Activity for Andy aka
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #287257 |
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— | about 2 years ago |
| Edit | Post #287257 |
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— | about 2 years ago |
| Comment | Post #287257 |
It's a dc output that is unspecified but about 1 kW. The input supply is around +/-250 volts DC. There is no 3-phase connection. Output is variable between 300 volts and 600 volts. Thanks for asking. (more) |
— | about 2 years ago |
| Comment | Post #287257 |
No, I just don't want to use a transformer. Sure there's current sensing and voltage sensing and specialty drivers but these are beyond what I'm asking about. Basically, how would you configure the single ended power supply version to work with split input supplies is what I'm asking. (more) |
— | about 2 years ago |
| Edit | Post #287257 |
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— | about 2 years ago |
| Edit | Post #287257 |
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— | about 2 years ago |
| Comment | Post #287258 |
+1 for the obvious work round. I'm going to edit my question (with full respect to your answer) asking for non-transformer solutions. (more) |
— | about 2 years ago |
| Edit | Post #287257 |
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| Edit | Post #287257 |
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| Edit | Post #287257 | Initial revision | — | about 2 years ago |
| Question | — |
Buck-boost converter fed from split input supply I have recently finished designing a buck-boost converter for a job that uses a split (+/-) input power supply. Load power is taken equally from both positive and negative input supplies and, the load is connected to 0 volts (mid-rail of the split input supply). $$$$ For a single rail supply,... (more) |
— | about 2 years ago |
| Comment | Post #287232 |
Not trying to be picky here; just trying to clarify when you say \$R_{23}\$ is the resistance at point X seen from the input, wouldn't it be easier on the eyes if you said it's the resistance at point x with the input disconnected. Maybe I have misinterpreted. Similar story for \$R_{13}\$. (more) |
— | about 2 years ago |
| Comment | Post #287232 |
OK cheers. I thought afterwards it must be. (more) |
— | about 2 years ago |
| Comment | Post #287232 |
OK I understand R23 and R13 now. Just k I didn't follow. (more) |
— | about 2 years ago |
| Comment | Post #287232 |
There are no mainline on stream calculators that get it right except 1 (mine). They all show daft maths that appears to have been copied from the originator to all the rest hence, why I did my own. In your answer, I'm not sure what R23 or R13 is meant to mean and ditto k. I appreciate the work but yo... (more) |
— | about 2 years ago |
| Comment | Post #287197 |
R1 + R3//(R2 + RL) = Rin in mathjax is \$R_1 + R_3 || (R_2 + R_L) = R_{IN}\$ = `\$R_1 + R_3 || (R_2 + R_L)= R_{IN}\$`. Thanks for trying Olin
(more) |
— | about 2 years ago |
| Comment | Post #287196 |
Yes, both are connected and both must see their respective impedances when they look into the taper-pad port (no tricks etc.). I'm probably going to post another unrelated question to give the site a boost. Maybe you should think about posting the odd question (that you can answer) given you are site... (more) |
— | about 2 years ago |
| Edit | Post #287196 |
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— | about 2 years ago |
| Edit | Post #287196 |
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— | about 2 years ago |
| Edit | Post #287196 |
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— | about 2 years ago |
| Edit | Post #287196 |
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— | about 2 years ago |
| Edit | Post #287196 |
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— | about 2 years ago |
| Edit | Post #287196 | Initial revision | — | about 2 years ago |
| Question | — |
Deriving resistor values for a taper pad attenuator A taper pad is a resistive attenuator that maintains impedances on both ports and provides a specific amount of gain-loss (\$A{12}\$): - Image alt text I have derived formulas for each resistor (that I know to be correct) and have checked with micro-cap using DC analysis: - Image alt text ... (more) |
— | about 2 years ago |
| Comment | Post #286988 |
It's basically homework with no attempt to solve (more) |
— | about 2 years ago |
| Edit | Post #286945 |
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| Edit | Post #286945 |
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| Edit | Post #286945 |
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| Edit | Post #286951 |
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— | about 2 years ago |
| Edit | Post #286951 |
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— | about 2 years ago |
