The following circuit, derived from the DAC0832 digital to analog converter (DAC) datasheet is one approach to making the conversion from stepwise digital information to a voltage. It's called an R 2R ladder and is part of the circuit used in the two DAC0832s on the board. R and Rfb are about 15K ohms, which makes 2R about 30K ohms. The actual values are not as important as the fact that the resistors are very closely matched to each other.
The "1" and "0" indicate the positions of MOSFET switches in the converter. A switch will connect to the "1" side if the corresponding bit is on, and to the "0" side if the bit is off. A switch connected to the "1" position will send a portion of Vref-derived current to Iout1, whereas a switch connected to the "0" position will send a portion of the current to Iout2.
To see how the R 2R ladder fits into the scheme of things, consider the following. The left side is a copy of one of the DAC0832 sections taken from the board's schematic. On the right is a simplified block diagram of the same thing, derived from the datasheet. Rfb is drawn in the manner shown to indicate it is internal to the DAC but can be accessed outside and be connected the Op-Amp:
Iout1 of the R 2R ladder is connected to the inverting input of the Op-Amp for more on Op-Amps). Iout2 of the R 2R ladder is connected to the non-inverting input and to ground. One end of the internal Rfb feedback resistor is connected to the output of the external Op-Amp. The other end is internally connected to the R 2R ladder's Iout1 as shown above. Thus, it is connected from the output of the Op-Amp to the inverting input.
CIRCUIT # 2
The current through R1 does not influence the current referenced at the inverting input. Although one end is connected to the -5V reference, the other end simply goes to ground, so the current through R1 never even makes it to the inverting input.
All of the other resistors do contribute to the current at the inverting input and to the output of the amplifier. The current through R2 splits into two paths. One is through R3 which goes directly to the inverting input, and the other is through the network of R4 and all of the other resistors to ground. The method to determine their composite value is more easily seen if the R4 part of the drawing is reorganized. A careful study will show that only the drawing has changed. The circuit is the same:
First consider the two 30K resistors on the bottom. Recalling the parallel resistor equation from How To Read A Schematic,
Rparallel = 1/(1/30K + 1/30K) = 15K ohms
A tip: the parallel value of resistors of the same value is the value of one of the resistors divided by the number of resistors.
The 15K parallel combination adds to the 15K above it since it is in series with it. Again, from How To Read A Schematic,
Rseries = 15K + 15K = 30K ohms
This 30K combination is in parallel with the 30K to the left of it, which produces 15K again which adds to the 15K above to produce 30K, and so on. The final value of the network is 30K. It would be a very good idea to print the picture, calculate the values and write them down as an exercise to clarify the process. There are other, more sophisticated ways to analyze the circuit, but this is straight forward and will do just fine for now.
Reference
http://www.learn-c.com/experiment8.htm
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