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Topic: RESOLVED: help with PNP transistor circuit, please (Read 619 times) previous topic - next topic

TomGeorge

Lonely
Ohhhh..sorrryyy...... :'(
Can you draw what you mean please?
Tom.... :)
Everything runs on smoke, let the smoke out, it stops running....

ted



TomGeorge


hmm..
You apply 5v to the 33k, hence the voltage on the gate Vg goes to 5V with respect to gnd, at this moment the source is at gnd. So Vgs is 5V.
The N-CH MOSFET begins to conduct, current flows and a voltage develops across the load, Vs (source) with respect to gnd.
NOW the voltage at the gate Vg is still 5 V but Vgs decreases as Vs increases and the MOSFET turns OFF.
The MOSFET will not turn fully ON,  Vs cannot = Vd (drain) so load never has 12V across it.
Tom... :o
Everything runs on smoke, let the smoke out, it stops running....

ted

#19
Mar 18, 2018, 04:04 pm Last Edit: Mar 18, 2018, 04:13 pm by ted
The post #3 says try, If it is not working , add Tom's analysis and fix it, maybe just reduce 12V or add R3.


MarkT

#21
Mar 18, 2018, 04:26 pm Last Edit: Mar 18, 2018, 04:30 pm by MarkT
When my circuit is active (+5V on the base of the NPN transistor) the NPN transistor closes and I effectively have 12V across Emitter-Base on the PNP transistor. I'm guessing that this is why it stops working as this is way more than the max 5V allowed.

Now, I just can't get my head around how I could switch 12V (or even up to 40V) with a PNP transistor that only allows 5V Emitter-Base voltage.

Any help would be very much appreciated!

Thanks,
Chris

Go and measure the voltages and you'll see you are wrong. R2 protects the PNP base from excess voltage.  The base-emitter junction will be at 0.7V or so since the current is limited to a low value.

The 5V specification on Vbe is _reverse_ voltage, nothing to do with this circuit at all.

I'd change R1, R2 and R3 all to 1k.  Then it might work a lot better as you won't be under-driving
any of the transistors - they should saturate properly.  R3 was way too high to do anything(*), R2 was
preventing saturation if the output needs any significant current, R1 probably doesn't matter but why
not use all the same for simplicity?

(*) R3 is to speed turn-off of the PNP transistor from the saturated state, but with such a high value
it probably made no measurable difference.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

ted

Tom
I forgot to mention that this button is not a current switch but voltage switch, I mean the current passing through it is very small, so even if MONSFET is not completely closed (10k resistance ? ) drop voltage willbe very small on it.

jremington

Quote
I forgot to mention that this button is not a current switch but voltage switch, I mean the current passing through it is very small, so even if MONSFET is not completely closed (10k resistance ? ) drop voltage willbe very small on it.
It is a very bad circuit and should not be recommended to beginners.

ted


jremington

#25
Mar 18, 2018, 07:45 pm Last Edit: Mar 18, 2018, 07:46 pm by jremington
Tom already explained the problem. Google "high side switch" to see what competent designers do.

GolamMostafa

#26
Mar 18, 2018, 07:50 pm Last Edit: Mar 18, 2018, 07:52 pm by GolamMostafa
Let us if  the transistor Q2 of the following circuit (of OP) enters into saturation under the given circuit parameters.

Fig-1: The transistorized switch circuit of OP


Fig-2: VCE(sat) voltage table of 2N3906 transistor

Fig-2 indicates that the 2N3906 transistor requires 1 mA (5 mA) current as its base current to enter into saturation. Let us see, how much we get for the Q2 of Fig-1.

==> IB + IR3 = IR2  //assume Q1 is saturated with VCE ~= 0V.
==> IB = (12-0.7-0.0)/33K - 0.7/330K  
==> IB = (113-0.7)/330k
==> iB  = 0.34 mA << 1 mA (practically Q2 is OFF? and +12V is intermittently reaching at the load)

I would recommend to change all the resistors (R1, R2, R3) to 2.2K and then we have:
== IB = (11.3 -0.7)/2.2k = 4.8 mA

The Q2 will happily enter into the saturation zone. The +12V will now reach to the load when the Q1 is made ON by a static signal or a pulsed signal.

ted

I am talking about my circuit, I just curious, will  this concept damage arduino ?

yesyes

Thanks for all your suggestions.
I now understand that the 5V Emitter-Base voltage is the max reverse voltage. That also explains why in some datasheets it is stated as -5V.


In the end, this seems to be the solution:

I would recommend to change all the resistors (R1, R2, R3) to 2.2K and then we have:
== IB = (11.3 -0.7)/2.2k = 4.8 mA

I've had these resistor values way too high. The values were taken from an example I found online. This will teach me not to blindly use what I find but do my own calculations. ;-)

I have finally found the time to solder 2k2 resistors in parallel to the existing R1 and R2 33k resistors as a test (I left R3 as it is for now as that's probably not even necessary). It seems to work more reliably now. I'll keep it running like this for a while and report back. If it turns out to work fine, I'll completely replace all R1s and R2s with 2k2 (there are 6 of these circuits, one for each button).

Thanks again for all your help!
Chris

Location: Berkshire, UK
My Astro and DIY projects website: http://yesyes.info/

MarkT

Glad its working.  BTW that -5V specification on base emitter reverse voltage is very common, since
the emitter is very highly doped and the base-emitter junction behaves a bit like a zener diode in
reverse.  Its all too easy to destroy a transistor by exceeding the base emitter reverse voltage, although
its not a risk in simple switching circuits as the base normally cannot go below the emitter voltage
as the emitter is at ground (for NPN).

-5V as a reverse voltage specification for Vbe in transistors and for LEDs seems suspiciously universal,
I suspect manufacturers don't both testing beyond 5V in either case, or there's the same underlying
reason (very heavy doping).
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

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