Yes, you will have some noise left, even if the Zener removes most of it.
Zeners only have the rated voltage over them at rated current. So with a simple resistor-Zener circuit, varying the input voltage will cause the Zener current to vary too, which varies the output voltage.
In AC, the usual circuit having a resistor R and a zener in series forms a voltage divider with the diode's dynamic resistance.
Check zener datasheet for Rd, "Dynamic resistance" or "Differential resistance" of the zener (it depends on current).
Example: BZX84C-6V2 at 5mA Rd=6 to 10 ohms
Therefore, output voltage variation will be input ...
Zener is used to clamp excess voltage (noise) in the circuit. So I guess the answer to your question is no, Zener won't have noise and can effectively remove the circuit's noise. But, you have to select a Zener with voltage rating according to what you want.
3 pin vs 2 pin
the 3rd pin is a dummy plastic pin. The supply is double insulated.
3V to 4.5V
their specs are vague but the 0.05A current and power consumption will reduce with voltage but also increase lifespan somewhat.
I read they are rated for only 200h continuously. (due to vibrator contact arc erosion, I expect)
However if 4.5V* 0.05A ...
LSCS is one of the best suppliers of these Taiwanese chips.
There is greater stock and wider variety.
The suffix differences are significant but acceptable in your case.
All the G's are Green meaning RoHS meaning lead-free.
I could suggest + or - which differ by only +/- 50 mV. "+" gives slightly more capacity at the expense of charge cycle ...
The two are either the same part but renumbered due to expanded features of other variations, or slightly similar.
If you are working from scratch then look at the datasheets for both and decide if they are drop in identical or similar enough you can adapt it.
You have to make sure your using the datasheet for the specific manufacturer version of the dw01 ...
"with a current within 0.05A" doesn't make much sense. Whatever the case, "supply the pump with exactly the power it needs" must take into consideration the current needed for a very brief time to start the motor. That will be quite a bit more than the current needed to run the motor.
Typically what happens is that the motor lurches in an ...
My guess is that with high voltages if they are being switched synchronously by a pair of FETs, the lumped caps in parallel with the FET Coss helps to balance the voltage changes when switched off with an inductive load.
A reverse polarity protection diode is meant to be used in conjunction with a fuse, so that if power is connected backwards the diode will conduct and blow the fuse.
Sometimes this diode will be a zener or TVS (transient voltage suppressor) diode so that it also gives some protection against overvoltage, but very often is is just a plain old rectifier diode ...
Typically as drawn D1 would be a zener or avalanche diode and is used for semi-sacrificial overvoltage protection. In this case we are overcoming the reverse bias limit of a diode to perform it's function .
As comment stated you believe the documentation is wrong, in case of series diode it would protect reverse voltage , but the voltage drop and resultant ...
If it is time to be useful for a doubt not addressed here, let me know, meanwhile I will try to answer your two questions Q1 and Q2 as it goes.
Recommended solution for Voltage Dropout:
I understand you would prefer to focus on the relay solution, that I will discuss ahead, but a short word about the solution to mitigate the short voltage interruption - ...
Many Thanks to Bravale, Antonio51, and User_1818839.
I tested the circuit and it provides +489 volts and -244 volts. The caps are 470uF at 400 volts. For the positive side, I used 2 of these caps in series and each is paralleled with a 100K, 5W resistor for balance and discharge. For the negative supply, I used a single electrolytic (470uF, 400V) paralleled ...
This is likely due to overheating of the power supply. A lot of voltage regulators shutdown on overheat and auto-enable when cooled down. If the reason for overheating does not go away, they will overheat again and turn off again, producing the flashing effect.
The most common reason for overheating is overload. Are you sure you are not drawing too much ...
The op-amps are connected as a rail splitter to take the floating 10V supply and effectively make it +/-5V.
Loud hum in old audio electronics is typically because of old dried-out electrolytic capacitors in the power supply, such as the 470uF/25V C0002 in your schematic. If you have an ESR meter check that it is within spec, or simply replace older large ...
Maybe a power-path prioritizer is what you might need: -
It uses internal MOSFETs so there is little volt drop. One of two inputs is automatically selected. Please read the data sheets to ensure it meets your voltage and current requirements. There are other types that are more powerful should they be needed.
You would want to avoid having one power supply feeding back into the other one when both are connected at the same time for some reason. This might happen because of some voltage variations, even with identical supplies.
What you want is called 'supply or-ing' or just combining two supply rails. I take it they dont need to be selectable/switched?
For high ...
For parallel resonant heating circuits, you need to include a matching inductor in the inverter output in series with the parallel LC tank, something like as shown in the following image:
Image source: https://www.uihm.com/en/Induction-Heating-Technology/Base-details-of-High-Frequency-Induction-Heating.html
Having an inductor on only one side is in fact ...
