2N3904 emitter looking funky - Page 1
Forum section: electronics / beginners

author: sy
author_rank: Regular Contributor
timestamp: December 11, 2023, 07:42:30 am
content: Hello, I am trying to rapidly switch a transistor (2n3904) on and off by using a square wave input.The square wave is 3Vpk-pk running at 200Hz with a 50% duty cycle and 1.5V offset.When I probe the emitter terminal (in orange) there seems to be some exponential decay. Is this a result of frequency response? I am new to this and not sure what I should be looking at to fix up the signal

author: Andy Chee
author_rank: Super Contributor
timestamp: December 11, 2023, 08:14:08 am
content: If you're measuring the emitter, you should use an emitter resistor to ground, and probe the emitter-resistor junction.

author: iMo
author_rank: Super Contributor
timestamp: December 11, 2023, 08:26:33 am
content: You cannot see that signal at the emitter, because the emitter in your schematics is grounded (and your o'scope probe is grounded as well).And in case you see a signal there in your real circuit - there has to be some parasitic impedance or resistance between the emitter and the "ground".And the decay comes from some parasitic capacitancies in your circuit (and in the transistor as well).

author: gamalot
author_rank: Super Contributor
timestamp: December 11, 2023, 08:44:28 am
content: Could it be a very low quality breadboard? I once bought 2 pieces from a local store in Australia, and their contact resistance was very high!

author: iMo
author_rank: Super Contributor
timestamp: December 11, 2023, 08:50:09 am
content: Some breadboards have got split ground and Vcc (blue and red segmented) power lines - double check that.So your emitter is floating perhaps..

author: ArdWar
author_rank: Frequent Contributor
timestamp: December 11, 2023, 08:57:57 am
content: There must be some misunderstanding here regarding your actual schematics and/or measurement setup. Not even the worst breadboard in the world will give ~1 volt drop at that level of current (and if itdid, it probably won't be that steady).

author: iMo
author_rank: Super Contributor
timestamp: December 11, 2023, 09:11:36 am
content: Your transistor with the floating (not grounded) emitter and the o'scope probe..

author: sy
author_rank: Regular Contributor
timestamp: December 11, 2023, 12:05:21 pm
content: I have separated the collector resistor to have an emitter loading resistor as suggested.My signal now looks more square, but I was wondering why the pk-pk voltage drop is around 1V?If the collector emitter voltage (Vce) is about 0.3V across a silicone diode, and I have a voltage divider between 33Ohm and 100Ohm, should I be expecting a ~2Vpk-pk?My scope channel is 1MOhm input impedance and AWG output impedance is high-z.I've gone ahead and measured the longest rail of the breadboard and it gives 0.1Ohms resistance so I dont think it would load the circuit down that muchProbing from ground to the collector (between R3 and pin3), the square wave jumps to 2.62V-2.92V. Does this mean my transistor is giving a 1V drop across Vce for some reason?

author: iMo
author_rank: Super Contributor
timestamp: December 11, 2023, 12:30:04 pm
content: That is because the transistor is a current controlled/acting device.The collector-emitter current is aprox 9mA, therefore at 100ohm emitter resistor you get 9mA*100ohm=900mV voltage.The aprox 0.3V voltage drop is at the collector's 33ohm resistor, therefore the collector voltage jumps from 3.0V to aprox 2.7V, therefore the collector-emitter voltage will not be 0.3V in your setup.You have to provide a larger base current to open the transistor more (Ic=beta*Ib) in order to saturate the transistor and thus get small collector-emitter voltage.

author: iMo
author_rank: Super Contributor
timestamp: December 11, 2023, 12:51:19 pm
content: Look at this - various base currents..The simulation shows in order to get less than 0.4V between collector-emitter you would need the base resistor have smaller than 4k7 (aprox).You may calculate the base current as your homework

author: sy
author_rank: Regular Contributor
timestamp: December 11, 2023, 01:03:36 pm
content: Quote from: iMo on December 11, 2023, 12:30:04 pmI see, so that means if the transistor is not fully saturated then there will be a greater voltage drop across Vce?This was my thought process in terms of calculations:Ib = (3V-0.7V)/47k = 48uADC current gain for this transistor is 300Ic = 300*Ib = 14mABut already it seems off since you said there was 9mA flowing from collector to emitter.Say for example I wanted 15mA from collector to emitter, how would I do the calculations to choose a lower value resistor at the base to provide more current and open the transistor more?And also, what parameter are we looking for in the data sheet to know that the transistor is fully saturated?Thanks for all the help thus far @iMo

