I have a voltage regulator circuit powered by a CT. I rectify the output AC voltage to DC voltage and provide that as an input to the regulator L7805CV. My input voltage is about 7.20V When I connect the regulator, the input drops to 4.6 V and the output is only 3.3 V instead of 5V. When I try it with another regulated voltage of about 15 V, the output is fine to 5V and there is no drop from the input side. I tried the regulator circuit with appropriate capacitors at input and output side still makes no difference. Why is there a drop only when I connect it to the CT?
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1\$\begingroup\$ Show your circuit. How are you measuring the voltages? What do you consider to be 'appropriate capacitors', in particular what is the value of the input filter capacitor? \$\endgroup\$– Spehro 'speff' PefhanyCommented Dec 14, 2016 at 7:06
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1\$\begingroup\$ How much current does the transformer provide? \$\endgroup\$– Ignacio Vazquez-AbramsCommented Dec 14, 2016 at 7:07
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1\$\begingroup\$ I am measuring the voltages with a multimeter. I used .47uF for the input side and 100 nF in the output pin. @SpehroPefhany \$\endgroup\$– harishCommented Dec 14, 2016 at 8:23
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1\$\begingroup\$ CT details and current in primary of CT are required. \$\endgroup\$– Andy akaCommented Dec 14, 2016 at 10:22
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2 Answers
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The dropout of a 7805 is around 1.6V with no load, but this is not guaranteed you really need more voltage, but let's work with that. It also draws around 5mA with no output current.
To reduce the ripple voltage to (say) 0.1V you need a capacitance of 0.01s*0.005A/0.1V= 500uF. Your 0.47uF is woefully inadequate, and your measurements are suspect because they have a shipload of ripple.
That's assuming you have no other parts, however you have not entered your circuit schematic, so that's going to be it.
answered Dec 14, 2016 at 8:36
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\$\begingroup\$ I have no load connected to the circuit. It's only the CT, Bridge rectifier and the 7805 regulator in the circuit. \$\endgroup\$– harishCommented Dec 14, 2016 at 9:19
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\$\begingroup\$ This actually helped. I increased my capacitor value to 1000uF as it was only available now. I have the output pin to 4.3V. I will try making the capacitance value higher even more and maybe I will end up having the 5V required. \$\endgroup\$– harishCommented Dec 14, 2016 at 9:29
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1\$\begingroup\$ Probably not- you probably have to change the turns ratio in your CT. \$\endgroup\$ Commented Dec 14, 2016 at 9:35
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A Voltge transformer produces 10% higher voltage with no load and the peak is 1.4*RMS and the '05 Regulator has a TYPICAL dropout in datasheet of 2V. (old BJT technology).
This means you are expected to provide 7V Typ and 1.5times this for no load or 10.5V.
Voltage transformers have about 10 % conduction loss and achieve rated AC Vrms at rated load.
CT's have higher conduction loss if only rated for low secondary currents since the turns ratios tend to be high and only specified for lower currents at a fixed burden R load.
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\$\begingroup\$ I will try with the CT connected to a higher load so that I would get a higher voltage above 10.5V and will post the results. \$\endgroup\$– harishCommented Dec 14, 2016 at 8:37