Duplicate

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'm trying find "same or better" equivalent replacement 747 opamp. footprint / pinout can different, don't mind. using parametric search on arrow, while reading datasheet 747 found lt1355 might replacement. however, quite expensive; $16.89 1 single ic (dip package). searching lt1355 on octopart shows arrow cheapest one, price steep me. i'm fine around $8/pc. what other dual opamps can serve replacement 747? usually asking products/parts off-topic since can benefit wants better 741, i'm making exception. there literally hundreds of suitable 741 replacement opamps. the parts list waaaay expensive. have special properties few designs need. you should not looking 741 replacement per sé more "generic opamp". there usable opamps had less money. lm358, 10 1 euro (shipping included!) on ebay other examples: mcp602, ca3140, tlc271 most of these have versions 1, 2 or 4 opamps in 1 housing. you have compare datasheets see suits ...

i using tps61021 boost converter ic increase input voltage 1.2v 3.3v. have designed boost converter application called ti webench. input voltage: 1.2v output voltage: 3.3v min current: 300ma the pcb have designed, measuring 3.3v, efficiency pathetic. below 25%. in bom, ti webench software recommends using 10uf 10v capacitor, using 10uf 6.3v. not able understand why efficiency low? datasheet inductor . below in tabular format have written current values have measured multimeter. fig.1 measurement of current @ different load resistance. fig.2 ti webench application output fig.3 schematic design of boost converter pin = 1.2v × 570ma = 684 mw pout = 3.3v × 105 ma = 346.5 mw efficiency = pout/pin = 346.5/684 = 50.7% how on earth did come 18.42%? similarly, 64% , 70% other 2 loads. these numbers sound pretty me @ these voltages. did webench give efficiency estimates specific configuration?

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 ...

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...

i have apc ups 9 ah battery. would ok replace 18 ah battery? i realize modifications needed since won't fit in case. in theory shouldn't matter how bigger ah rating of battery long regulate current , voltage same. takes long transfer coulombs of charge. bigger battery of same chemistry have larger capacitance , may have lower esr.

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)\$.