author: iMo
author_rank: Super Contributor
timestamp: December 11, 2023, 01:49:44 pm
content: Quote from: sy on December 11, 2023, 01:03:36 pmYes.The base current calculation to saturate the transistor depends on the wiring you actually follow.Imagine you want to have 133ohm in your collector and 0 ohm in emitter (simplest example).When fully on the current via the 133ohm resistor will beIc = (3V-0.2V)/133ohm= 22mA (aprox), where 0.2V is the Vce saturation voltage (example)In order to create such an Ic you would need at leastIb = Ic / beta = 22mA / 300 = 73uA, for switching purposes we use say 5x more = 0.35mA (in order to saturate the transistor fully under all conditions, with high power apps it is even 10x)Resistor for the base (aprox):Rbase = (3V-0.7V)/0.35mA = 6500 ohm (aprox), where 0.7V is Vbe for silicon transistor (aprox).With the emitter resistor the calculation has to incorporate the voltage drop at the emitter resistor as well.In the datasheet for switching purposes at lower frequencies the most important params are the max Vce, max Ice, beta/h21e at the specific Ice current, sometimes the Vce_sat at the specific Ice current. Also the SOA (safe operating area graph) when messing with high voltages and currents is good to look at.

author: macboy
author_rank: Super Contributor
timestamp: December 11, 2023, 02:17:14 pm
content: Q1 can source current to the output, so the output turns on rapidly.There is nothing to sink current from the output, except for the the probe's 10 Mohm to ground. This, combined with some stray capacitance of the breadboard, and the capacitance of the probe, results in the exponential decay you see. Just adding a resistor from emitter to ground will solve the problem by providing a current path from the output to ground, when the transistor is turned off (and when it is turned on of course). Try different values to see the effect. Don't go so low in value that excessive current is driven through the transistor and resistor. 10 ohm is too low, 100 ohm is probably fine... up to 100 Kohm or more will likely work too, but try it and observe the difference.edit:Haha, I just saw the grounded emitter and output - or what I assumed was the output, the yellow highlighted node... didn't expect that so my mind must have filtered it out.My comment above is only valid if the ground connection on the emitter/output is completely omitted, so the output is just the emitter of the transistor.

author: Yuu
author_rank: Regular Contributor
timestamp: December 11, 2023, 08:47:38 pm
content: I recommend reading a book on analog analysis/design.Quote from: sy on December 11, 2023, 12:05:21 pmYou cannot assume \( V_{CE} = 0.3\, V \). Look up regions of operation for the BJT. In the forward active region, \( V_{CE}\) can vary quite a bit while maintaining relatively constant \(I_C\). The only thing you can assume is that, when the transistor is on, you have about a 0.6V drop across the base-emitter junction (or across the emitter-base junction in the case of a PNP BJT).To figure out why you have a 1V drop from emitter to ground, analyze your circuit. Using Ohm's law when the input is high, we can derive a set of equations and then solve:\[ V_B = V_E + 0.6,\]\[ 3 - V_C = 33 \cdot I_C,\]\[ V_E = 100 \cdot I_C,\]\[ 3 - V_B = 47000 ( I_C/\beta) = 47000 (I_C/300).\]I get \(I_C = 9.4\,mA\), \(V_C = 2.7\,V\), and \(V_E = 0.94\,V\). The emitter voltage lines up with what you have which is about \(1\,V_{pp}\).If you wanted \(I_C = 15\,mA\), you could divide that by \(\beta\) (to get the base current you need) and also solve for emitter voltage and then add \(0.6V\) to get \(V_B\). Then do Ohm's law to find required base resistor.Usually when biasing BJTs, I think about what signal swing range I need and bias accordingly. Textbooks talk about biasing schemes such as biasing so that \(V_E = 1/3 V_{CC}\) and \(V_{CE} = 1/3 V_{CC}\) and so on.Also, it should be noted I've approximated \(I_C \approx I_E\). You can be more exact with your analysis although I suspect variance in beta will probably throw off calculations from reality more than exact \(I_E\).