Recent content by raithrovers1

I agree with Averagesupernova. You are drawing 12A of current from the battery approximately. Your battery will drain quickly and the voltage will drop depending on the state of charge. The http://www.calibrepower.co.uk" is probably telling you that the input battery voltage is low. You also...

The power supply is rated to supply up to 1000mA therefore it can supply any amount of current up to 1000mA. You only need 500mA of power so you will be OK. The important thing is that the voltage is the same and therefore you should have no problems. http://www.calibrepower.co.uk"

Why do you want more current? The LED has certain requirements. It will have a maximum current rating and voltage rating. Most Led's will only tolerate 2V across them and a current of approximately 20mA. They also require a current limiting resistor in series with to drop the voltage required...

This circuit charges up C3 at first as Inv2 output is 0V. Once C3 is charged there is no discharge path due to the diode. The diode seems to be where you are going wrong. The diode allows the capacitor to charge up but in theory separates it from the battery once charged. The rest of the...

You need to take a step back and think about it. Power supplies incuding batteries supply a voltage but it is also called a potential difference. For example - a 12V battery only has voltage between its own positive terminal and its own negative terminal. The potential to supply power is only...

A lot of things can confuse people when first learning electronics so don't let it get you down. Binary, logic 1's and zero's are all different terms for ON and OFF. A logic gate is operated either ON or OFF. Each logic gate has different voltage levels as to what is classed as ON. It usually...

I had made a few assumptions. 1. that this was the actual charger supplied by Black and decker. 2. that the output voltage was 14.8V First of all the output of the transformer is 15.8V. Times this by 1.414 for full wave rectification. = about 22Vdc. Therefore the battery it is charging is...

The LED requires about 20mA to operate. Therefore with the 100ohm resistor in series the voltage across this circuit for the LED to light (when the battery is charging) is 3.7V. This means that the voltage across your burned out resistor while it is supplying current to the battery should be...

It would depend on device but I am sure that if it is specified that it requires DC then it must have DC. Most devices contain integrated circuits, processors etc. These require a steady DC supply to work correctly with reference to a steady ground point. You would then be supplying 5 volts for...

If life was only that easy! You need to get the battery data sheet which will give you the data. Battery discharge graphs are curves and are not linear therefore you cannot simply divide 50 by 2 to get 25 hours. The battery you mention is 50Ah but at what C rating is this? For example if the...

It is difficult to see how wind turbines on a large sclae will ever replace any other power generating scheme due to it relying on the weather and storing large amounts of power. However if it was done on a small scale ie. homeowners and private business fitting turbines, then once the...

No you wouldn't ignore the diode. In the first case the ideal diode is a short circuit. Current will always try to find the path in a circuit with the least resistance. The least resistance you can get in an ideal circuit is a short circuit so all of the current will flow through the short...

Congratulations on taking Electronics 1! I am sure you will enjoy it but it will give many days of sore heads and head scratching. Just keep in mind that it will make sense in the end no matter how hard you are struggling. To answer you first question. I think you are confusing yourself a bit...