Whats with the extra transistor and resistors? - Page 1
Forum section: electronics / beginners

author: gsck
author_rank: Newbie
timestamp: January 08, 2022, 01:29:51 am
content: I was interested in how the TL866 programmer worked, so I had a look at some schematics I found online.Looking at the pin driver circuit I got confused with the Vpp switch, it is completely different to the Vdd having 2 extra resistors with all different values and an extra PNP transistor.Why is it different to the Vdd switch, surely it's could just be the same circuit but with the higher voltage being passed through?

author: Cerebus
author_rank: Super Contributor
timestamp: January 08, 2022, 01:42:04 am
content: Ask yourself, if the PNP transistor supplying Vpp is operating from a higher voltage than the microprocessor (or whatever) supplying the control signal to it, how do you turn it off? What is the highest Vb that the controller apply to the PNP? what will the Vbe of the PNP be under these circumstances? What does that mean about the ON/OFF state of the PNP?

author: GerryR
author_rank: Frequent Contributor
timestamp: January 08, 2022, 12:09:06 pm
content: Vdd switch responds to a negative going input signal (sinking); Vpp switch responds to a positive going input signal (sourcing).

author: Cerebus
author_rank: Super Contributor
timestamp: January 08, 2022, 12:58:34 pm
content: Quote from: GerryR on January 08, 2022, 12:09:06 pmNo, they are both sourcing. Yes they respond to different control polarities, but as they are almost certainly connected to a microcontroller so no-one cares about the control polarity because changing that is truly trivial in software. The question was "Why is it different to the Vdd switch, surely it's could just be the same circuit but with the higher voltage being passed through?", and you haven't answered that. There's a reason that one needs an additional transistor to operate at all, and that reason can be discerned by answering the questions I posed (which one hopes everyone realises are there to guide someone towards finding the answerthemselves, not just handing it to them on a plate, becuase that's how you learn to figure this stuff out for oneself).

author: GerryR
author_rank: Frequent Contributor
timestamp: January 08, 2022, 01:40:33 pm
content: The output of both switches may be the same polarity, but as I stated, the Vdd switch needs a sinking signal, and the Vpp needs a sourcing signal, to provide the output. They are different because they are used in different circumstances. And I didn't hand anything to the OP on a "plate" because he still has to figure out how each circuit operates.You need to calm down a little; the tone of your response to my post I find to be rather annoying!

author: Cerebus
author_rank: Super Contributor
timestamp: January 08, 2022, 05:01:21 pm
content: Quote from: GerryR on January 08, 2022, 01:40:33 pmYour use of sourcing and sinking was (1) unusual terminology for the circumstances, (2) potentially confusing for abeginnerwho has asked a question in thebeginnerstopic. So I pitched in to clarify what you were saying.QuoteI didn't say that you did, I didn't evenimplyit. I was just clarifying why I wasn't providing an "answer on a plate" because sometimes people find the round-about didactic style of asking questions instead of providing pat answers frustrating.QuoteNormally I'd be quite conciliatory here because I meant no slight, but frankly I've had enough of people getting high handed about being addressed in a perfectly neutral tone. You didn't answer the question, you didn't even chime in with any further useful steps towards the answer. You did provide something extraneous that might be confusing and wanted a bit of clarification. And you've got upset because someone pointed that out in a perfectly neutral tone. Clearly there was nothing un-calm or intemperate in what I said to you, the same cannot be said about your reaction with which you've diverted the thread from helping gsck to dealing with your hurt feelings.

author: Kim Christensen
author_rank: Super Contributor
timestamp: January 08, 2022, 05:46:58 pm
content: Quote from: gsck on January 08, 2022, 01:29:51 amTypically, VDD is 5V (Could be 3.3V) and VPP is around 12V.In the case of the VDD-Switch:To turn the PNP transistor off, you'd need to make VDD_X high (5V) so that no base current flows. (near zero volts across the Emitter-Base of the PNP)But in the case of the VPP-Switch:If you left out the NPN, and connected the base of the PNP directly via a resistor, then VPP_X would need to rise to near 12V for the PNP to turn off. (It's emitter would be at 12V)But the logic level IC (powered from 5V) driving VPP_X can only go up to 5V, so you can see the problem with that.

