electronic devices

Question

Every time I try to run the circuit the fuse breaks and the display turns off.

also I need the display to be supplied by the five voltage output from the power supply I designed

I attached the multisim file

jw0752 · Answer

Hi Anwar, If C2 is really 530 Farads as depicted in your schematic it is going to look just like a short to the power supply. You will have to add a resistor between the power supply bridge output and C2 to limit the current to a value that is less than the failure level of the fuse. At five volts and one amp the would be a value of about 5 ohms. John

michaelkellett · Answer

530F is a strange value for the capacitor - it's way, way too big (possibly a typo ?)

If the current charging it was 10A it would take 53 seconds to get only 1V across the capacitor - can't be charged through a 1A fuse.

You are using a simulator - so look at the current in the transformer secondary.

A sensible value for the capacitor would be 470uF (about 1 million times smaller). (Biggish electrolytic capacitors are made in a limted range of values - 220, 330, 470, 560 uF are common, I've never seen a 530uF capacitor).

Out of 22600 aluminium electrolytic caps Farnell offer 1410 at 470uF, 288 at 560uF, 40 at 510uF and none (0) at 530uF.

MK

colporteur · Answer

Can you separate the power supply from the circuit? My lab professor in electronics hammer into us the concept of incrementally dividing a system down in the troubleshooting process thereby isolating the problem area. The example used was an AM/FM radio. The user complaint no audio. Starting at one end and working through a system is a very inefficient troubleshooting process. If you divide the radio in half you begin to eliminate half the circuit at each point until you arrive at the end that is not working. In this case you have designed and built a power supply and circuit. The project works on paper. The real test is when built. By isolating the circuit load from the output of U3, you split troubleshooting the power supply from troubleshooting the circuit.

jw0752 · Answer

If C2 is 530 uF then no resistor between the bridge and C2 is needed. Let's assume that your output voltage to the segments is 5 Volts, and that the voltage drop across the segments is 2 volts. 5 Volts - 2 volts is 3 Volts so that is the voltage that will drop across the resistor that we place in series with the segment. In this case you have chosen 100 ohms. If we apply Ohm's law to determine the current we get 3 Volts divided by 100 ohms which is 30 mA. I would consider this high for a a seven segment display. To make the display last longer 330 ohms might be better. You can also check the data sheet on the seven segment display to get more accurate junction drop voltages and recommended segment currents. John

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