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i attempted make myself 2-in-1 programmer can program both at89cx051 , at89s52 on same board without having plugging/unplugging of port. the main issue strobe pin aka reset (rst in circuit). (pin 1 of db25 connector). tested pc , got no results. what seems work if connected rst vcc via 1k resistor. works because though output @ 3.8v instead of 5 (which vcc at), puts at89s52 in reset mode expected, i'm not getting 12v @ pin 1 of small at89cx051 socket. if don't make manual connection, 1.4v measured @ rst, after added 10k resistor between vcc , rst. without rst being forced ground, should 5v i'm not. i have tested transistors manually leg leg using diode setting on multimeter , turned out fine. resistors going base of transistors 2.2k , used 1k resistors in place of diodes. other resistors 10k. resistor going buffer (74hc125) 0 ohm. all capacitors 0.1uf. why getting such awkward voltages reset pin? using weak of resistors or something? update so ended ma...

i have old audio amplifier (jvc a-gx2b) produces crackling noises on 1 of outputs. so, opened , discovered 2 largest capacitors seem leaking : can replace capacitor same characteristics (4700uf 50v), or should instead take specific capacitors audio? if yes, ones ? in addition, have interest replace capacitors of amp (not 2 largest)? 1: wear safety gloves when handling toxic leaky capacitors, remove , clean safely isoprop. , mark note polarity (-) 2 : measure dia , lead pitch dxl [mm] sort parts this 3: audio or better low esr, highest ripple current caps. < 50mohm, > 3aripple 4: better caps run cooler lower esr , rated higher temp. may not fit. 5: japanese caps have better reliability i suggest 1st choice panasonic @ digikey .. cost of shipping may more cost of caps. as suggested: cap voltage must same or more - capacitance tolerance doesn't matter, , may larger value. - large e-caps may replaced @ same time.

i have no experience of dealing pcb manufacturers except time once used pcbway's service. know if there standard pcb board sizes 1 needs create when getting custom pcb made? my basic idea not incur cost loss of material pcb manufacturer faces. example, let's use board of 10*10 inch manufacture , submit custom board request dimension of 7*7 inch, increase cost of pcb? sometimes pcb makers have per-panel price. tell size using..or more importantly, area have work with-- usable area (not panel size) @ least 10" x 16", maybe 10.5" x 16.5" or bit more. small panel. large production believe larger panels more typical. typically phenolic single sided boards use larger panel sizes. this reference lists typical standard pcb panel sizes follows: panel borders (maybe 0.5" on each side) take away area. it not particularly important unless pcbs (or self-panelized pcbs) getting size few panel- materials not biggest part of pcb cost in prototype...

can sample , recover signals non infinite "length" using nyquist shannon theorem? example if have signal $$ x(t)= u(t+5)-u(t-5) $$ , know sampling period t less 10s (t<10). can recover using nyquist theorem? if not , recover it? edit: have taken fourier transform of signal , : $$x(ω)=2i(\frac{1}{iω}+πδ(ω))sin(5ω)$$ can lead finding period $$\frac{2π}{5} $$ , see if f>= 5/π can recovered. may not possible because used theorem although on finite signal no need think "finite length". though continuous time signals defined infinite time interval, in practice analyse in finite interval only. x(t) defined in interval [-5 5]. the frequency domain representation of such signal infinite bandwidth sinc function. hence not possible define particular sample rate per nyquist theorem, reconstruct without losing information. can sample definite sample rate, implicitly band limits signal. sampled signal after reconstruction through dac , lpf, not pe...

i know in magnitude spectrum of single square dc pulse (0-8v) lobes 0 @ frequencies $$\frac{1}{t},\frac{2}{t},\frac{3}{t},...etc. $$ t=pulse duration. if square pulse goes -2v 6v, formula 0 points of spectrum still same, ie. $$\frac{n}{t}$$ n=1,2,3... etc. , t=pulse duration? what difference? applying offset signal change spectrum @ dc (frequency of 0hz). if single square pulse goes 0v 8v, value of dc \$\mathcal{f}\{ f(t)\}|_{\omega=0}=0\$. if square wave goes -2v 6v, \$\mathcal{f}\{f(t)\}|_{\omega=0}=-2\cdot\delta(0)\$.

according miller effect, within amplifying devices such transistors have inverting voltage gain higher one, there should increased input capacitance increased voltage gain of amplifier. i has been said effect limiting amplifier @ higher frequencies. comes incorporation of method of increasing transistor's bandwidth. 1 of them dominant pole compensation. according wikipedia: "when capacitor introduced between input , output sides of amplifier intention of moving pole lowest in frequency (usually input pole) lower frequencies, pole splitting causes pole next in frequency (usually output pole) move higher frequency." i part when states 2nd pole should moved higher in frequency (that 2nd roll-off), why heck want decrease 1st pole , consequently first gain roll-off begins @ lower frequencies before. what point of it? i mean, shouldn't desired gain vs. frequency curve flat possible, long possible - higher frequencies? addition of capacitor betwee...

is ib current correct went kvl through first transistor your forgetting re. you need use effective resistances r4 , r6, value \$r4_e = r4 * (1+\beta) = 40,000 * 101 = 4,040,000\omega\$ \$r6_e = r6 * (1+\beta) = 1,000 * 101 = 101,000\omega\$ using base circuit looks this. simulate circuit – schematic created using circuitlab you solve using various methods, shortcut realise that, ideal vbe, voltage @ top of both emitter resistors identical. such can treat both legs being in parallel. circuit can further simplified to.. simulate circuit now should able calculate current through circuit as.. \$i_{rb} = (2.6 - 0.6)/(2,260 + 98,540) = 19.84ua\$ that makes base voltages both.. \$v_b = 2.6 - 19.84ua * 2.26k\omega = 2.555v\$ so \$ib1 = (2.555-0.6)/4.04m\omega = 0.484ua\$.. , \$ib2 = (2.555-0.6)/101k\omega = 19.356ua\$ however: these numbers make sense if transistor not saturated. question states \$i_e \approx \beta * i_b\$ indicates ...