# Transistor saturates no matter the size of base resistor

Problem

I have a circuit as depicted below, and after calculating the base resistor, I wanted to see when the base current would be too little to saturate the transistor. I kept increasing the resistance but even at $560k$ where the resistor had decreased the current to $7.8\mu A$, the transistor is still saturating. Shouldn't there be a base current threshold?

What I am trying

What I would like to be able to to, is to switch on and off a diode from an arduino pin. The supply voltage is 5 volts, the forward voltage drop across the diode is 3.4V and the transistor is a 2N3904.

According to this blog post, I should be able to find the base current by

$$I_B = \frac{I_C}{h_{FE}} = \frac{20mA}{100} = 0.2 mA$$ with min $h_{FE} = 100$ from the datasheet

And then the base resistor value by

$$R_B = \frac{V_{port}-V_{BE}}{I_B} = \frac{4.2V-0.7V}{0.2mA} = 17.5k \Omega$$

But the size of this resistor value seems to be irrelevant.

simulate this circuit – Schematic created using CircuitLab

• How are you determining the transistor is saturating – sherrellbc Jul 20 '14 at 16:30
• Saturation is usually defined in a datasheet as the point where Vce <= 0.2V for your load current. So use a DVM to determine that voltage and find the Rb at which it happens to be 0.2V. As the answers say, that's a very different thing from when the transistor is on etough to light the LED. – Brian Drummond Jul 20 '14 at 16:53
• Apparently a transistor is a voltage-controlled device. From: "How Does A Transistor Work? No, How Do They Really Work?" amasci.com/amateur/transis.html – fuzzyhair2 Jul 20 '14 at 18:19
• @fuzzyhair2 Yes, there is some merit to that view, but it's not so useful an approach for switching circuits. – Spehro Pefhany Jul 20 '14 at 18:30
• DVM= digital voltmeter. And voltage-control of BJTs, that's if you want to know the physics behind them, lift the hood and learn how BJTs work inside. For most design tasks you want a simplified model which conceals the physics: close the hood, ignore the engine internals, and just drive. – wbeaty Jul 20 '14 at 23:49

The typical gain of a 2N3904 is 200-300 at a couple mA collector current (more as it warms up due to not being saturated)

Even with 560K, that's half a mA or so, which will give plenty of light from a modern LED, but you should be able to see that it's not as bright as when a 10K resistor is used.

Do not use the hfe for this calculation if you want the transistor saturated hard on, use a forced beta of something like 20 to 50, if the typical hfe is 200 or so and the minimum 100 or so. If you use, say 30, in your equation you get a resistor value of 8.8K, so you might use 10K or 8.2K.

The reason is that you won't likely have a guaranteed hfe for the current you're using, and the hfe decreases at temperature extremes. It's still only "wasting" a few percent of the LED current, so no big deal.

To prove this to yourself, take a voltmeter and measure Vce of the transistor when it is on. If it is something like 50-100mV it is saturated.

• Hm. My answer could be completely wrong thing and I may have a miss-interpretation of saturation. I thought a transistor was saturated when the configuration was such that an increase in base current had no impact on the output current. That is, you have saturated in the input with feeder current, so to speak. – sherrellbc Jul 20 '14 at 16:58
• Thank you for your elaborate answer. So i'm reading this correctly, there is no "right" base resistor to choose? So in designing this, I just calculate Rc with a forced beta, as you suggested, and then find a resistor value for Rb that will give me the correct Ic? And to ensure that the transistor is saturated, I would then measure Vce? – Attaque Jul 20 '14 at 17:22
• @Attaque, that is true if you are wanting the transistor to saturate. You are basically assuming the gain is much lower than it actually is. You design around this low-gain value and ultimately the actual higher gain value will force the transistor to saturate. The voltage from collector to emitter is a derivative of the current through the device - namely the on-state resistance. 200mV or less is a typical value. – sherrellbc Jul 20 '14 at 17:29
• Ah, okay. I measured around 100mV with the 10k and 8k resistors. But just to make sure, I do want the transistor to saturate if i'm using it as a switch right? That's when the transistor is fully on? – Attaque Jul 20 '14 at 17:52
• Ah yes of course, sorry. The minimum LED voltage is 3 volts and the power supply is 5v 3A (one of these ebay.co.uk/itm/…). I have a 80 ohm resistor in series with the LED now, which give me 18.4mAh through the LED. The brightness is very good. Thank you for your help – Attaque Jul 20 '14 at 20:44

From your comment I have to say the LED turning on does not indicate saturation in this configuration. Saturation of the transistor is defined as the case when increasing the base current is not followed by a further increase in collector current (i.e. the collector current has saturated to a maximum). Actually, saturation is a state of the transistor used to characterize certain configurations, but from my experience it's implied when you speak of the current saturation. More formally, the following statement is no longer true

$$I_{c} = I_{B}\beta$$

It depends on your particular model of LED, but some can be driven with current as low as 5mA or less and still emit light.

Also note that you are only assuming 100 as your gain coefficient for that transistor. You have cited that from the datasheet 100 is listed as the minimum. As such, it is quite likely the gain will be much higher than that.

• Ic = Ib / beta is never true. – EM Fields Jul 20 '14 at 20:02
• @EMFields, You're right. My mistake. – sherrellbc Jul 20 '14 at 20:04

The most obvious thing you're doing wrong is using the minimum h[sub]fe[/sub] in your calculations. That is, starting from a given value of the base resistor R, you find the LED current I by solving I = h[sub]fe[/sub] x (V[sub]out[/sub] - V[sub]B[/sub])/R ,right? So using the minimum h[sub]fe[/sub] only tells you the minimum I which your resistor will produce.

From the data sheet, the maximum h[sub]fe[/sub] of a 3904 is 300. Using your own equation for a base resistor suggests an LED current of 2 mA, which is certainly enough to light the LED.

The second most obvious thing you're doing wrong is actually more serious: you don't seem to know what "saturation" means. Look at the second graph in Spehro's answer. You'll notice that the collector voltages involved are very low, less than .15 volts. Granted, the particular numbers involved depend on exactly which transistor you're using, and exactly what current you're interested in, but his graph is obviously appropriate for your concerns.

Once you understand what saturations means, the graph ought to suggest a very quick test to determine if your transistor really is saturated: short it out and see what happens to the LED brightness. If the 2N3904 really is saturated, it will have a voltage of about .1 volts across it, and the voltage across the LED/limit resistor will be about 4.9 volts. Shorting out the transistor will increase this to 5 volts, with a very small increase in current (and therefor a very small increase in LED brightness). In your case, you will see a large increase in brightness, and this will tell you that the transistor was not actually saturated - it was just producing small amount of light from the appropriately small current produced by your choice of base resistor.

Before you do this, though, you need to address the third (possibly) obvious problem with your setup: your LED resistor. Let us, for the moment, assume that your LED has a forward voltage of 2.5 volts. Then, with 5 volts across the LED/resistor combination, there will be 2.5 volts across the resistor. This will produce a current of 33 mA. This is not a killer all by itself, but you stated that you were expecting 20 mA. Unless you are using an LED with a forward voltage of 3.5 volts. This is certainly possible, and occurs with superbright greens and blues, but I'd have thought you'd notice the reduced brightness from your LED. If you really are using an LED with a 3.5 volt drop please disregard this comment.