Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now

# Tag Info

0

Microwave transformers are cost optimised. They run at high magnetisation current so that the core material is "very well used". Removing the shunts, as you have done, is a good start. Adding more primary turns may be difficult with the windings that you have added, but, given that you have the now removed shunt volume to work with, this suggests that ...

0

A paper written by D. Ludois, J. Lee, P. Mendoza, G. Venkataramanan suggests that adapting a MOT transformer for other use could benefit from removing the magnetic shunts and the filament winding and adding about 10 turns to the primary. There may be information on the internet that shows how to do that. Look at: Reuse-of-Post-Consumer-E-Waste-for-Low-Cost-...

0

(not a direct answer) I don't use the equations. I figure it out from datasheet plots and excel as I did for the diode. How to model a real diode 1N4148? Then I verified on Falstad Simulator with my parameters Since Ri load line at BJT saturation or FET conduction has 50% factory tolerance , your choice of datasheets matters with overdrive levels. ...

3

The output is very likely a push pull stage which has a (relatively) low output resistance. What you effectively have is this (assuming you are using a CMOS device) simulate this circuit – Schematic created using CircuitLab The output when high will have M2 on and M1 off. The output of this stage when high will have a typical resistance to 1.8V of ...

0

Before illumination, the charge density at any point is zero in the semiconductor. Once illuminated, the generated excess carriers (electron hole pair) start diffusing into the semiconductor. If both the carriers diffuses in the same manner, the semiconductor will be having zero charge density at each and every point. But electron and hole has different ...

0

The negative sign in the equation $E = -L\frac{di}{dt}$ implies back emf i.e. the voltage induced in the coil due to current flowing. Given that you might want to calculate the current that flows in an inductor for an applied voltage, the formula is $V = L\frac{di}{dt}$. Note that I used the voltage symbol "E" for back emf and "V" for applied voltage ...

-2

I think you can simply use a diode bridge and a capacitor to get 12V DC from your transformer and connect to the battery. Since the 12V battery usually charges to 14.4V maximum DC voltage, your AC power supply will not damage the battery by over charging it. If the bat voltage is low, it will be charged up to a little over 12V and that's it. For better ...

0

Load power is positive V=LdI/dt while generated power is negative including back EMF while coasting a motor with generated output. Dont confuse these conventions for L phase current polarity.

0

8 Ohm speakers can be 4 Ohms DC so a big cap to block DC reduces wasted power as they show in MAX9788 spec. TO generate 5.5V max use only good batteries. e.g. 3x C cells 4.5V or 4 x1.2V = 4.8V NiMH. or DIY discrete design below. The driver should be a push-pull FET bridge going direct to the battery. A Piezo Speaker or something >32 OHms would work ...

0

In your case I would use a simple two cell 28650 and use a buck converter to keep a steady voltage of 5V. Using those cells is better and can deliver more current to your circuit - way better than a 9V battery. regards

2

If you use a buck switching regulator instead of a linear regulator like the 7805, you can get more current at 5 V than your battery puts in at 9 V. But you can't get more power out than you put in. So to get 850 mA at 5 V, you will need something like 500-550 mA at 9 V, which is more than it's reasonable to expect from a 9 V battery. You could either use ...

0

Are there any obvious problems with the circuit, that may lead to the observed drifting behavior? The minimum supply for an LPV821 is 1.7 volts and you are expecting it to work above 1.6 volts. Your AC current is 1 uA and this produces a voltage (across the 20 ohm resistor) of maybe 28 uVp-p. Given that your op-amp is running without feedback AND it has ...

1

In the case shown, with both diodes near the power supply but with a very long wire between them, there will be no delay. However, in the case shown below, there will be a delay: simulate this circuit – Schematic created using CircuitLab The reason there's a delay in this case and not in the other is related to the fact that information must travel ...

4

But what assumption should we make about the voltage on the collector on the NPN BJT? You don't need to make any assumption about the collector voltage of the NPN. You have enough information to find its base current, and you know its $\beta$. From this you can answer the question without even calculating the collector voltage. (Although you should ...

0

Yes, D1 will emitt light a very short time earlier than D2. The current flowing through D1 will have to load the the very long cable's capacity before the same current will flow through both LEDs.

0

No, for an ideal diode (or LED) there is no capacitance or inductance and the current through both is equal. (and in the ideal world wires are made out of superconductors) For a real world diode there is a slight amount of parasitic capacitance and inductance that might cause a slight difference if you're looking at very small timescales, for most ...

3

the LDO may look like this. The only path for the 16uA is thru the feedback-loop divider resistor chain. simulate this circuit – Schematic created using CircuitLab Normally the opamp, itself wired in non-inverting operation because the PNP power dissipation device (that bipolar with emitter wired to Vunreg) is inverting, implements a negative-...