Yes. The motors will only pull the current they need, as shown by their voltage vs current graph (or the values you listed). You can connect them directly to the 3.7~4.2V battery only and their strength will change as the battery drains. You can use a series resistor like you would with an led if you want to limit the power drawn. Or you can connect them to ...
What you can try :
add a fan blowing on your FET
add a heat sink on top of your FET (can be combined with fan)
if the PCB itself is hot (at least under the FET), then IF PLANE FOR IS NOTHING BUT GROUND, you can put your PCB on a big metal plate acting as a big heat sink for the PCB itself. Be carefull : if you have anything else than ground on the bottom ...
By splitting the supply like that you are creating a virtual ground. Note that your (audio?) input signal would have to be referenced to this virtual ground as well for your amplifier to work. If the signal is referenced to the ground of the power supply (CC) then it is still out of range for the amplifier.
The circuit on the left is stable like you pointed ...
But I don't understand, why? How does the voltage follower stabilize
ground? And how in the first place is the ground not stable in the
configuration on the left?
It's quite simple. The circuit on the left cannot sustain DC currents more than a few hundred microamps for a few seconds before a significant error is introduced in the "middle" DC ...
The wire to the first pump (the MCU-controlled one) can be observed with a Hall sensor, and that could trigger a relay turning on/off the second pump (High-voltage one).
There's a diagram in the UGN5303 datasheet for such usage of Hall sensors (current monitor).
Or better than that, if the systems are not isolated, both pumps could be switched by the MCU at ...
With this answer, I want to focus on the part that was not answered yet:
why Power supply rejection ration (PSRR) is never specified for DC/DC?
Is this value meaningless?
It is neither meaningless nor unspecified but depends on external components such as feedback loop, L, C, load.
In a small-signal, ac model, the PSSR can be shown to be
Do I have to put diodes between two PSUs?
No diodes are needed, diodes are used with power supplies to prevent the reversal of current, this happens if one voltage on one side of the diode would be lower than the other side (like if one supply went down or was disconnected). As long as both supplies are powered at the same time from the same 12V supply, it ...
You're overthinking this. You do not need to worry about connecting the power supply inputs in parallel. As long as your 12V source is high enough quality to handle the input ripple and load transients (and of course, the total power draw), there is no issue there, and you certainly do not need diodes or anything else for that matter.
it is only power ...
It means that the MCU cannot detect an undervoltage condition and reset itself if the supply voltage drops below the minimum operating voltage, but does not go low enough to be considered to be fully powered off, the internal state of any circuit in the MCU can't be guaranteed and it may not wake up from shutdown mode.
If the supply does not dip below ...
I can confirm that the DA-2 will likely work if you avoid combining the positive and ground wires. For my unit, I connected one +12V line to the input of my RGEEK-300L (LD-A300W) DC-DC power supply and one +12V line each to pins 3 and 4 of the ATX P4 power cable feeding the CPU. I then tied the sense and remote lines to the same ground as the DC-DC PSU. ...
I found a way to solve it amazingly and easly and cheap. I bought an oscilloscope that haven't received yet but by experimental I added a 0.01uf between the leads of the thermocouple and now it works without any issue. Also now the 3 thermocouples readings are exactly the same. Before I had like 0.50 C difference between each one and now is perfect.
Looks like they just want to ensure the output can go down to 0V so they sink a bit of current from the output via R134.
This will take care of any leakage in the Sziklai output stage, and will discharge C104 reasonably promptly when you crank the output voltage down.
Alright, I guess this counts as an answer. I'm an idiot. I checked so many things, checked that CRT connector til I was blue in the face. Missed a not so obvious check though. The connector going to the board had a bad contact in the female side that I'd never have seen if I didn't try the separate transformer trick and still got nothing. I then checked the ...
Your first problem is your choice of a Darlington follower for your output stage. The voltage across a Darlington cannot be less than two diode drops, which would amount to about 1.5 volts. So the output can never be greater than about 7.5 volts for a 9-volt supply - and this is an absolute limit. In your case your getting about 7 volts swing, which is a ...
Oh great, a 'manual' consisting of just a schematic with no component values. :(
Here is a part of it:-
Your wiring looks OK. VCC and IN(x) should go to the Rpi. JD-VCC and GND should go directly to the power supply via separate wires. To prevent a possible ground loop you should not connect GND to the Rpi.
It might be more convenient to connect Rpi 5 V to ...
