# How can I keep a dummy load constant resistance when resistance changes by temperature?

A dummy load by definition converts the energy to heat. Since this raises the loads temperature, which raises the loads resistance, the load will not have a stable resistance. This makes it hard to measure, experiment, or determine which circuit works best, since over time, the resistance of the load is always changing.

For example: I'm using a 20 W 50 Ohm resistor as a dummy load in a simple 5 W QRP transmitter. As I key it, the resistance seems to go up. How can I try different circuit combinations to see what works best?

# Update

I'm testing CW. However, for testing purposes, I've tied the key down to ground, since if I have to hold the key down, I can't work the oscilloscope. I power the transmitter on (= 100% duty cycle), work the oscilloscope, and power it off, taking between 30 - 90 seconds total.

The peak-to-peak voltage consistently rises during this time. I don't know it's due to heat, but it's the best explanation:

• Peak-to-peak voltage rises consistently as the transmitter is on
• If I turn the transmitter off and then turn it back on quickly, peak-to-peak is higher than if I leave it off for a few minutes
• Putting the dummy load in the fridge before power up seems to lower the peak-to-peak - but, after a few minutes of transmitting, it climbs back
• The temperature should stabilize around whatever duty cycle you are testing. How do you know your measurement problems are related to heat? Commented Jul 31 at 13:17
• @clvrmnky webmarc Updated Commented Jul 31 at 17:58
• The tempco of the resistor is insignificant here, it's probably well within ratings so only a few percent change. What you're seeing here is the effect of the transistor heating up, causing big changes in bias point and gain. Measure the base DC voltage (RF off) with it cold and warm and see the difference. There are various techniques for stabilising a transistor amplifier against self-induced thermal changes. Commented Jul 31 at 18:24
• Flip the script: Power the resistor from something else that you turn off when you turn on your transmitter. Thus keep it at a stable hot temperature. Commented Aug 1 at 1:23

First, I imagine that you've already looked at lowering the power output, but just in case... do that. :-)

You won't be able to keep it (the temperature) perfectly constant, but you can definitely keep it close enough by making the resistance/heat-sink MUCH larger than the source.

One way to accomplish this is by using, say 10 x 500Ω 20W resistors in parallel. Now, each resistor is only dissipating 0.5W (assuming perfect balancing).

If you drop the resistor assembly in a container of mineral oil, now the heat can be conducted away from the resistors for quite some time before temperature becomes an issue.

BTW, you can drop your present resistor assembly in mineral oil and get some relief; especially if you put the oil in the freezer for a bit first.

As pointed out by Jim, you'll want to try to source non-wire-wound resistors to avoid de-tuning the dummy load with inductive reactance.

• While this is a good answer, i would like to add that the type of resistor does have an effect. many common resistors are wire wound. these can and do act like a coil adding inductance. Try to find a carbon resistor instead.
– Jim
Commented Jul 31 at 19:41
• Thanks for pointing that out, I've incorporated your point into the answer. Commented Jul 31 at 21:01

Since this raises the loads temperature, which raises the loads resistance, the load will not have a stable resistance.

Well, first of all, that's why loads for high powers have massive cooling ribs.

Then, this is, just like an incandescent lamp, a self-stabilizing control loop: resistance increases, with higher resistance, dissipated power drops, temperatures starts falling again. Since thermal systems are typically well-dampened this isn't going to oscillate much, but converge to a static resistance.

For example: I'm using a 20 W 50 Ohm resistor as a dummy load in a simple 5 W QRP transmitter. As I key it, the resistance seems to go up. How can I try different circuit combinations to see what works best?

First of all, check whether your power resistor is a good RF load at the frequencies you care about. High-powered resistors can be wirewound, and well, a wound wire is an inductor, not a pure resistor.

Then, resistors come with a temperature coefficients, in $$\alpha_R=\Delta R/T$$, i.e. the datasheet tells you how much your resistance changes with a Kelvin in temperature change, and not only the 20 °C nominal resistance (and tolerance of that)! It will also tell you the thermal resistance, $$R_{\text{therm}}$$

Takes all the guesswork out of this to read the documentation! Afterwards, it's really just solving the very simple equation

\begin{align} R(T) &= R_{\text{nom., 20 °C}} + (T-20 \text{ °C}) \cdot \alpha_R\\ T(P) &= T_{\text{amb.}} + R_{\text{therm}}\cdot P \\ P(R) &= \frac{R(T) - Z_0}{R(T) + Z_0}\cdot P_{\text{source}}, \end{align}

where

symbol entity
$$T$$ Temperature of your resistor
$$R(T)$$ Resistance of your resistor at temperature $$T$$
$$T(P)$$ Temperature of your resistor when power $$P$$ is sunk into it
$$P(R)$$ Power that your power source of given source impedance $$Z_0$$ sinks into a resistor of resistance $$R$$
$$R_{\text{nom., 20 °C}}$$ the nominal resistance of your resistor at "standard" room temperature 20°C. In your example, that was 20 Ω.
$$\alpha_R$$ Temperature coefficient of your resistor, most commonly given in $$\Omega/\text{K}$$.
$$R_{\text{therm}}$$ Thermal resistance of resistor to ambient. How much warmer does the resistor get (in $$K$$elvin) per Watt of power?
$$Z_0$$ source impedance (in most cases, will be the e.g. 50 Ω transmission line impedance, which you connect your load to)
$$P_{\text{source}}$$ power of the power source
$$T_{\text{amb.}}$$ the actual ambient temperature. Might be 20°C, might not be.