author: GerryR
author_rank: Frequent Contributor
timestamp: January 08, 2022, 06:04:33 pm
content: Quote from: Cerebus on January 08, 2022, 05:01:21 pmNo high-handedness here, my feelings haven't been hurt and I'm not upset. I was "annoyed" at the tone of your answer. If you go back and read the OP's question, I did answer it. He was confused by the two different configurations and wanted clarification as to why the extra parts. I told him why; I'm not sure you did, but only the OP can state whether he has been answered to his satisfaction. Even the tone of your answer here ("but frankly I've had enough of people getting high handed.....") is annoying; you may not mean to, but you come across very terse.Appologies to the OP for this slight diversion. I hope your question has been answered satisfactorily.

author: gsck
author_rank: Newbie
timestamp: January 08, 2022, 06:27:44 pm
content: Having stayed up all last night learning about transistors, I assume for the VPP driver circuit they have 2 transistors to allow for much more current.As for sinking and sourcing, when I looked it up most of what I found was in relation to PLDs but from a guess I think it applies here as well, in the VPP circuit does the NPN act as a current sink to the PNP?

author: gsck
author_rank: Newbie
timestamp: January 08, 2022, 06:31:43 pm
content: Quote from: Kim Christensen on January 08, 2022, 05:46:58 pmAh this helps a lot, thanks.

author: Cerebus
author_rank: Super Contributor
timestamp: January 08, 2022, 06:34:23 pm
content: Quote from: GerryR on January 08, 2022, 06:04:33 pmI'm sorry to say that although you may think you'd answered the question I can't see how you could have. Look at Kim's answer above (he having let the cat out of the didactic bag). Now think about my questions in light of that as the answer. The NPN is needed as a level shifter, not merely to invert the polarity - which is what it sounded like you were saying. Do you see any way that what you said explains that? I can't, and if I can't I'm pretty sure that someone approaching this from a beginner perspective can't. Now, I'm not trying to "have a go" I'm trying to explain why what you said doesn't appear (at least to me) to have been helpful. Not being funny, but do you actually understand the circuits - because Igenuinelycan't tell from what you've said whether you do or not? That's not meant to be an insult, it's not meant to be a put down, it's simply trying to get over that if you do understand it you have failed to explain yourself well.QuoteThat tone was quite deliberate, you did not misinterpret it as you did the first. I've had my fill of self-opinionated muppets to deal with in the last few days, you got caught in the tail end of that. So yes, at the moment I'm being grouchy, but you're still being oversensitive to being corrected or clarified in a perfectly neutral tone.

author: Cerebus
author_rank: Super Contributor
timestamp: January 08, 2022, 06:52:04 pm
content: Quote from: gsck on January 08, 2022, 06:27:44 pmKim's right on the money. The PNP transistor is going to saturate (that is turn on as far as it's able to) once the voltage across the base and emitter is somewhere around 1 volt. Once in saturation it will carry as much current as it is capable of carrying and is being demanded of it. Having the NPN on the base won't increase the current carrying capacity, the NPN transistor could only see a fraction of the current passing through the PNP, the amount of collector current divided by the PNP transistor's current gain , probably around only 1% of it (1/ß or 1/hFEdepending on which units you've learned); 99% of the current passed will be straight from emitter to collector.We'd normally call that NPN transistor alevel shifter, shifting its input voltage range (the output voltage range of the control signal) to a range suitable for controlling the base of the PNP transistor. It also acts as a logical inverter, but that function is merely incidental here.QuoteAs will be abundantly clear from above I think that describing it in those terms here is unhelpful and ambigious, even if strictly true. One wouldn't normally use the terminology quite that way in this scenario - I least I wouldn't and I'd never previously encountered anyone else who would. In transistor switching circuits like this, that is circuits that are intended to be either fully on or fully off, it's much easier to think in voltage terms. Yes, one has to account for currents at some point, both for meeting the required load and rating transistors, but for figuring out the basic ON/OFF operation it's much easier to think in voltages.

author: GerryR
author_rank: Frequent Contributor
timestamp: January 08, 2022, 07:00:36 pm
content: Quote from: gsck on January 08, 2022, 06:27:44 pmThat is 100% correct. Sinking and sourcing terminology is very common in switching circuits, though obviously some people seem to have been able to avoid it. If you get into industrial control systems, you will not be able to avoid the terminology. Even control modules sold by companies like Rockwell International, Mitsubishi and others, specify them as "sinking and sourcing." Learn the correct terminology, you won't be sorry.