0

simulate this circuit – Schematic created using CircuitLab Class D audio amplifier would be a good choice. The one that you can eliminate the blocking capacitor at the input. Your current requirement seems to be a close match for what an D class amp is made for. But, the audio amp would amplify the voltage in certain relation. It compares input ...

0

The peak current will be maximum at about half way on the dimmer, when the thyristor is switching near the peak of the AC waveform. You'll get approximately that voltage divided by the light bulb resistance (which changes with voltage as it heats up so the actual maximum may be a bit lower than 50%). You can perhaps estimate it from the known hot resistance ...

0

Actually, this could be a tricky question, because a light dimmer will typically have a pulse-width-modulated output, so assuming a sine wave, driving the cap and lamp would not be correct. If you had a pure sine wave and simply wanted the Magnitude of the current, then you can compute the Reactance of the cap, and add that in series with the lamp ...

2

Bottom of page 17 of this Farnell document defines it as “GPA – General Purpose Amps (Inductive Load)”.

2

To really know, you'd have to have a few hundred dollars to get access to the CUL 61058-1 document. However, it probably stands for General Purpose Amps GPA. The rating is the same as a regular amp, they probably redefine it to differentiate between DC amps and AC amps or to mean both. Source: http://www.state-elec.com/honeywell/pdf/MICRO-SWITCH%20V15.pdf

0

For reasons of power dissipation and voltage load regulation , the most common current shunt chooses the R value to be 50mV at max current. (More or less) Then choose the Pd rating for the resisitor to be 2x pwr of 0.05V*I for 50% reduction of 120’C rise in temp. of the resistor. This is also used in high side current sense IC’s which use analog parts ...

0

Yes. There is a trade off between range and precision of measurements in any ADC setup. Since current sensors only give us sensitivity in V/I and we care only about I, sensitivity is inversely related to precision. This also makes sense as more sensitive sensors ought to reflect smaller changes in current.

5

You're using a linear regulator which simply "burns off" the excess voltage. The current does not change and remains the same so you can draw up to 1.25 A at the output of the regulator. So after the regulator you're limited to 5 V, 1.25 A so 6.25 W. When you draw that 6.25 W there is 12 V - 5 V = 7 V at 1.25 A meaning 7 V * 1.25 A = 8.75 W dissipated in ...

3

In a linear voltage regulator as the LM117, all the voltage drop × current is turned into heat. That's about 9W in you case. You can draw 1.25A@5V from the 5V output. If you wanted to draw more current on the 5V side than it is supplied on the 12V side and produce less heat, you had to use a switching regulator. There are some manufacturers which produce ...

1

Quoting from Wikipedia Switching converters (such as buck converters) provide much greater power efficiency as DC-to-DC converters than linear regulators, which are simpler circuits that lower voltages by dissipating power as heat, but do not step up output current. Buck converters can be highly efficient (often higher than 90%) Most converter ...

1

The conservation of energy and the regulator efficiency are the key to understand your problem. simulate this circuit – Schematic created using CircuitLab One thing is certain: $$P_{out} = \eta \cdot P_{in} \\ and \\ P=U \cdot I$$ Thus as an example if your load draws 3 A at a regulated voltage of 3 V and your regulator as a 90% efficciency and a ...

1

For each converter, $P_{out} = P_{in} \times efficiency$ Therefore, $P_{in} = \frac{P_{out} }{ efficiency}$ The total battery current is $I_{in} = \frac{\sum P_{in}}{V_{in}}$

0

Thank you for the answers, it really made me think and I find solution of @Aaron and @Oldfart well fitting for one relay. Only, what I find disturbing is, when I have 16 charging capacitors by only 5V, 50mA power source, which means that the current will be divided. My conclusion is to use extra power source with suitable power rating as @AndersG said (I ...

1

look at the number next to the pointer, if all the digit positions are filled that would be the maximum reading on the display. the decimal marks the boundary between one thousand multiplier an the next, betweem mA ad uA or between A and mA I understand that the current is probably 220 mA (thus 0.22) but what is the signification of 02.2 (see the 3rd ...

1

Depending on the type of your power supply, the effects will vary. If it is a smart power supply with CC and CV modes, the current will be limited to 1A (in CC mode) and the voltage will be reduced. In the case of a dumb power supply, either a safety mechanism will be triggered (a fuse or a breaker) or it will continue to work at a undesired state. The ...

2

There are a number of things that can happen. The supply blows a protection fuse. A designed in circuit breaker trips. The supply let's out smoke and dies. The supply output voltage is dragged down to some much lower output voltage and delivers less current correspondingly. The overload makes the supply temperature rise to a point where a thermal switch ...

