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Topic: [SOLVED]BJT cascade problem? (Read 606 times) previous topic - next topic

jremington

#15
Jul 05, 2019, 01:40 am Last Edit: Jul 05, 2019, 01:41 am by jremington
Quote
is that exact same circuit is working on breadboard
It seems rather obvious to me that the working circuit is not exactly the same as the non-working circuit.

And none of us has a good idea what you are actually trying to do with either version.

surepic

#16
Jul 05, 2019, 05:25 am Last Edit: Jul 05, 2019, 05:46 am by surepic
@Paul__B +1 for "Do not put resistors from base to emitter in logic circuits to form a voltage divider." Will dig deeper into this this weekend for sure. Just followed the suggestions one by one took 4-5hrs and only disoldering base to emitter resistors from both npn and pnp solved the problem on PCB board.

On breadboard its working with and without rbes. May be breadboards trace resistance and capacitance are doing something that problem is not noticeable. Most of the time its exact opposite I have to physically solder on board for things to work as they should but this time everything was upside down.

Thanks everybody for suggestions and pics of PCB are attached (don't throw stones on me for flux residue :)  didn't clean yet). This board was built only to test one block of the project.

Test run of servo 30mins flawless.

MarkT

Do not put resistors from base to emitter in logic circuits to form a voltage divider.  Insofar as they are needed, they should go on the left-hand side of R1 and R3.
This advice is simply wrong.

Absolutely put them in, they are used to speed up switch-off (clearing stored charge), and prevent amplification of leakage currents in multi-stage circuit like this one.  Its fairly common to see 1:1 dividers on the base to
give really good turn-off speed.  So long as the drive voltage is several volts there is no issue with voltage loss
since a base requires about 0.65V only.

BJTs are not MOSFETs.  You seem to be confused between them.  I've seen it said about the voltage divider
thing for MOSFETs, but often the values are 150 ohms and 10k and it for those values it makes no practical
difference which place the resistors go as the divide ratio is 0.985 and logic level MOSFETs are commonly
rated for 4.5V gate voltage.  Dropping 5V to 4.925V is a non-issue.  And you can use 100k gate-source
resistor for a MOSFET if you want, completely finessing the issue.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

surepic

I wrote that by adjusting values of restistors i got perfect falling edge from pnp output. Started from 10*rb and then went down to 3-5 times which showed better results but yesterday the only thing that make the motor actually spin was eliminating them by disoldering them. Dont even want to investigate further cos of too much time spent on unexplained issue.

ReverseEMF

#19
Jul 05, 2019, 06:01 pm Last Edit: Jul 05, 2019, 06:19 pm by ReverseEMF
Problem is described in post #1. I rearead it and cant think of any info that i can add.

And forgot to ask if u could explain this:"Do not put resistors from base to emitter in logic circuits to form a voltage divider.".

Thanks.
Actually, I classify this as "Uncalled for Hysteria".  And, actually, this is more of an issue for a MOSFET input, than a BiPolar input.  Let me esplain:

BiPolar input:
Using the component labels in your Spice Diagram, let's call R1 10k and R2 100k.  And, lets say the Base Emitter voltage is 0.7V when 5V is applied to R1 [as shown].  That means there's 0.7V across R2, for a current of 7µA.  And, there's 5 - 0.7, or 4.3V across R1, for a current of 430µA.  So, the current entering the Base of the transistor is 430 - 7 = 423µA.
Now, Paul__B would have you place R2 on the other side of R1.   Lets see what that looks like:  the current entering the Base of the transistor is (5-0.7)/10k = 430µA.  And, the current flowing through R2 doesn't even find the Base.  So the difference is a matter of the 7µA that was subtracted by R2, in the first scenario.  Hardly significant, since it will make little difference in Q1's Collector-Emitter saturation voltage [i.e. the transistor will still be turned-on just fine].
MOSFET Input
In this case, there is more of the "Voltage Divider" effect Paul__B was on about.  Mainly because a MOSFET is voltage controlled, whereas a BiPolar Transistor is current controlled.  See, in the case of the MOSFET input, R1 and R2 form a voltage divider, and as such, the voltage on the Gate will be a bit lower, than it would if R2 were on the other side of R1.  And, that's the crux of Paul__B's objection.  But, it's a non-issue, because he forgot this only applies to a MOSFET input.  With a BiPolar input, any voltage difference between the two configurations is minuscule [a matter of microvolts!].  Which is why the current was only different by a few microamps.

And, here's the math:  With R1 and R2 in the configuration shown in your Spice Diagram, the voltage at the Gate will be R2*5V/R1+R2 = 100k(5)/110k = 4.5V, which could, very likely, make a significant difference in the channel resistance, depending on what the Gate Threshold Voltage is.  With R2 on the other side of R1, ALL of the 5V is applied to the Gate.  And that's how we like it!

And Mark_T's assessment is even far more educated, so hats off to him*!!

* Or her -- ala "Michael" of Star Trek Discovery, or Currer, Ellis and Acton Bell, or George Eliot.
"It's a big galaxy, Mr. Scott"

Please DON'T Private Message to me, what should be part of the Public Conversation -- especially if it's to correct a mistake, or contradict a statement!  Let it ALL hang out!!

herbschwarz

Will we ever know if the defect on the PCB was a cold solder
or a solder short or something else?
Herb

surepic

#21
Jul 06, 2019, 01:26 am Last Edit: Jul 06, 2019, 01:27 am by surepic
Thanks ReverseEMF and MarkT for taking time to explain the base-emitter resistor with examples.

@herbschwarz i will solder back rbes with ratio 1:3 and will update. I consider myself good at soldering so cold solders are definitely not present here. Even vias im filling for extra conductivity. But as i start following all suggestions yesterday and was doing one by one i missed to update here few things that could solve the problem but werent tested till i took rbes out. First i reduced pulldown resistor of pnp to 2.2K grom 4.7K then filled vcc via (0.384mm) with solder and for extra conductivity added piece of wire (cut from resistor leg) and only after that disoldered rbes and tested. That via was passing vcc to the motor via headers as you can see on the picture of the board that i posted. For me is really hard to explain something like this problem for others to easily get into that.

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