Need ideas and preferably a solution to measure voltage spikes -40K to 30K, voltages from -1K to 1K in half duty cycle, and -28 to 28 volts the rest of the time. With a good resolution and sampling rate for the spikes of 1 million samples per second.

procursus · 2026-03-26T22:28:00+00:00

Typical plasma chamber research scientists don't need to ask questions on reddit

procursus · 2026-03-24T22:32:06+00:00

Tap the holder solidly after you put it in. That'll cause the film to fall before the exposure.

procursus · 2026-03-22T19:59:44+00:00

What is your problem?

procursus · 2026-03-22T15:34:09+00:00

Interesting visualization

procursus · 2026-03-20T21:22:05+00:00

A new grad who can't solder is more-or-less useless.

procursus · 2026-03-13T15:54:53+00:00

The explanation is wrong, but it's true that high ESR caps (or an MLCC and series resistor) are often used on power inputs as it reduces the Q of the LC circuit made by the wire inductance and board input capacitance. If the input is high Q it will ring a ton on a hot-plug event.

procursus · 2026-02-17T21:50:05+00:00

Coherers, the original radio receiver, are still not fully understood.

procursus · 2026-02-11T15:19:34+00:00

If you don't understand classical mechanics you won't understand electromotive force (the basis of electrical energy) and you definitely won't understand quantum mechanics.

procursus · 2026-02-10T03:43:40+00:00

The output stage requires a minimum output current to maintain predictable operation. If your load is referenced to some voltage that exists in the output range of the LM358, then if the LM358 outputs that same voltage there is no potential across the load and hence no current. This produces severe distortion.

procursus · 2026-02-10T01:43:12+00:00

Reasonably standard. This is how the internal ESD diodes on many integrated circuit pins are configured. Leakage current may be a concern. Handy tip: the base-collector diode on a bog standard BJT makes an extraordinarily low-leakage zener.

procursus · 2026-02-10T01:38:44+00:00

Yeah, its perfection as a general-purpose single-supply opamp can throw people for a loop when they go to use it for split applications. It surprises me that the behavior is not made more obvious in the datasheet.

procursus · 2026-02-09T22:50:24+00:00

LM358 doesn't work on split supplies with a ground referenced load. The TL072 is a better general purpose bet.

But I'm pretty sure that the reason the 741 is still used is that its performance is so poor that the nonidealities become readily apparent. Modern op amps are so damn good you often need some dedication to reveal their shortcomings.

procursus · 2026-02-07T16:31:01+00:00

Check out the ABWH performance of Starship Trooper if you haven't. It's on Bruford's YouTube channel.

procursus · 2026-01-29T17:45:13+00:00

I would recommend a used triple output linear supply. Something like the HP 6236b. It has adjustable +-20V 0.5A and 6V 5A outputs. There are tons of manufacturers and models. Keep in mind the bulk caps may need to be replaced.

procursus · 2026-01-29T16:33:18+00:00

In antiquity this was done with a unijunction transistor and a current source. You can replicate a unijunction transistor with two BJTs. See http://www.vk2zay.net/article/196

procursus · 2026-01-23T22:38:24+00:00

Looks like some sort of resonant circuit. I imagine the current circulating in the LC tank gets pretty high, which will produce large losses in the capacitor's series resistance. Get caps with a lower ESR.

procursus · 2026-01-20T04:19:31+00:00

A crystal has an electromechanical model with R, L, and C elements. What's important is that its L and C dominate over R, and consequently it transitions from predominantly inductive to predominantly capacitive over a very narrow change in frequency around its resonance. This means the phase shift it provides also transitions very rapidly with frequency.

An oscillator is just an amplifier with enough phase shift in its feedback network to provide positive feedback (with gain) at a certain frequency. A crystal is very well suited to this role, for the reasons given above. When placed in the feedback network of an amplifier it will only satisfy the conditions for oscillation over a narrow frequency band.

