That literally is not a gross oversimplification if you are at all familiar with semiconductor physics.
> BJTs can be thought of as two diodes (P–N junctions) sharing a common region that minority carriers can move through. A PNP BJT will function like two diodes that share an N-type cathode region, and the NPN like two diodes sharing a P-type anode region
> In the discussion below, focus is on the NPN BJT. In what is called active mode, the base–emitter voltage VBE and collector–base voltage VCB are positive, forward biasing the emitter–base junction and reverse-biasing the collector–base junction. In this mode, electrons are injected from the forward biased n-type emitter region into the p-type base where they diffuse as minority carriers to the reverse-biased n-type collector and are swept away by the electric field in the reverse-biased collector–base junction.
https://en.m.wikipedia.org/wiki/Bipolar_junction_transistor#
In the real world there is stray capacitance and inductance on components that can give undesired results. The physics is based on best case theoretical scenarios.
There is no gate in a bjt.
You can in many cases leave the **base** floating without issues.
Adding a pull down resistor can speed up transistor switching, as when VBE approaches 0, the diode will sink exponentially less current, as opposed to a linear sink with a resistor.
Another commenter mentioned that the collector base leakage current sometimes may be large enough to allow for a significant about of collector emitter current, however this often is not an issue.
“Most of the time it’s ok so we can just not add this part that will make sure no undesirable effects happen” I’m sure nothing bad will happen and our product will definitely be safe and won’t be recalled
Lol there are so many commercial circuits which leave the base of npn transistors floating.
Here are the circuit diagrams of the lm324 and lm358, arguable some of the most mass produced ic’s ever.
https://components101.com/article/difference-between-lm358-and-lm324
Notice how there aren’t pull up/down resistors at the input of the bjts?
Also notice Q2 and Q3, along with Q1 and Q4, when Q1 and Q4 have the no voltage applied between their base and emitter, they inherently leave the base of Q2 and Q3 floating, as there is no path to the supply voltage. According to you this will make the circuit unreliable.
But the circuit is not unreliable. Why? Because you don’t know what you are talking about.
I hate that I have to say this, but an OpAmp isn't a digital logic gate. If my AND gate produces an analog signal, its unusable. Likewise my OpAmp if it won't.
Also, that's not a Darlington pair as they don't have a common collector.
Right off the bat you’re insulting my knowledge of semiconductors without even asking how much I know which tells me you only have a basic understanding of BJTs and are overcompensating. Yes calling a BJT a diode is a vast oversimplification, they are LIKE a diode because they have a p-n junction but they are vastly different
It wasn’t an insult, but you are certainly not helping your case. I didn’t say a bjt was a diode, I said the base emitter junction forms a diode. Did you completely gloss over the wiki article I linked?
Please explain how the pn junction formed between the base and emitter junction in an NPN bjt is vastly different to a p-n diode.
If you draw the band diagram of the pn junction it will be exactly that of a pn diode.
> An ordinary BJT is a low input impedance current driven device. If there's no connection to the base, then base-emitter diode forward conduction/leakage will quickly take the base potential down to the emitter potential. The transistor will be off.
https://electronics.stackexchange.com/questions/527665/transistor-with-floating-base
During instances like this, when you use a resistor for a particular application other than purely resisting the flow of current(pull down, pull up, current sensing, etc) , what goes into determining what the actual resistance value needs to be? Is there a standard?
It depends on your application. If you're using an FET, the gate (acts as/is) a capacitor, so the pull-up or pull-down resistor value would be chosen based upon how fast you need the FET to shut off.
For a BJT, you're generally using a pull-up or pull-down to keep stray (antenna or otherwise induced) charges from making your circuit act weird...5k to 100k, depending on how much power is available (battery life, etc).
Or... The intent may be to use the BJT in the non-saturated region, where it acts like a current amplifier, and what looks like pull-up and pull-down resistors are actually called bias resistors, to set the Base voltage at a known value (0.7v-1.0v typically), but that's a whole different conversation.
