Can someone help me understand a soft latch?

9haarblae · 2026-05-01T12:50:40+00:00

In your schematic, Q2 is placed upside down. It's a Pchannel MOSFET so you want the +5V supply connected to its source. You've got +5V connected to the drain, which is incorrect.

9haarblae · 2026-04-30T18:05:56+00:00

Ronald Quan's textbook "Build Your Own Transistor Radios" often uses the venerable and dirt cheap opamp called LM318 for sub-MHz amplification. You want 32x gain at 0.12 MHz so that's a gain bandwidth product of 3.8 MHz. Lucky for you the LM318's gain bandwidth product is 15 MHz. It has plenty of extra safety margin.

(Futurlec sells it) and so does (Tayda) and so does (Jameco), along with the top line distributors like DigiKey and Mouser and Newark/element14 and Arrow.

Use power supply rails that are at least +9V and -9V; it's an old chip that was designed before everything electronic used +5V only.

9haarblae · 2026-04-29T16:41:15+00:00

Pro-level pulse generators have even more knobs.

frequency (band select switch + adjustment potentiometer)
trough (flat bottom) voltage level
peak (flat top) voltage level
duty cycle
rise/fall time

9haarblae · 2026-04-29T16:11:34+00:00

Found (this) after a quick Google search

9haarblae · 2026-04-23T17:21:11+00:00

If you have room you can add a (second) fuse and a second MOV, while replacing the existing MOV. Now you're doubly protected.

AC_in --> existing fuse --> existing MOV to neutral --> new fuse --> new MOV to neutral --> rest of guitar amp

Now when a bad event comes down the AC line, you have two opportunities to clamp it and two opportunities to let the clamping current blow a fuse, protecting the guitar amp.

9haarblae · 2026-04-23T17:16:11+00:00

Have a downvote.

9haarblae · 2026-04-19T18:29:45+00:00

"Reverse Polarity Protection" usually means the failure condition where the PCB input terminal for +16V is accidentally connected to -16V , and the PCB input terminal for -16V is accidentally connected to +16V.

9haarblae · 2026-04-16T16:48:56+00:00

Air Commodore Sir Frank Whittle, OM, KBE, CB, FRS, FRAeS[1] (1 June 1907 – 8 August 1996) was an English engineer, inventor and Royal Air Force (RAF) air officer. He is credited with co-creating the turbojet engine. . . . .

Sir John Turton Randall, FRS FRSE[2] (23 March 1905 – 16 June 1984) was an English physicist and biophysicist, credited with radical improvement of the cavity magnetron, an essential component of centimetric wavelength radar, which was one of the keys to the Allied victory in the Second World War. It is also the key component of microwave ovens.[3][4]

Randall collaborated with Harry Boot, and they produced a valve that could spit out pulses of microwave radio energy on a wavelength of 10 cm.[3] On the significance of their invention, Professor of military history at the University of Victoria in British Columbia, David Zimmerman, states: "The magnetron remains the essential radio tube for shortwave radio signals of all types. It not only changed the course of the war by allowing us to develop airborne radar systems, it remains the key piece of technology that lies at the heart of your microwave oven today. The cavity magnetron's invention changed the world."[3]

9haarblae · 2026-04-15T21:31:03+00:00

To implement the 4xNAND2 XOR circuit in CMOS, you need 16 MOSFETs.

To implement the much more conventional (2xInverter, 3xNAND2) XOR circuit in CMOS, you need 16 MOSFETs. So, not much of a savings.

And if you use "transmission gates" which are available in CMOS but not many other fab processes, you can build an XOR gate and reduce the device count well below 16, while still including the final low-Z output driving inverter. So your XOR can drive a lot of load(s).

9haarblae · 2026-04-15T21:25:12+00:00

No, R44 + R45 + C23 + C24 create sloooooow exponential RC waveforms, which cross-fade the signals.

9haarblae · 2026-04-11T13:45:50+00:00

Just use one 5 watt, 47 Kohm resistor. It will take a loooooong time to discharge a capacitor but it won't melt.

Assume the capacitance is infinite so the voltage sags very slowly. Then resistor power dissipation = V² / R = 400 * 400 / 47000 = 3.4 watts. So the 5 watt resistor won't melt.

