Digital Ramp ADC Analysis difficulty

I'm trying this problem on allaboutcircuits too. However they havn't been as helpful as this board is for other problems (mostly mathematics).

Homework Statement

I have been given an assignment which is part of, what I consider, the most difficult part of electronics engineering. Circuit analysis. It is to analyse the attached circuit, which is a fuel level meter for a car. It is a design for an ADC to convert an analogue voltage (v_fuel) ranging from 0 to 3V (3V being a full tank) into a 4 bit binary number, which is then sent to a 7 segment display.

(a) Calculate suitable resistances for R1 to R5 so that the binary values from the counter produce voltages from 0 V to approximately 3.75 V in approximately 0.25 V steps at point K in the circuit.

(b) Explain and sketch graphs of what is happening at points J, K, L.

Homework Equations

Resistors R1 to R4 are related in that R3=2*R4, R2=2*R3, R1=2*R2.
OA1, OA2, OA3 are all HCT logic family op-amps (TL074).
Vp = 5V, Vm = -5V.

The Attempt at a Solution

I understand that the resistor network, and, OA1, OA2, OA3 are all together a DAC. OA2 is simply inverting and isolating, OA3 is definitely a comparator and I understand that more than any other part. I know OA1 is a summing amplifier, but I am uncertain how it relates to the resistor network. I think that it is shifting the voltage range from 0V -> 5V, to 0V -> 3.75V. I am having extreme difficulty approaching this problem, I have all the lecture notes and books on this topic. Should I treat the summing amplifier as an inverting negative feedback op-amp, and find a fractional gain of it, then apply the descending values of resistance? (b) should be straightforward once I understand (a).

Attachments

• cct.png
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vk6kro
The counter and the opamps are there to generate a stepped ramp which rises to eventually equal the fuel gauge voltage. This causes the current value of the counter to be loaded into the register.

You could start by assuming all the outputs of the counter are high. This puts all the resistors (R1 to R4) in parallel, with 5 volts on one side of them and an opamp input on the other.

Now, work out the size of the resistor which would give the required voltage at point J.

This is the combined parallel resistance of R1 to R4. You know their relationships to each other so you can work out the value of each of them.

As you know, the resistors are in the ratio R (MSB), 2R, 4R, 8R. To get actual values you will therefore need to decide the value of R. What are the constraints?

The value of R must not be too small, or the counter outputs would be heavily loaded: this could do damage, or at least make the output voltages droop, leading to inaccuracy. The total current with all bits high must also be less than the maximum available from OA1, when delivering the required -3.75V into 10kohms load.

The resistances should not be arbitrarily high however, or the speed of operation could be unduly slow. Errors due to bias current and leakage would also increase. At a guess, R of the order of tens of k ohms might be suitable, but you should make calculations based on the data for the devices actually used.

In a practical situation with discrete components there would also be the issue of preferred values, but possibly this is not necessary for your assignment.

Ah thank you for the help lads, but I had already worked this out. The next problem that has halted me is this:

(e) What binary value will appear at the outputs Q3 to Q0 of the register REG1, when v_fuel is 1.1 V?

All of a sudden the comparator doesnt make sense! I think too much tea and stress is blurring my logic, I'd be grateful if someone can confirm it:

Take Vstep to be the +voltage into the comparator (OA3). Take Vo to be the output from the comparator into the diode.
if Vstep > Vfuel : Vo = -5V therefore '1' at REG1
if step = Vfuel : Vo = 0 therefore '1' at REG1
if step < Vfuel : Vo = +5V therefore '0' at REG1

So if Vstep >= Vfuel, then the shift register moves and stays at '1'. And it will only allow the data through once, as it operates only on the rising clock edge? Is this correct?

So this would mean that when Vfuel = 1.1V, then binary 0100 (3) will appear at Q3 to Q0 on REG1. As step 3 is 1V.