Q:

BCD to 7 Segment is producing too little voltage

I am trying to use a BCD to 7 segment decoder to power my 7 segment display. This is the decoder I'm using (http://www.mouser.com/ProductDetail/Texas-Instruments/SN74LS47N/?qs=sGAEpiMZZMtYFXwiBRPs05jHvQJpC1NM).
Right now, I just have it hard wired to feed ground into it's binary inputs. I believe this should result in a zero being displayed. However, nothing is being displayed. I used a multimeter to check the outputs and found that only around .14 volts were being supplied out of the chip. I have verified that the supply voltage supplied to the chip is 5 volts and the chip ground is correct. I tried swapping the chip, and that didn't help. If I directly wire the 7 segment display, it operates normally.
When I change the inputs, I find that the correct outputs are high and low, but the high only goes to .14 volts.
Did I wire something wrong? Or maybe I misunderstood the operation of the chip? I admit I'm a novice when it comes to this stuff.

A:

The chip you are using has open collector outputs. You won't measure any voltage on the output of the pins as this are used to sink not source.
If you think of the segments of your 7 segment display of LEDs, the output of your chip will actually function as the ground. If we were to look at the internal of your chip then it would look like this

simulate this circuit – Schematic created using CircuitLab
When your output is 'high' it turns the transistor on and allows current to flow from the COM pin of the 7 segment display, through the LED and to ground. This however is only the case for a common anode 7 segment display. If you have a common cathode 7 segment display then your LEDs are going to facing the opposite direction which means when the transistor turns on your will have both the anode and cathode of the LED connected to ground, hence why you won't see the LED turn on.
Edit: After reading that you've used a common cathode 7 segment display then this is what your circuit actually looks like

simulate this circuit
All you've got here is a bunch of grounds so there's not going to be any current flowing through the LEDs of your 7 segment display. In response to the comments of the guys above, what they're suggesting is to add some pull up resistors to the output. What this means is that when the output is low and the transistor is off, the LEDs will be powered by the 5V that flows through the pull up resistor

simulate this circuit
The easiest way however is to just buy a common anode 7 segment display

