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Hero999
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Posts posted by Hero999
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Rather than messing around with PDFs, the address to the original image is at the top of each page.
It doesn't look like it'll work very well. The peak value of VCC will be around 15*sqrt(2) - 0.7 = 20.5V. Q1 and Q2 are source followers with a voltage drop of over 5V for even a small current drain and 0.5V of headroom doesn't allow for much ripple.
If it's just a headphone amplifier, this could be fine but it's much easier to replace that op-amp and MOSFETs with an LM7815 and LM7915 which should give you superior regulation and current limiting too.
A properly designed audio amplifier doesn't need a well regulated amplifier. You could probably use a 12V-0-12V transformer, with an appropriate rectifier and filter capacitor. -
I got an energy meter free from the energy company but the batteries ran out and now it's hardly used.
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A thermal circuit breaker would do.
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Let's get this straight: you're replacing red LEDs with blue ones right?
If so, it's no surprise it'll be dimmer: the human eye is less sensitive to blue light (although this is somewhat negated by the fact blue LEDs are very bright) and the higher voltage drop means less current will flow, given the same series resistor. You can increase the brightness by reducing the series resistor but check the LED datasheet to be sure you're not exceeding the current rating. -
hello audioguru
1)I need 5 dc for output voltage.
2)I will find current after calculation of voltage.
That's not the correct way to design it. You need to calculate the current first, before you can build a power supply. If you build a power supply which doesn't provide enough current, the voltage will fall below 5V and the circuit won't work. If you over engineer the power supply so it can provide a lot more current than you need, you would've wasted your money and it'll take up more space than necessary. -
I don't use that software.
Could you please post a PNG, GIF or PDF so I can see it. -
Are you sure it's 100Vpp and not RMS?
100Vpp is the same as 35VAC RMS. Most transformers are specified in RMS and it's unlikely you'll be able to find one with a 35V primary as it isn't a standard mains voltage in any country. -
The feedback is normally talken via an opto-coupler to give isolation between the primary and secondary side.
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There's nothing special about an op-amp. It's just a differential amplifier with a very high gain.
In the circuit posted above, all the op-amp does is try to keep the voltage on the inverting and non-inverting inputs equal by adjusting its output voltage. The voltage across R2 will be nearly equal to the voltage on the LM358-2.5 which is 2.5V. R2 forms a potential divider with R1 with a division factor of 9 so the voltage across R1 is 9 times higher than R2. The voltage accorss RL is the same as the voltage across R1+R2.
Putting two op-amps in the same feedback loop is a bad idea. -
The AC component of the signal remains the same. The high pass filter blocks the DC component and there appears to be some DC bias on the output which isn't shown, otherwise it would be centred around 0V.
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The buffer has a high input impedance and low output impedance and is used to avoid the effects of loading.
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I agree with the above: you should test the 555 timer circuit first to be sure it works on its own.
If the relay is already being driven from a transistor, there's no need for a diode.
Have you connected the 0V rails of both circuits together? There needs to be a common 0V reference for the circuits to interface with one another.
If a pulse is applied to pin 2 for longer than the length of the delay, the output will stay high until the input goes low again. If this is a problem, you should AC couple pin 2 to the trigger signal via a small capacitor (100pF to 100nF) and bias it to +V with a pull-up resistor (10k to 1M).
See the application note linked below for more information:
http://www.doctronics.co.uk/pdf_files/555an.pdf -
Did you use a back-EMF protection diode?
How much current does the relay take? You may need to add a transistor to drive the relay. -
Your circuit uses more components and is inferior.
It doesn't have to have a gain of 1. It's possible to use a higher gain.
See the circuit attached which has a gain of 10, enabling a more accurate 2.5V reference to be used, rather than a zener diode. It also has current limiting and a capacitor is connected to the output to improve the transient response.
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Most capacitors in this value range have poor tolerance/stability. Ceramics are probably the worst as far as stability is concerned. The capacitance drops as the bias voltage increases and they're often piezo electric, i.e. when they vibrate, they generate electricity. Electrolytic capacitors have very poor tolerance, typically -20%/+80%.
Polypropylene or polyester is probably best for this application as 5% tolerance is widely available and it has good stability. -
That's true but there's no point in trying to aim for any decent level of accuracy, when typical capacitors have a tolerance of 20%.
sorry i didnt see this one . i was looking for something with maths. but that mean if i wnat 10s. then i have to use 1Mohm and 10uf but with the app says with this one will give me 11s.
im ok with these numbers :) . thanks a lot.... -
It does say on the datasheet, see figure 11 on page 10.
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It's on the datasheet.
http://www.ti.com/lit/ds/slfs022h/slfs022h.pdf -
Yes, you're right. The inverting input needs to be connected to the output so the Darlington pair is in the feedback loop.
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Because it takes time for a capacitor to charge and discharge and time for a field to grow/decay in an inductor.
Think of an inductor as a flywheel, you can't change the speed of a spinning flywheel suddenly, it takes time to speed up and time to slow down.
A capacitor is the opposite of an inductor. -
Here's a hint:
37% = 100% - 63%
So if you have a 1uF capacitor and a 1M resistor in series, the time constant will be 1s.
If you start at zero and connect the circuit to 100V it will take 1 second for the voltage on the capacitor to reach 63V, that's 63% of the steady state value.
On the other hand, if you start with the capacitor fully charge to 100V and connect the resistor in parallel, the voltage will decay to 37V in 1 second. -
The circuit will give poor regulation because the gain is low.
Use a single op-amp and put the pass transistor in the feedback loop.
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I don't know. The pictures you've posted are far too small to see anything.
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I can assure you a black light will work for PCB exposure - I've done it before with good results.
What are these?
in General
Posted
The look like capacitors.