# Op-amp plus diodes: resistors' values calculation

I am tasked with following: R1, R3 and R3 are to be calculated so that the diodes work in forward bias. Assuming the diodes have a voltage drop of 0.7 V and the op-amp is ideal I have calculated that R1 = 270 kΩ.

I am quite blank on the other two resistors. Can anybody push me in the right direction?

simulate this circuit – Schematic created using CircuitLab

• There are a few open issues with the question. You list that R1, R3 and R3 are to be calculated, it seems you meant R1, R3 and R9. As connected the diodes will always be forward biased, perhaps you meant properly biased? The circuit is setup as a low Vcc level detector. But you did not include the required trigger level for the detector, that is what will determine the value of R1. You also you do not include the LED's forward voltage (Vf) or what the LED current needs to be, that will help determine the value of R9.
– Nedd
Jan 11, 2023 at 2:23

How did you get your value for R1? I would be rather surprised if only one precise value worked (it would mean that in practice it never works, as there is always some tolerance). I would rather expect that the values "just" have to respect some inequalities.

Some hints :

1. what are the possible values for Vo? Which ones do enable D4 to be forward biaised? How do you obtain those values for Vo?
2. based on 1), what can you say about the voltages at the + and - entrences of the op-amp?
3. Based on 2, you should be able to get some constraints on R1 and R3
4. check what constraints you have on R9
5. double check that if the resistors are all with range, you get all resistors forward biased (the steps 1 to 4 give you necessary conditions, but this isn't a proof that respecting those conditions will be enough. But I expect that based on those conditions, you should be able to proof that all diodes are forward biased)
• You are wrongly assuming that the diodes are 0.7V then the resistance of R3 can be anything. But real diodes all have a forward voltage from 0.55V to 1V, and the forward voltage changes when the temperature and current changes. Where can you find an opamp that works when the supply voltage drops to 2.5V? Jan 10, 2023 at 2:39
• @Audioguru "Assuming the diodes have a voltage drop of 0.7 V". Also, where are you getting the supply voltage dropping to 2.5V from? Jan 10, 2023 at 7:04
• I assume that the 3.6V is from a Lithium-Ion battery voltage that is running down. Jan 10, 2023 at 17:12
• @Audioguru : from the phrasing "I'm tasked", I suppose it is some homework that OP has to do (therefore I only gave clues on how to proceed to solve the problem, instead of solving it myself). So I think there is no point on making assumptions about application or what exactly the voltage source it (or what the real foward voltage is). Jan 10, 2023 at 20:09

D1 and D2 set up a reference of about 1.4 volts on the "-" op amp input.
The two diode reference voltage will vary a bit depending on the diode current and temperature.
The value of R3 is not very critical but it should be low enough to pass a current that keeps the diode's forward voltage above the knee of the VI curve, the diode's forward voltage is more linear above that point up to a certain level. However R3 should not be so low a value that it wastes excess power, puts the diode into the non-linear part of the curve, or brings the current any where near the diode's max forward continuous current spec (Ifm). Using a chart of the diodes forward characteristics, (see datasheet for a 1N4148 ), determine a diode current Id near the assumed 0.7 Vf point then calculate an appropriate R3 value to achieve that.
The general equation for R3 will be:

R3 = (Vcc - Vf(d1) - Vf(d2) ) / Id

R9 will determine the current through LED D4. But you should also know Vf of the LED, the desired current level for the LED, and what the minimum level is for Vcc.
The general equation to find R9 will be:

R9 = (Vcc(min) - Vf(d4)) / I(d4)

.