I am designing a tachometer with 30 LEDs. I am using a 7805 voltage regulator with my circuit. The freq/voltage converter worked fine but when I connected the signal out to three LM3914 LED drivers to drive a total of 30 LEDs my voltage out of my 7805 dropped down to 3.4 volts. If my supply voltage to my regulator is 12v why is the regulator not regulating my voltage to 5 votls? Is 30 LEDs drawing too much voltage for this?
I appreciate any help that you can provide.
Thanks
Actually, I'm not using a heat sink for my regulator. Do you think it's necessary? Ok...I think I may have found the problem. The spec sheet for the LM3914 pg. 14 states that several LEDs could be a problem and that I should put a resistor in series with the power source that leads to the LEDs along with a capacitor to ground. This should limit my wattage.
Thanks for your help.
Actually, I'm not using a heat sink for my regulator. Do you think it's necessary? Ok...I think I may have found the problem. The spec sheet for the LM3914 pg. 14 states that several LEDs could be a problem and that I should put a resistor in series with the power source that leads to the LEDs along with a capacitor to ground. This should limit my wattage.
Thanks for your help.
If the 7805 gets too hot to touch, it will require a heatsink, a lot depends on the incoming supply and the current drawn by the load. With a 12V supply and the maximum 1A the 7805 is dissipating 7W of heat, and would require a substantial heatsink.
No. It gives a regulated current output, not a voltage output. It will operate on a supply as low as 3V, but its max dropout voltage of 1.5V needs a supply at least 1.5V more than the LED voltage drop.
If the LEDs are 2V with 20mA programmed for them, with your "recommended" 8V supply the IC must dissipate more than 1.2W which will make it extremely hot. With a 5V supply the IC's heating will be halved. Of course you can't make dissipation disappear, it will be transferred to the regulator and its heatsink.