# Confusion with resistor and diode voltages when heated in a series circuit

Below there are two circuits supplied by a 10V supply. The only difference is the one on the left is 10k in series with 1k; and the one on the right is 10k in series with a diode. Confusion 1 about circuit (A):

If the resistor is heated its resistance increases. And if the current would remain the same we would say that the voltage across the resistor would increase. But the current will not remain same because the total circuit resistance increase will decrease the current. Will the voltage across the 1k resistor increase or decrease? Which one will dominate? Increase in resistance or the decrease in current?

If the temperature of the resistor increases resistance increases and this yields current decreases according to Ohm's law V = R * I. R is increasing and I is decreasing. In a practice which one dominates and what happens to the voltage across the 1k resistor?

Confusion 2 about circuit (B):

If the diode is heated its DC resistance decreases. If the current would remain the same the voltage across the diode would also decrease. But since the circuit current will increase I'm again facing the same difficulty here. The diode DC resistance is decreasing which immediately should increase the circuit current. Again something is very cloudy for me to decide what happens to the voltage across the diode when it is heated in practice. Which one will dominate? Increase in current or decrease in DC resistance? How can it be explained step by step manner?

• 10k@10V limits the current to 1mA. At this current, self-heating of the diode is irrelevant, the power is at maximum 600mV*1mA=600µW. – Janka Jun 4 '18 at 0:58
• i dunno why u guys wanna change my question and answer something else .. – cm64 Jun 4 '18 at 1:04
• This usually means your question isn't well-defined and takes a lot of abroads into the brush. – Janka Jun 4 '18 at 1:08
• I will make one other observation. If you want to create a voltage reference, the resistor circuit is better when the input voltage (10V) is well-regulated. But if the input voltage is variable, it may be better to use the diode as a voltage reference in spite of the temperature effect, because the diode is not very sensitive to input voltage. You can also look into temperature compensation techniques. – mkeith Jun 4 '18 at 1:10
• @cm64 - "i dunno why u guys wanna change my question and answer something else" I don't think they're doing it deliberately. Obviously you understand the question, but I also interpreted it differently to you, due to the diagrams. Where they say "HEAT", they don't say "External heat applied here" and I thought they were trying to tell us which components would get hot by self-heating in the circuit. Your question also says "If the resistor [/ diode] is heated [...]" without saying the important word externally, so mis-interpretations could continue. Glad it was answered in the end! – SamGibson Jun 4 '18 at 1:13

## 1 Answer

In circuit A, let's rename the 1k resistor R1. The voltage across R1 can be calculated using the resistor divider formula as follows:

V = 10 * R1 / (10k + R1)

So that is why it is a bit confusing. R1 is on the top and bottom. But it works out mathematically that as R1 goes up, V goes up. And as R1 goes down, V goes down. You can graph it or plug in a few values to see this. Or solve for R1.

Note that resistors can have positive or negative temperature coeficient. So it is not a given that heating up R1 will make the resistance go up. It may remain unchanged, or even go down.

Now let's talk about circuit B. For ordinary silicon diodes, with constant current, the forward voltage goes down as temperature goes up. But it is not correct to think of the diode as having DC resistance. Diodes don't behave like resistors. The way the diode behaves, when current is flowing through it, is that the voltage is relatively constant, only changing a little bit as the current changes. That is just how diodes behave.

So what happens is, as the temperature goes up, the voltage of the diode goes down, and the current will go up slightly. But this increase in current will not be anywhere near enough to make up for the voltage decrease due to temperature change. So the temperature effect wins, and the diode voltage goes down.