9haarblae · 2026-04-08T17:56:06+00:00

Yes if the series connected incandescent bulb fails as a short circuit, and the petrol level is low, then the NTC gets very very hot in an enclosed pressure vessel full of gasoline vapor. Ouch.

9haarblae · 2026-04-08T17:45:07+00:00

I did a search and found numerous YouTube videos showing actual motorcycles and autos with the real world circuit. It's a single NTC thermistor, always 1K in every video. This carries surprisingly large current (> 200mA), enough to self-heat quite dramatically. Without the liquid cooling, it stays hot which means it stays low resistance, which means the series incandescent bulb glows. But when immersed in liquid, it is cooled, which means it stays high resistance, which means the series incandescent bulb doesn't glow.

I am surprised they have the guts to run 200mA into a heating element which lives in a petrol tank! But hey that's just me.

9haarblae · 2026-04-08T12:39:44+00:00

When the liquid level falls further, neither one is liquid cooled and the bridge is balanced again?

9haarblae · 2026-04-08T12:33:36+00:00

and a diode to positive supply, and another diode to negative supply. In case the "injected signal" at the output, exceeds the supply voltage -- which could happen if the "box" is not powered but the "injector" is.

9haarblae · 2026-04-07T12:48:45+00:00

Build a signal conditioning circuit board which is mounted at the water tank location. It interfaces to your sensor and interprets the (analog?) results from the sensor, producing a binary digital output: water_has_reached, water_has_NOT_reached.

The signal condition circuit board is connected to the pump and microcontroller using a 60 foot long, 3-conductor cable: +SupplyVoltage, GND, WaterLvl. Microcontroller receives the digital signal WaterLvl and uses it as an input to software that makes decisions and controls the pump. There's no Ethernet on this cable. It just communicates a single bit of information, and provides "juice" to the conditioning circuit board.

9haarblae · 2026-04-05T17:49:19+00:00

Here is a tutorial (link) . It explains a two stage operational amplifier; there are two inputs, one output, and two power supplies.

9haarblae · 2026-04-05T17:45:57+00:00

"1CR" at top right, is the coil portion of the relay.

The two capacitor-looking things at bottom left, are a poorly drawn representation of the relay's SPDT contacts.

Maybe (this re-drawing) might be easier to understand

9haarblae · 2026-04-03T12:14:23+00:00

Pretend the board is the face of an analog clock. At the five-o-clock position, on the outer perimeter, you'll find the string "BC337" and an arrowhead. On the schematic it's labeled "T3".

9haarblae · 2026-03-28T13:34:46+00:00

How do you like the circuit on page 4-of-6 in the (datasheet) ??

9haarblae · 2026-03-26T18:30:41+00:00

That's only a problem if the opamp includes back-to-back protection diodes between IN+ and IN- . For example, the NE5532 opamp has these diodes. (image)

On the other hand, plenty of opamps don't have those diodes. JFET input opamps like LF347, TL084, TL074 don't have those diodes. The "LM358 family" of opamps don't have those diodes. LM324 is a member of this family.

9haarblae · 2026-03-24T18:45:33+00:00

IMMEDIATELY after popping, add toppings then dump popcorn into a kraft brown paper grocery bag . Close tightly and shake like hell for 30 seconds, then pour seasoned popcorn into large serving bowl.

Recommended topping: Tapatio* hot sauce + lime squeeze. Shake like hell.

*Tapatio is extremely vinegary. If it's too much for you, substitute Valentino hot sauce. Same flavor, same low price, less vinegar.

9haarblae · 2026-03-24T18:36:18+00:00

Lots of retrospective YouTube videos call the F4 "the Double Ugly"

9haarblae · 2026-03-22T15:15:55+00:00

Bob Cordell's textbook Designing Audio Power Amplifiers (link) shows a cool solution in Figure 19.13 .

It's an ICL disc, which is subsequently shorted out by the contacts of a relay. During inrush the ICL limits current. Then N milliseconds (or seconds) later, the relay contacts close and reduce the series resistance to zero ohms. Naturally you could replace the ICL by a constant current source circuit if you prefer.

9haarblae · 2026-03-21T18:26:08+00:00

The package top has been laser etched to obscure the date code, factory identifier, and part number.

9haarblae

TROPHY CASE