To answer your direct question no, raising the source by 1.5V will not change Vgs to 4.8. Vgs is the voltage between the gate (g) and the source (s) so if you move the source higher you have to move the gate higher yet. (Clarification from OP in comments)
Yes pulling the source of the FET down to increase the Vgs would work but would certainly be unusual. ...
Yes, this will work, but can be cumbersome and somewhat non-ideal.
Is the VTH of the FET really 4 V ? How well does it turn on with 4.8 V on the gate ?
Now when the Teensy puts out a 0, the FET will get 1.5 V. This may make the FET partially on (leaky).
You would be better off to get a real level shifter device. Perhaps a 74LVC245 would be suitable.
This sounds a bit like a shopping question, however it's pretty easy to buy DC-DC converters for 300V more or less output. Just shop at your favorite distributor.
They tend to be a bit pricey.
Nobody (most likely) is going to want as low a current as you want, so probably something in the 1W range would be the minimum.
To monitor the current on the low side ...
I contacted the author of the article mentioned in this similar question and confirmed that they failed to measure noise in 18650 with their instrumentation, so this noise will be at most the same internal noise as the instrument used.
So the noise of the 18650 will certainly be less than a LDO with a noise of about 3 uV RMS.
In my case, it is better ...
R5/R6 are the feedback network, dividing the output voltage by 2. This feeds the base of Q1 which compares this voltage to the 22V zener used as reference. So the output voltage is set to (22V+0.6V)*2 = 45.2V.
When output voltage is too high, Q1 passes more current, bringing down Q2's base and reducing output voltage.
C4 is the Miller compensation cap for ...
From page 14 of the datasheet.
Your temperature rise is much higher than expected so I'd check that you're not saturating the inductor. A 65°C temperature rise above ambient suggests that you're running at 7.5 A.
The DC resistance is listed as 40.8 mΩ. 7.5 A will cause a dissipation of \$ P = I^2R = 7.5^2 \times 0.04 = 2.25\ \text W \$.
It is not possible to draw 5A with a 7824. In addition, it is not possible to regulate with just 1.8 V more. These regulators have a dropout limit of 2 V (typical.) Incidentally, series regulators are the worst choice as the excess voltage has to be dissipated as heat.
Perhaps you can proceed more skillfully and not regulate the voltage, as this should be ...
If a voltage regulator was the only feasible solution I'd be looking at a low-drop-out buck regulator like the one below using the LT8638S (for example): -
The example circuit above is for a 12 volt regulator but it looks like it should be good for a 24 volt output at up to 5 amps without much trouble.
I'm not recommending that device in particularly (...
Pin 16 of an HD38800 microprocessor is a "body bias" or "back bias" pin (Vbb). The HD38800 microprocessor is a PMOS device, (a family of devices that is now largely obsolete, having been replaced by CMOS). Vbb biases the substrate upon which all the PMOS devices sit.
The BJT Q1 appears to be regulating the voltage of Vbb. It is peculiar, ...
The 5V pins on the GPIO header are connected directly to the USB C power connector. There's a voltage regulator on board that provides lower voltages for the Pi circuitry. As far as I know, the device doesn't use 5V directly however it is sent to the USB ports, so don't forget to include anything plugged into those ports in your total power calculation.
I suggest plug type is 5.5x2.5 barrel jack. Fairly popular and useful kind of connector you can see everywhere, even disassemble from old equipment or broken power supply. To avoid soldering, you can buy this type of plug. Be cearfull not confusing it with 5.5x2.1 jack. (2.1mm wount feet (see the picute)).
With power supply, it is a ...
You can buy a UK housing, with UK plug-pins, for the innards of your EU wall-wart
These things are often glued or welded together so you'll likely have to very carefully apply a rotary tool or mini-hacksaw to your EU wall-wart to extract the PCB without damaging any electronics parts.
You can re-use the locking cable, even if you have to chop it off ...
A device will draw the power it wants. Just because it has current availability, doesn't mean it will want it.
Use a 8 amp buck converter with CC and CV control.
These have protection. But fuse the pin to your amp limit.
Now if you buy a 5v shelf product, you may find you can get voltage sag, or brown out, so it's worth checking the loads and voltage.
If your doing DIY POE (not according to spec and the device won't be plugged into other POE devices that aren't yours, there are some cheaper devices sourced from china that don't follow the spec).
If you are following the spec (and you do want POE compatibility, like connecting to POE switch, then the device will need to negotiate for more power with the ...
This brings down the overall efficiency to 374/462 = 0.81. Am I wrong in this calculation?
Nope. It's maths.
Basically, you have two cascade-connected converters (i.e. PFC's output is the main regulator's input). This means that the output power provided by the PFC pre-regulator is taken by the main regulator, and some portion of this power will be ...