q2)=Ic(Q4) in such case, something somethnig Kirhoff somethnig.

author: MathWizard
author_rank: Super Contributor
timestamp: January 19, 2022, 08:08:13 pm
content: Ok so like a good current source the Vce will still be set by the external conditions. With the base of Q5 disconnected, Ltspice says about 3.8V was there, so what does that number mean? What u might expect if u put a DMM there ?In this example, if Q1 Ic + 2Ib are entering Q3's collector, is it ok to say that if 2Ib was the base current of Q5, then the diff pair would both have the same current Ic+2Ib, and in the sim Q5's current is about 2x Ib 1. Then I could get a Vbe.Last year I would have tried a matrix eqn but they usually go nowhere when there's Vce's everywhere, or I'm doing it wrong.

author: magic
author_rank: Super Contributor
timestamp: January 19, 2022, 08:49:46 pm
content: 3.8V means that the output shifted ~0.5V when Q5 base current was removed from it, so you could calculate what SPICE thinks about output impedance of that node.If you inject the missing 2·Ib from a current source to the right side, you should in theory get equal collector voltages on both sides and you likely will in SPICE.But Q5 is not a current source, it will set this node's voltage to whatever is required to pull the missing 2·Ib from Q5. If Q5 is a Darlington pair, you will see 5V-2·Vbe.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 19, 2022, 10:20:12 pm
content: A DMM assumes reality, in which case you'll most likely measure something close to -0.5V, or 4.8V, as the transistors will not be nearly so well matched (in SPICE, they are exactly identical unless marked otherwise). Even if the transistors are monolithic matched (which will be within, some percent or fraction thereof, I think? mfg tolerances aren't perfect, merely pretty good). Well, the latter might be good enough to see something intermediate, or at least have a good chance of seeing such.A DMM can read an intermediate voltage, if the current difference happens to be just equal to the current flow through its (typ 10M) resistance. Very unlikely to read anything but saturation on a hi-Z range (typ 1G+?).Tim

author: emece67
author_rank: Frequent Contributor
timestamp: January 19, 2022, 11:41:59 pm
content: .

author: MathWizard
author_rank: Super Contributor
timestamp: January 20, 2022, 01:26:29 am
content: Ok thanks folks, so no worries if I can't calculate it then in some cases. I was trying a model of a TL431 with 3904/6, maybe that will make more sense now, but I haven't done much formally on multi-BJT circuits like all the current mirrors.It used some BJT's of different area's, I'll have to go over the physics and look that up in LTSpice. I haven't tried any mult-MOSFET calculations yet, but the internal model of MOSFETs are fresh in my head.I haven't tried anything on the breadboard in ages, maybe this weekend.

author: magic
author_rank: Super Contributor
timestamp: January 20, 2022, 09:26:11 am
content: You could start with something simpler.I ended up never building it (maybe I should), but it worked in sim. It's 1.25V and hardwired for unity gain, although that could be changed by separating R1,R2 ("VREF") from the rest ("VCC"). Input impedance and "bias current" is much worse than in the real thing, for reasons that should be obvious when you look at the published schematic (which is mostly accurate).If you know how this works and how to tune it, you will hopefully know how to upgrade it to full TL431 with 2.5V threshold and high input impedance.I see I ended up using single transistors for Q1,Q2 and different collector resistors. It's also possible to use equal resistors and parallel a few transistors for Q2 or a combination of both approaches. The relation of R3 to R1 and R2 depends on the ratio of collector resistors and/or transistor count.

author: MathWizard
author_rank: Super Contributor
timestamp: January 23, 2022, 01:04:13 am
content: Ok I'll try that too. I found out how to change emitter area in LTSpice, it does not pop right up in my search anyways, so in the BJT instance in the schematic, it will say like Q3 and below it the BJT model # like 2N3904. All you have to do is type m=10 after 2N3904 , with a space, and for programing reasons it would make a 2N3904 with 10x the current for a given Vbe.With that TL431 model, I find I don't really know how to work backwards, considering all the current runs through the Kathode resistor. But I'm probably supposed to be making more assumptions, and setting outputs/currents/etc. But thats a lot of work, I can't do it yet.I did find a better BJT equation, that does include the (1+Vce/Va) term *the [exp(Vbe/nVt)-1], so I have to mess with that a lot more (if I want my calc's to match the sim's better). I don't like ignoring base currents, or just setting all Vbe to 0.7V, so I run in to too much trouble really.I worked through a few current mirror types today, and was able to calculate the main stuff, by setting a current, rail voltages, and then working through it, and finding a resistance(s).I still think if I was given the resistance+rail voltages, I'd have a much harder time with it. Like trying to figure out band-gap references, still get too bogged down. Nevermind trying matrix eqn's on this, how do they do them, with actual eqn's for Vbe, Ic like Lambert functions, before calculators let alone computersIn the last circuit of this p11, a bootstrapped current source w/ high PSRR, Q3 shuts off in equilibrium,https://inst.eecs.berkeley.edu/~ee140/fa19/dis/01/current_mirrors.pdfI need to actually complete what happens if I ignore base currents /B->inf, and all or most Vbe=constant. Maybe then I could get a few values, others, not much yet besides a list of equal things.What other multi-BJT pdf's should I look at to make more sense of these circuits (besides going over more basics)

author: magic
author_rank: Super Contributor
timestamp: January 23, 2022, 09:25:28 am
content: You can omit the M= and simply write "2N3904 3".You may be interested in this:How to make a bandgap voltage reference in one easy lesson- from the inventor of one of the most popular bandgap cell topologiesTL431 teardown- it's not exactly the same as in the datasheet in terms of resistor and transistor ratios IIRC, but same topologyA different TL431- this also worksQuote from: MathWizard on January 23, 2022, 01:04:13 amThis is not a constant current source, it's PTAT (proportional to absolute temperature) current generator or "constant gm bias generator" as it's also known in the context of IC design, where it is used to increase bias current to compensate for loss of gm with temperature and maintain well-defined gain and bandwidth.Its PSRR is questionable too because Early effect affects gain of the two mirrors differently. Particularly in old IC technology, where PNPs really suck. Linear Tech developed a much more advanced mirror circuit for some of their voltage references which eliminates Early from the equation.Other PTAT generators exist.Ignoring Early, it's pretty trivial. The top mirror has gain of exactly one, so both columns conduct equal current. Vbe of Q1 is less than Q2 (by a temp-dependent logarithm of area ratio) and that voltage must be dropped by R1, determining the operating current for any given temperature. The rest is convincing yourself that the ideal setpoint will be reached from any other state by feedback.

author: Marco
author_rank: Super Contributor
timestamp: January 23, 2022, 09:52:53 am
content: Quote from: MathWizard on January 19, 2022, 08:08:13 pmJust want to add that to first approximation, assuming it's a voltage controlled current source was what steered you wrong. Assume near infinite beta, assume perfect symmetry, but don't assume the early effect isn't there.With near infinite beta, with perfect symmetry and with early effect obviously the Q1 and Q2 collector voltages have to be equal.