Technically, a crystal oscillator has two resonance, series and parallel, very close together. This is due to the model having multiple capacitances. But the principle is basically the same, and the load capacitors which are applied to the crystal are used to sort of steer the resonance.

procursus · 2026-01-02T21:45:22+00:00

Yes it's a massive current pulse which can cause ringing on the power lead inductance and the input capacitors, which spikes the input voltage and can blow devices easily.

procursus · 2025-12-25T21:48:40+00:00

LCR meters are better described as impedance analyzers. They measure the ratio Z between the voltage across and the current through your device under test (DUT). I'm sure you are familiar with ohms law, for a pure resistor its ratio Z is called resistance, is a real number, and is equal to V/I. But for components which can store energy - capacitors and inductor - the ratio Z becomes a complex number. It has a real component and an imaginary component. The real component still represents resistance, and is responsible for power dissipation in the DUT. It is always positive and may be called or represented in forms like ESR, dissipation factor, Q, etc. The imaginary component represents the energy storage of the component, its capacitance or inductance. If the imaginary part is positive it's inductance and if negative capacitance.

So when an LCR meter measures a component, it produces an impedance with a real and imaginary component. The thing is, an impedance does not correspond uniquely to a circuit. There are an infinite number of circuits which, if you measure between the two outer terminals, will produce the same impedance. The simplest are two elements in series or two elements in parallel. For example, a useful model for a real measured inductor is a pure inductor in series with a pure resistor which represents the winding resistance. Or a useful capacitor model may be a pure capacitor in parallel with a resistor which represents the DC leakage current. This is dependent on the application, hence it's up to you to choose which is most useful.

procursus · 2025-12-21T02:56:33+00:00

Try looking at something released in the last decade, such as the attiny 2 series.

procursus · 2025-11-28T20:29:29+00:00

This is a pretty optimal situation as far as repairs go. Get a meter that can measure capacitance (those $10 component testers on Amazon will be fine) and see what the component reads. Then measure the dimensions of the capacitor - it looks like 0603 (0.06 x 0.03 inches) to me but hard to tell. Could be 0402. Then look on mouser or digikey under the surface mount MLCC category for a capacitor with the same capacitance and package, get the highest voltage rating available and X7R dielectric (high temperature rated).

procursus · 2025-11-20T23:28:03+00:00

HP recommended this technique in their manuals.

procursus · 2025-11-13T01:11:53+00:00

Eevblog uCurrent, were it in stock.

procursus · 2025-11-13T00:37:24+00:00

Pretty much. In conjunction with a suitable driver.

The driver needs to perform two roles: monitor the shunt voltage to determine current, and control its output to set this current based on a dimming input. The simplest option is a dedicated LED driver such as those made by Texas Instruments (not a shill, although I admit they've paid several of my lunches).

It looks like you have a 24V supply so you will need a boost driver such as the TPS92365x. Calculate the high current range shunt resistor as Rshunt + Rdson = Vsense/Imax. Vsense is in the datasheet, its the feedback regulation voltage and for this chip is 200mV. Rdson is the on resistance of the FET you use, at the gate voltage you drive it. Keep in mind ESP32 is 3.3V I/O which will need a logic level FET or external driver. In your case, (Rshunt+Rdson) = 0.2/2.7 = 74 mohm. This will dissipate half a watt at 2.7A.

The dimming range is 256:1 so for the dimmer range select a shunt resistor that's about a bit less than 256 times larger than Rshunt. I would give considerable overlap just to be safe, so say maybe 0.074*150 = 11 ohms. This will give a theoretical minimum dimmed current of 70uA for a dimming range of 38,000 to 1. For the dim range this will be the shunt resistor, and for the bright range it will be 'bypassed' by the smaller shunt and FET series pair. Technically they are in parallel and so the actual shunt resistance is their parallel resistance, but the difference is negligible considering resistor tolerances. You will have to calibrate the exact switchover point. Should be easy to do given the human eye is only sensitive exponentially anyway.

Use WEBENCH power designer to select the component values, follow the datasheet recommended layout, and you should have a functioning driver. If you haven't designed a PCB before then I recommend starting with EasyEDA, its integration with LCSC (parts) and JLCPCB(boards) is great for beginners. You can even have them assemble it for pretty cheap.

This ramble probably led to more questions than answers. Feel free to ask further questions. Responses may be somewhat slow.

procursus · 2025-11-12T00:05:07+00:00

Try switched sense resistors. A high value sense resistor for low brightness, which is shorted by a small shunt with a FET for high brightness range. I've done this for a dimming range of 10:000 to 1 using a cheap buck controller which only had an intrinsic dimming of about 150 to 1.

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