# Learning transistor based circuits

simulate this circuit – Schematic created using CircuitLab

In one of our electronic circuit tests, this question was asked: What is the output voltage of given circuit?

This circuit looks like a common emitter amplifier circuit to me, but when I tried to calculate the output, I am getting 0.047 V.

My questions are:

• Why is this circuit not amplifying, is it required to add capacitors?
• How to select suitable resistor values of Rb, Rc and Re in any transistor based circuits?

I know some of the formulas but I don't know how to implement:

$$I_b = V_{in} - \frac{V_{be}}{R_b}$$ $$V_{be} = 0.7 \mathrm{\;assumed}$$ $$I_b = 0.372\mathrm{\,mA}$$ $$I_c = \beta I_b$$

Considered β = 100, so:

$$I_c = 0.037\mathrm{\,A}$$ $$\mathrm{Gain} = \frac{V_{out}}{V_{in}}$$

• $47mV$ seems quite reasonable to me. Keep in mind that this is an inverting amplifier. – Dzarda Aug 27 '14 at 10:03
• EDIT: 47mV is of course complete bogus, since this is a BJT. – Dzarda Aug 27 '14 at 12:45
• I formatted the formulas using MathJax, please let me know if anything is incorrect. – JYelton Aug 27 '14 at 16:25
• If you ground Q1_c the I_R1 is V/R = 15V/10k = 1.5 mA. | /I_R2 ~= V2/R2 = 10V/25k =~~ 0.4 mA. (Slightly less as Olin says due to Vbe <>0. | For Beta = 100. Ic = Ib x Beta = 0.4 mA x 100 = 40 mA. BUT we saw above that the absolute max R1 can carry is 1.5 mA when the transistor is a hard short to ground. So the transistor has enough base currentto support Ic = 40 mA BUT 1.5 mA in R1 is enough to take Q1_c to ground so the transistor is turned on hard in saturation. As Olin said. As Olin said (again) what exactly Vce is in saturation varies with specific transistor. .... – Russell McMahon Sep 15 '14 at 8:40
• .... If you drive the base VERY hard Vce may fall to very close to zero. Long ago I drove a transistor with Ib ~= 10 x Ic !!!. This turned it on VERY hard so Vce was a few millivolts. I was used to switch a divider string on and off so this made sense to do. (Nowadays I'd probably use a low Rdson MOSFET instead). – Russell McMahon Sep 15 '14 at 8:41

You should be able to see immediately from inspection (without doing any calculations) that the transistor is saturated. Generally you figure the C-E voltage of a saturated transistor is 200 mV or less unless the current is unusually large, which in this case it's obviously not due to the size of R1 and R3. The answer for any real electrical engineering purpose is therefore "200 mV or less", which we can see immediately from inspection. Part of designing good circuits is to make sure that this 200 mV uncertainty doesn't matter. If your professor wants a more accurate answer, then he's being academic and unrealistic, and you can tell him I said so.

Now let's do the math to back up what we already know from inspection is happening. Let's say the B-E drop is 700 mV, so there is 9.3 V accross R2, which means the base current is 370 µA. R1, R3, and V1 form a Thevenin source of 7.5 V and 5 kΩ. That means the collector current can't be more than 1.5 mA (C-E drop 0 which can't happen, but is useful to get the guaranteed not to exceed current). (1.5 mA)/(370 µA) = 4, which is the gain required for the transistor to saturate. You can easily rely on a 2N3904 to have well more gain than that at 1.5 mA collector current. The transistor is clearly well into saturation, which is why it was easy to see from inspection without actually running the numbers.

So now the question is what will the C-E drop be for the transistor at 1.5 mA collector current and well into saturation. Again, the basic electrical enginnering answer is "less than 200 mV". If you need a more accurate answer then two things are going on. First, the robustness of your overall circuit design is suspect if this really matters.

Second, you have to look in the datasheet to see what details they tell you about this particular transistor. This is where things get a little tricky since there are various variants of the 2N3904 out there, and different manufacturers will spec it a little differently. I just grabbed the Fairchild datasheet to use as a example. I wouldn't assume that this level of detail applied to 2N3904 from other manufacturers without checking. This is yet another reason it is better if your circuit works with anything in the 0-200 mV range. You don't have to worry about which variant from which manufacturer you are using, and purchasing can get the cheapest generic 2N3904 they can find that week.

On page 2, there is actually a spec for Collector-Emitter Saturation Voltage, which is a maximum of 200 mV for IC at 10 mA and 300 mV for IC at 50 mA. This is typical of datasheets in that they don't explicitly tell you what the part will do at all possible operating points. However, since our collector current is well below 10 mA, we can safely count on the C-E drop to be 200 mV or less. Note that's what we already knew from 3 seconds of inspection in the first place.

