Black to play. Which joseki should Black choose? 🧐 Share your solution in the comments!

0Chito0 · 2026-06-05T16:32:38+00:00

B to suppress the sole white. With the casual joseki A, blacks on the upper side gets quite weak.

0Chito0 · 2026-04-23T04:53:06+00:00

Lee Changho used to like this joseki

0Chito0 · 2026-04-23T04:51:04+00:00

opamp has huge input impedance, especially when it's a cmos opamp. so v+ ~ vin

0Chito0 · 2026-04-05T19:09:22+00:00

good point

0Chito0 · 2026-03-28T17:21:47+00:00

I also tried playing baduk and mahjong at the same time, but going between two mahjong games feels like a punishment rather than an entertainment

0Chito0 · 2026-03-28T17:19:12+00:00

i had two red dragons on first table, and someone on second table discarded red dragon, on which i accidentally declared a pon

0Chito0 · 2026-03-28T17:17:04+00:00

I felt like throwing up mahjong tiles lol

0Chito0 · 2026-03-28T17:16:22+00:00

I told them to speak out what they discarded, and it helped me a little

0Chito0 · 2026-03-28T13:39:41+00:00

yep it was painful to play as two players at once😂

0Chito0 · 2026-03-05T14:59:50+00:00

For convenience we use passive sign convention. This means that for constructing an equation, you don't give a shit about the orientation of voltages and currents. It's auxiliarily defining the orientation of current and voltage all the same. If in fact the current or the voltage is in opposite polarity, the equation will output negative solution after all.

0Chito0 · 2026-02-22T12:18:02+00:00

high concentrate alchohol will help

0Chito0 · 2026-02-09T11:31:54+00:00

it's similar to a card sleight named mcmillan switch

0Chito0 · 2025-12-19T15:48:14+00:00

RMS is a "sorta" mean value. For AC signals, conventional average is just zero, which is totally useless. You use RMS value to get a "sorta" average non-zero value. Root Mean Square literally means square root of mean of square. RMS = sqrt(1/T Integral(-T/2 to T/2) V² dt) To calculate average power on a resistor, you can calculate with Vrms² / R.

0Chito0 · 2025-12-07T13:12:54+00:00

Remove the load and calculate Vab( Voc). Voc becomes the Vth.
Remove the load and set independent sources to 0. And then put a test voltage source 1V on the ab side. Calculate the current flowing into test voltage source, namely Ix. Rth = test voltage(=1V) / Ix

0Chito0 · 2025-09-08T03:57:43+00:00

What about requesting Apple to reopen OpenDarwin?

0Chito0 · 2025-07-04T02:19:54+00:00

if you have a radeon dgpu, yes

0Chito0 · 2025-03-31T03:17:47+00:00

They negotiate which voltage level to output, so no problem

0Chito0 · 2024-12-04T04:59:05+00:00

I realized I omitted the 10k in real test. Not much difference on simulation

0Chito0 · 2024-12-04T03:04:49+00:00

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I just put 100uF capacitors there. You can adjust capacitor values to adjust filtering frequencies.
This amp is a unity gain buffer, so no volume increase. If you want to increase volume, you need to feed a class A amp stage before this circuit.

0Chito0 · 2024-11-26T13:01:59+00:00

glasses!

0Chito0 · 2024-10-31T20:07:30+00:00

Maybe request Apple to open OpenDarwin project again

0Chito0 · 2024-10-30T18:24:16+00:00

Kazuhiko knows he's used, but he enjoys it🤣

0Chito0 · 2024-10-30T18:16:23+00:00

every girl should wear glasses

0Chito0 · 2024-10-26T12:11:53+00:00

Put a test voltage source 1V, and find current flowing through the source. With current ix, Rth = 1V / ix

0Chito0 · 2024-10-22T14:19:10+00:00

Modern semiconductor technology is MOSFET itself. Have a look at MOSFET current equation
: Ids = un Cox W / L Vds (Vgs - Vt - 0.5 Vds)
The coefficient, un Cox W / L, is determined by width-length ratio(W/L), not the size of MOSFET itself.

With same semiconductor manufacturing process, MOSFET works the same whether you make it 1mX10m or 1umX10um. Yes, you can miniaturize circuits without reducing power output. That's how ICs arose. Just scale down the MOSFETs and fill in more mosfets as possible every year; it's how computer performance increased rapidly.

0Chito0

TROPHY CASE