So, the rest is school math: you insert $$T(P)$$ into $$R(T)$$ (eq. $$\eqref{a}$$), then you insert $$P(R)$$ into that (eq. $$\eqref{b}$$):

\begin{align} R(T(P)) &= R_{\text{nom., 20 °C}} + (T_{\text{ambient}} + R_{\text{therm}}\cdot P-20 \text{ °C}) \cdot \alpha_R \label{a}\tag{1}\\ R&= R_{\text{nom., 20 °C}} + (T_{\text{ambient}} + R_{\text{therm}}\cdot \frac{R - Z_0}{R + Z_0}\cdot P_{\text{source}}-20 \text{ °C}) \cdot \alpha_R \label{b}\tag{2} \end{align}

bit of rearranging, move constants to the right end, and give them names, so that we don't end up with sore wrists from writing:

\begin{align} R&= \frac{R - Z_0}{R + Z_0}\cdot \underbrace{ P_{\text{source}} \cdot \alpha_R \cdot R_{\text{therm}}}_{=:C_1} + \underbrace{R_{\text{nom., 20 °C}} +(T_{\text{amb.}} -20 \text{ °C}) \cdot \alpha_R}_{=:C_2} \label{c}\tag{3}\\ R&= \frac{R - Z_0}{R + Z_0}\cdot C_1 + C_2 \label{d}\tag{4} \end{align} OK, the variable of interest in the denominator sucks, so extend with $$\frac{R+Z_0}{R+Z_0}$$: \begin{align} R\cdot (R+Z_0) &=(R-Z_0) \cdot C_1 + C_2 \cdot (R+Z_0)\\ R^2 + R\cdot Z_0&=R\cdot C_1 + R\cdot C_2 + Z_0 \cdot (-C_1) + Z_0 \cdot C_2\\ &=R\cdot (C_1 + C_2) + Z_0\cdot(C_2-C_1) \end{align} OK, that looks a lot like a quadratic equation! Let's bring everything to one side, by subtracting all there is on the right side of the equation sign from the equation: \begin{align} 0&= R^2 + R\cdot Z_0 -R\cdot (C_1 + C_2) - Z_0\cdot(C_2-C_1)\\ &= R^2 + R\cdot(Z_0 - C_1 - C_2) - Z_0\cdot(C_2-C_1) \\ &= R^2 + 2\cdot R\cdot\frac 12 \cdot( Z_0 - C_1 - C_2) - Z_0\cdot(C_2-C_1) \\ Z_0\cdot(C_2-C_1) &= R^2 + 2\cdot R\cdot\frac 12 \cdot( Z_0 - C_1 - C_2) \\ Z_0\cdot(C_2-C_1) + \left(\frac 12 \cdot( Z_0 - C_1 - C_2) \right)^2 &=\underbrace{ R^2 + 2\cdot R\cdot\frac 12 \cdot( Z_0 - C_1 - C_2) + \left(\frac 12 \cdot( Z_0 - C_1 - C_2) \right)^2}_{= a^2 + 2ab + c^2 = (a+b)^2} \\ Z_0\cdot(C_2-C_1) + \left(\frac 12 \cdot( Z_0 - C_1 - C_2) \right)^2 &= \left(R + \frac 12 \cdot( Z_0 - C_1 - C_2)\right)^2 \end{align}

See, a quadratic equation! These things can have two solutions, but you'll quickly notice one can't be physical (not a real, positive resistance).

\begin{align} R&=- \frac 12 \cdot( Z_0 - C_1 - C_2) \pm \sqrt{ Z_0\cdot(C_2-C_1) + \left(\frac 12 \cdot( Z_0 - C_1 - C_2) \right)^2 } \end{align}

Just insert the values of $$C_1$$ and $$C_2$$ as defined above, and you have hard numbers for at which resistance your heating resistor stabilizes.

The answer is simple: use equipment and materials made to do the job. Unless you use an RF resistor made for dumb loads or attenuators, you will get poor results.

Axial-type resistors are useless because they are not heat-stable and look more like a series of capacitors and inductors than resistors. Resistors made to terminate RF have extremely low inductance/capacitance, tight tolerance, and heat stability. Use the right equipment, and you will not have that problem.

Below is a 450-watt, 50-ohm RF Resistor. They do not have wires because wires are inductors. Take one of the RF Resistors and put it into the proper heatsink. Better yet, for \$20, buy a commercial 25-watt dumb load good from DC to 1 GHz

Good luck.

• stick it in mineral oil sammy. Commented Jul 31 at 15:48