| Edit | Post #286951 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Technique to reset pulse transformer core quickly > Is there any way to reset a core quickly without having ground / reference terminal pulled towards negative voltage Unfortunately not but, if instead of trying to send the complete pulse through the transformer, you send edge information (i.e. use a series capacitor to perform a kind of diff... (more) |
— | about 2 years ago |
| Edit | Post #286945 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Power amplifier for remote controller The data sheet gives a perfectly good example of using the nRF24L01 without an external PA. It has an internal PA capable of driving an antenna Is there a way to calculate the approximate range of this remote controller? You can use the Friis transmission equation to calculate the free-space li... (more) |
— | about 2 years ago |
| Comment | Post #282053 |
@#53586 welcome to the punishment planet (more) |
— | over 2 years ago |
| Comment | Post #286850 |
The fuse needs to be on the input supply side to the zeners and not the output to the sensors. (more) |
— | over 2 years ago |
| Comment | Post #286823 |
I hadn't really thought about an actual device but the TLV looks a decent choice. (more) |
— | over 2 years ago |
| Comment | Post #286823 |
I won't work using this type of chip with a massive impedance feeding the chip's Vcc pin. I've said enough on this subject now. Take heed or be prepared for disappointment. (more) |
— | over 2 years ago |
| Comment | Post #286823 |
The absolute maximum voltage the supervisor chip can survive is 7 volts. Two series batteries can be 8.5 volts when close to fully charged. And, if that wasn't a problem enough, you would need to use a PMOS FET in the positive lead else you could carry on discharging the battery through the MOSFET bu... (more) |
— | over 2 years ago |
| Comment | Post #286809 |
It draws 6 uA current and that will still deplete your battery. I'd be concerned about that but sure, in principle, that's the sort of comparator idea that fits in with my thinking. Check that it can handle supplies as high as 9 volts and, regards the resistor values, there should be enough informati... (more) |
— | over 2 years ago |
| Edit | Post #286809 |
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— | over 2 years ago |
| Edit | Post #286809 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Battery protection circuit > What kind of IC can be used to drive the MOSFET, so that when the voltage drops below 6.2 V the battery disconnects from the load? That reverse protection MOSFET works fine for reverse protection but, it cannot be used to turn-off the load when the battery is correctly connected. Look closely... (more) |
— | over 2 years ago |
| Comment | Post #286792 |
The op-amp (beyond its local feedback) needs to have a resistor in series with the output that feeds the first inductor L1 else, it's not a Hartley oscillator. Please learn how to draw schematics correctly. (more) |
— | over 2 years ago |
| Comment | Post #286793 |
Actually, to make this circuit work correctly, you need theoretically an extra resistor to ensure that the pi-network behaves like a 3rd order filter and not a 2nd order filter. This means that the op-amp MUST have output resistance (despite the circuit showing none). I mean, what's the point of the ... (more) |
— | over 2 years ago |
| Comment | Post #286745 |
Yes, you could make a converter using a BJT oscillator and a transformer connected to a rectifier and smoothing capacitor but, it would be low-powered, inefficient and have poor regulation so, why would you bother? (more) |
— | over 2 years ago |
| Comment | Post #286738 |
A BJT doesn't have a CE junction; it has a BE and BC junction. The yellow line you drew is too simplistic as is the graph you drew it over. Each curve in that graph has a differently sloped saturation region like [this](https://it.mathworks.com/help/examples/sps_product/win64/ee_npn_02.png). This mea... (more) |
— | over 2 years ago |
| Comment | Post #286724 |
R/L is actually \$4\pi\$. You have inadvertently chosen to work in radians/sec but you didn't convert Hz to rad/sec. Rookie error. Also, you didn't appear to understand that the basic relationship involved is this: \$Q=\dfrac{1}{R}\sqrt{\dfrac{1}{LC}}\$ and, Q is centre frequency divided by bandwidth... (more) |
— | over 2 years ago |
| Edit | Post #286714 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Complex frequency of a pole > if we don't apply a sinusoidal signal at the input s may become a complex number $$$$ All real-world signals exist only on the \$j\omega\$ axis. Consider that a complicated signal can be broken down into a fundamental sinewave and its harmonics. The harmonics are still sinusoidal an... (more) |
— | over 2 years ago |