2

The internal digital meter inside the multimeter is a 200 mV (0 to 199.9 mV) meter module. We know this because it is standard, popular specification and because the dial's most sensitive range is 200 mV DC. For current measurement the meter needs to switch in a shunt that will drop 200 mV at the specified range current. From Ohm's Law we can work out the ...

2

We've manufactured precision current sources for over 26 years and can offer some suggestions. For current sense use the Vishay VPR221Z series or the VCS331Z depending on your wattage. For best results heat sink the sense resistor despite the Tempco of 0.2ppm/°C. Use instrumentation grade amplifiers such as the INA103 for current sense and also for voltage ...

1

Although the negative resistance is veiled in mystery, in fact it is a quite simple concept. It can be easily explained by analyzing the voltage drops across resistances. The positive resistor subtracts its voltage drop from the input voltage thus decreasing the current while the (S-shaped) negative resistor adds its voltage drop to the input voltage thus ...

1

A slight modification to your design. A mosfet can have a lower voltage drop than a BJT due to small Rds-on (10mΩ is common). It's not about transistor current gain; you need to operate the transistor in saturation mode. This is much easier to do with a mosfet. No need for the upper transistor Q2. I added C1 as a current reservoir, like @Oldfart ...

0

Q2 is not needed and R1 could be larger, say 10k. I suggest you check out the ULNxxxx series of relay drivers. There you have say 8 drivers in one DIL package. Anyway, you will need an extra power source.

2

You can't in the way you try it. Your source is 250mW, your relay requires 500mW. You have not got enough energy. A transistor can amplify current but not out of thin air. It needs to come from somewhere. What you can try is to see what the relay hold current is. It is often much lower then the 'attack'* or "pick" current. In that cause you might store ...

0

There aren't really good options for limiting current in the current and voltage spec you have. One could limit the current with a DC motor through the supply and get a 48V 5A supply, but read up on the supply you purchase and how it handles overcurrent (or under voltage conditions) Another option outside of a fuse is a Positive Temperature Coefficient ...

0

The only way to clamp or cap current (same thing I believe) is to power the motor through an electronic speed controller (ESC). With a DC motor, limiting the current is pretty much the same as limiting torque. Once the current limit is reached, the limiting will reduce the voltage to whatever value is required to hold the current at the set level. It that ...

3

There's no reason to drive the GPIO pins high at all. Just configure them as open-drain — drive the pin low when you want to activate the corresponding "button", and tristate it otherwise. As long as the fob voltage is not higher than the MCU voltage, it will be fine. And running the fob at 3.3 V should be fine, too.

3

Another important point when using low valued sense resistors is using a Kelvin connection. In the upper picture, the thick trace will have an impedance (resistance and inductance). Also the solder joints (or whatever connection you use) to the sense resistor will have contact resistance. When using small sense resistor, you can certainly not neglect the ...

3

This circuit will have issues as you're over the edge concerning the input common mode range of the AD823. At the left side of R4 there's 15 V. The + input of the opamp will get: 10k / (10k + 1 k) * 15 V = 13.6 V Now let's consult the datasheet, on page 5 there are numbers for a +/- 15 V supply, not what you're using but it will gave an indication. Look at "...

4

But i want to understand from circuit perspective what i could do here to minimize various errors and get the best circuit performance ? The first thing to take note of is the input common-mode voltage range and, for this device it is limited to 13 volts on a 15 volt rail. Typically it might be as high as 13.8 volts but, given that the +in node will see ...

1

Since DC motors are also generators, the reverse voltage raises proportional with the RPM. So the higher the RPM, the higher the reverse voltage -> less current. Also the ambient temperature is most probably not stable, which leads to fluctuating current measurement results.

3

No-one can know what will happen to your module in that case. That's because from "Update 2" in my answer on a previous question about the same "XY-016" module, there is evidence of multiple different MT3608 clone ICs being used on those modules. The behaviour could be different depending on the specific clone that you have. Either don't risk it (which is ...

2

Using a battery that is not brand new so its loaded voltage is 8.0V and using a 3.2V blue LED, then the current is (8V - 3.2V)/220 ohms= 21.8mA. No problem when most 5mm blue LEDs have a maximum continuous current of 30mA.

2

Here is how this works ----- simulate this circuit – Schematic created using CircuitLab I picked a 180 ohm resistor, because at 20 ohms/volt we are guaranteed a 50 milliamp intersection at the left axis. Thus we scale 20 ohms/volt by 9 volts, finding a 180 ohm resistor provides that current. Note other LEDs will have lower forward voltage, moving ...

1

The battery will go flat sooner. Thanks for the comment, however will it impact the functioning of the device? E.g. if the device needs say 630 mA of current, will the battery be able to supply it with its 600 mAh capacity? We don't say 630 mA "of current" in the same way we don't say 65 km" of speed" although we can say it the other way around, "a speed ...

Top 50 recent answers are included