Electronic – Seven Segment problems,voltage and current

Actually this is a very common arrangement. Each digit has its anodes connected together, and the segments for all digits have their cathodes connected. The controller enables one of the common anodes, drives the segments, and waits, then turns off the cathodes and moves on to the next common anode. This arrangement and control method is called multiplexing, and the linked article will give you a lot of information. The part in question is a common anode display, but common cathode displays are just as common.

Multiplexing allows you to drive a large number of display elements with fewer wires than individual addressing, at the expense of a more complex control algorithm. You can go a step further by doing something called Charlieplexing which allows you to drive even more elements with less I/O if you can turn off the drivers (as opposed to just setting them high or low, which is the requirement for a multiplexed display controller).

Now that you know it's called a multiplexed display, you should have no trouble finding all kinds of example circuits or microcontroller code fragments to implement it.

My first assumption from the symptom is that you have a single cathode resistor, which would explain everything. However, you say this is not the case and that you have individual 100 Ω resistors on each anode line.

The answer then must be that whatever is driving the cathode can't handle the combined current. Apparently you are driving the cathode with just a microcontroller output, which has insufficient current sink capability. Do the math. You want 10 mA thru each of 7 segments. That means whatever is holding the cathodes low has to be able to handle 70 mA. A microcontroller pin is very very unlikely to do that. See the datasheet.

The solution is to add a transistor that can sink the current from each of the cathode connections:

70 mA is easy for any discrete transistor you will be able to get. Let's say the transistor has a minimum guaranteed gain of 50. That means the base current needs to be 70mA / 50 = 1.4 mA. Figure the B-E junction drops 700 mV. With 3.3 V drive from the micro, that leaves 3.3 V - 700 mV = 2.6 V accross the base resistor. 2.6 V / 1.4 mA = 1.9 kΩ, which is the absolute maximum allowed base resistor. Given that, I'd use 1 kΩ between the digital output and the base.

Note that this will invert the common cathode logic from what you had. The microcontroller output now has to go high to enable a digit.