Troubleshooting Bipolar Junction Transistors

Also drawn is the same circuit, if using an actual silicon rectifier diode, instead of the base-emitter junction of a transistor. The effect is the same. You may wonder then, about the value of the 100k resistor. How important is this, or what is this resistor doing? The 100k in this circuit, is doing nothing more than limiting the current drained by the battery. A diode, once above 0.6V, becomes a very low resistance, so it would quickly drain a battery if connected directly across the battery terminals. The resistor sets the drain. We'll use Ohm's Law to figure out the current.

Current = Voltage divided by Resistance

I = V/R

I = 9V / 100,00Ω

I = 0.00009A or 0.09mA or 90µA, which is a pretty low current, so our battery is OK.

Troubleshooting tip #1: The base-emitter junction is a diode. Expect approximately 0.6V from emitter to base for silicon transistors. Expect 0.1V to 0.3V for germanium transistors.

If the voltage is much less than 0.6V, the transistor is most likely OFF. There may be a short between the terminals externally, or internally if the device has been damaged. A low resistance check with an ohmmeter will check for those problems.

If the voltage is much greater than 0.6V, then the fault might be an open connection, again either externally (bad solder joint, broken trace, broken lead) or internally if the device is damaged. Use a low resistance check again, but this time to confirm that everything that is supposed to be connected is connected. This includes not just the positive voltage at the base, but also the negative, or ground, or 0V connection at the emitter.

If there is an emitter resistor, as in the next illustration, you still want to put the black lead of your meter at the emitter terminal, and the red lead at the base terminal. This is different that the usual "black lead always to ground" mentality that is good for most voltage readings. When troubleshooting, you can measure all the transistor voltages to ground, and then subtract the base voltage from the emitter voltage to get the same reading. It's quicker to do it this way.

Lets also throw in a pnp device example before moving on.

Once again, the measurement is between base and emitter, however a pnp device will have the positive voltage on the emitter, opposite of what we expect with a npn.

Reversing the leads with a digital meter would give a display of -0.6V (that's negative 0.6V). As long as you are aware of the lead's polarity being reversed, it's OK, but try to avoid unnecessary confusion during troubleshooting!

So we've successfully wired up a transistor as a diode. Doesn't do much but drain our battery, but we do have a complete circuit, and that is something. A lot of times in troubleshooting, we have to confirm that something is happening or not happening. Observing a good diode drop across the base-emitter junction is a good check to make during troubleshooting.

The Base-Emitter Current Controls the Collector-Emitter Current

An even better check is the collector-emitter voltage. This is the same type of check, but with the collector and emitter terminals, instead of the base and emitter. Checking the collector-emitter voltage will tell you if the transistor is: cutoff (OFF), saturated (FULL ON), or in-between those extremes ("forward active," or "on.") Once you see that the transistor is in one of those three states, the base-emitter voltage will tell you why the transistor is in the state it's in.

Let's connect the collector to something so we can actually observe this.

Here, the collector is connected directly to 9V. The 100k is still providing the voltage to the base, and the emitter has a 1k resistor to 0V.

Measuring the collector-emitter voltage, right at the terminals, just like the base-emitter voltage, gives us about 4.5V, which is in-between the extremes of 9V and 0V. This means our transistor is forward active, and if we had audio inputs and outputs, we should expect the circuit to work.

If the meter was reading less than 1V, I would think the transistor is "on" too much. If the meter reading was close to 9V, I would think the transistor was "off." For such a simple circuit, if every connection seemed OK, every measurement seemed right, we could adjust the collector-emitter voltage by changing the 100k resistor. To turn the transistor "less on" we could raise the 100k to a higher resistance. To turn the transistor "more on" we could lower the 100k to a lower resistance.

3 Types of Bias: Cutoff, Forward, and Reverse

Fiddling with the base resistor changes the transistor bias, or operating point. However, with just one resistor, from the power supply (9V here) to base, the bias will be difficult to set. The single resistor bias is only good for simple switching applications. For audio purposes, we'll need a very stable reference voltage to bias our transistor.

Notice in this illustration how we have 3 different voltages. There is 9V and 0V at each end of the power source, and then there's 1/2 of that between the two 1k resistors. Two 1k resistors are a fairly low impedance for this kind of voltage divider. Low impedance is good for a reference voltage. Why? Because we want to keep our voltage (4.5V here) steady, and we don't want the circuits that use this voltage to "load" it, and change our 4.5V reference to something smaller, like 4V or 3V or less.

The current through the divider will be 4.5ma. (9V/2000Ω) This is plenty of drive, and you probably won't see this much current devoted to a voltage divider in most battery powered units. 10k, 100k, and 1M voltage dividers are much more common, with draws of 450µA, 45µA, and 4.5µA respectively.

In the next illustration, I've put in two diodes, representing how a transistor does not work. If the base-emitter junction was just a diode, then the voltage at the divider would get pulled down as shown.

Transistors differ from plain diodes with their "transistor action." After replacing the diodes with a real transistor, the divider bounces back up to almost the unloaded voltage. We lose just a little voltage (0.02V here) to the base of the transistor. Why is this?

Once our transistor is ON, the base-emitter's impedance has an interesting change. Instead of a very low resistance, like a diode, the internal impedance, and the external emitter resistance, get multiplied by beta, which is the transistor's dc current gain. Textbook beta is almost always 100, although actual transistor betas will vary considerably, and they can vary due to many factors, but here we will simply assume that our ideal transistor has a constant beta of 100. We will ignore the small internal resistance for now, and just multiply the 1k emitter resistor by 100, to get an input impedance into the base terminal of 100k.

While 1k loads the divider heavily in the diode schematic, the "100k" transistor impedance is a very light load for the divider, and our reference voltage is maintained.