With this level of information in the datasheet, "200 mV or less" is actually the only correct answer. This is all the manufacturer promises the transistor will do. Now we know that almost certainly in our particular case the C-E voltage will be lower, but the absolute worst case spec is 200 mV. Claiming anything lower is actually wrong, and I would argue strongly with anyone that accepted a specific lower answer as correct in a engineering course. Second guessing the datasheet is irresponsible engineering.

So if I was grading the test, I'd mark any answer that gave a specific number below 200 mV as wrong. Even if you built the circuit and measured the C-E drop to be 89.3 mV, for example, I still wouldn't accept that as a valid answer to the question because it can't be counted on accross part variations within a batch, and accross parts from various manufacturers.

This is a area where theory and engineering differ, and is something engineering students need to learn. If your professor disagrees with this, and this is in a engineering course, then he's just plain wrong and needs to get out into the real world. And yes, you can (and should) tell him I said so.

• Here i did not understand, how did you find Vce? I am a beginner so taking allot of time to understand each and every steps explained,comparing with the theoretical calculation,some where i am going wrong with the formulas getting totally different answers. please help me to understand proper working of this circuit,and when i googled i found similar circuit as common emitter amplifier,so what exactly difference with the common emitter amplifier from this circuit?? how can i analyse circuit?? – yasmi Aug 28 '14 at 5:04
• I remember an old professor in our Elec Eng dept (I'm a Mech) that used to go nuts if a) like you say, students returned values like 89.3mV to something like Vce and b) students returned answers with a precision greater than that asked. He'd get "40.1249576V" as an answer to a question with 5% resistors. I want a University friend to start a course called "Real World Electronics", where caps are -20+80% and hfe is "in that region somewhere". :-) – carveone Aug 28 '14 at 10:03
• @yasmi: Too many questions at once. Pick what you really want to know. The basic answer to where 200 mV or less came from is from the datasheet. Take a look at the datasheet I mentioned. – Olin Lathrop Aug 28 '14 at 12:38
• @carveone: Exactly. All freshman engineering students at RPI in 1974 were required to take a course called Elementary Engineering. There was a lot of estimating, how close "close" really needed to be, etc. It was a real eye-opener for me. In one problem we had to decide the wire guage to use for something, and the numbers deliberately came out to a little smaller than one available guage, so we all naturally picked that one. Those were all marked wrong due to not leaving any margin for stuff to happen outside the problem statement. Great course. – Olin Lathrop Aug 28 '14 at 12:43

This isn't an amplifier as such. The transistor is acting as a switch. Let's have a quick look at the numbers using rules of thumb. (I'm assuming those voltages are DC with no AC component)

10V - 0.6V (Vbe) = 9.4V at the base. Divide by 25k = 0.37mA roughly base current. If we multiply that by (hand wave) a hfe of 100, we get 37mA into the collector before the transistor will drop out of saturation.

I personally go for 10 times that, so I'm positive the transistor will be fully saturated at 4mA into the collector.

So, without doing anything too hard like getting the output load involved, imagine the transistor is saturated. That means the collector is nearly at 0V. So 15V / 10k = 1.5mA. That's well below 4mA so the output will be close to 0V. Under 0.2 anyway.

• How do you calculate output voltage ?? and if i want to make it as common emitter amplifier what modifications i need to make in this circuit?? – yasmi Aug 27 '14 at 10:35
• Add an emitter resistor, for instance 1 KOhm for an amplification of 10. Note that you can't amplify 10V much when you have only a 15V supply, and that the output will be inverted. – Wouter van Ooijen Aug 27 '14 at 10:37
• Wouter gets there first! Add emitter resistor. If you can find a book called The Art of Electronics, a lot of this stuff is explained quite well there (in my opinion of course)! I realise I answered the circuit not really the questions you had below. – carveone Aug 27 '14 at 10:42
• I understood this circuit working,can you tell me,if i want to make it as amplifier circuit,how i have to make changes?? what all the parameters i need to consider to design transistor based circuits?? @carveone – yasmi Aug 28 '14 at 5:08
• That's kind of a big question for a comment box! I personally used the Art of Electronics book back in the day. These days there's lot of online resources but I don't know how many of them are good. I see allaboutcircuits.com which looks ok. Bit mathsy. This one looks a bit more fun and less mathsy: talkingelectronics.com/projects/TheTransistorAmplifier/…. Yes, it's tricky to get started and I find a lot of university stuff is a bit too theory based for me. – carveone Aug 28 '14 at 10:18