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Hades Applets contents visual index ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() | Binary Adder/Subtractor (4 bit)
Circuit Description
A four-bit adder/subtractor demonstration.
While it is perfectly possible to design a custom circuit
for the subtraction operation,
it is much more common to re-use an existing adder and
to replace a subtraction by a two-complement's addition.
The applet shows how this is done. Click the input switches or type the 's' bindkey to control the Sub/nAdd switch, and '1' .. '8' to control the A and B inputs. When the Sub/nAdd input is low (0), the XOR-gates act as non-inverting buffers, and the carry-input to the adder is 0. Therefore, the adder calculates a four-bit sum plus carry-out: (Cout,S3,S2,S1,S0) = (A3,A2,A1,A0) + (B3,B2,B1,B0) If the Sub/nAdd input is high (1), the XOR-gates act as inverting buffers, and the carry-input to the adder is 1. (Cout,S3,S2,S1,S0) = (A3,A2,A1,A0) - (B3,B2,B1,B0) | |||
Print version | Run this demo in the Hades editor (via Java WebStart) | ||||
Usage | FAQ | About | License | Feedback | Tutorial (PDF) | Referenzkarte (PDF, in German) | ||||
Impressum | http://tams.informatik.uni-hamburg.de/applets/hades/webdemos/20-arithmetic/40-addsub/add-sub.html |