Current sensing resistors are generally really small, <1ohm, because it's in series with the load, and you don't want to effect the voltage delivered to the load, and you don't want to dissipate power (heat) at that resistor.
Edited for clarity.
Thanks for the thorough explanation. That makes total sense. I think this is a start for my better understanding, because sometimes when I read schematics with different sensors and opamps, I see resistors sprinkled everywhere as if they were something people just throw in "for good measure" or without a rhyme or reason. And they so common that it seems like I'm just expected to just know why.
I think they are to avoid potential noise triggering when the switch is open and the wire is not otherwise terminated. Every floating wire is a radio antenna.
BJTs are not immune to noise. Yes they are better against noise than FETs, but it's still best to pull the control line to a known low state when the switch is not engaged.
Is it common practice to use pull down resistors with bjts? Unless the common emitter current gain (Hfe) is super high, I would be surprised if noise would actually be large enough to be able to deliver enough current through the base emitter diode junction to allow any non negligible currents to flow through the collector.
Even if noise isn't a problem, a secondary effect of pull-downs may be needed to sink Icb reverse bias leakage current. If this is not done, then it will be carried by the base-emitter-junction and can cause device turn-on.
This is a genuine real world effect which is well known and well documented but not always well-taught in courses.
Hm, that’s very interesting I never knew that. I still imagine it isn’t usually a practical consideration, unless hfe is very large.
Usually leakage current in silicon rectifier diodes (like the np junction formed at the collector base) are very small, so even after the common emitter current gain, it should still be on the order of micro amps at max right? Unless I’m overlooking something.
They're there to prevent the input to the transistor from floating up and triggering the gate. Yes, it will work without them, but with any static or magnetic fields present it can rise to the level of vcc.
The purpose of the pull down resistors is to ensure that the transistor remains in the "off" state, or a logical "0," when there is no input signal. By permanently pulling the base to ground in the absence of a logic high, they limiting unwanted current flow which helps to which helps to reduce power loss and stabilize the transistor's state.
Yes. People sometimes forget what happens outside of nominal conditions. At high temperatures, beta gets much higher, Vbe gets much lower, and collector to base leakage current becomes significant. At high temperatures quality bipolars can indeed generate self-sustaining collector current with an open base terminal.
In my experience (IC analog) most circuit design work is spent keeping the corner conditions functional.
>Electronics Learning Lab, by Forrest M. Mims
I was just thinking, damn, that looks like Mims, but I've never seen that one. Was just about to ask what the title was. I've got several of his. Going to have to track this one down.
Mechanical Make and break switches (on the microsecond time scale) “bounce” or “ring” when the contactors make contact due to mechanical momentum slamming together until the mechanical position settles (can be seen if hook an o-scope up to location of R3&R4 if resistors are removed). Because electrons move Faster (speed of light) than mechanics parts the electrons carry current on the atomic scale which can be seen as “noisy” transients. Use a capacitor to offer a low impedance path (to create a transient filter) to “suck” these transient currents to ground preventing a voltage from building across R3 or R4 and falsely turning on Q1 & Q2 as the currents are generated. Originally, Adding R3 and R4 in the first place improves the noise immunity of Q1 and Q2 because with the resistors present at the gates of the transistors NOW real (not environmental transients like AM radio) powerful DC voltages supplied by your switches turn on transistors as intended (ie: are not turned on by some random un- engineered chance or event).
Yes, this. There's two common ways of thinking this through. First, as above, the noise is considered as AC and conducted to ground through the low capacitive reactance (impedance, resistance in an AC circuit). Second, the capacitors can be thought of as DC filters, in conjunction with the resistors, as RC circuits, being able to calculate the time constant or period that the signal will be smoothed over. As with most sciences when there are two schools of thought, both are correct.
>Because electrons move Faster (speed of light) than mechanics parts the electrons carry current on the atomic scale which can be seen as “noisy” transients
Electric fields move at the speed of lights. Electrons move at like an inch per minute.
there are a few reasons you may want to do this for high voltage circuits.
1. Reducing base current
2. Reducing power dissipated Rb (R1 or R2)
the idea of this circuit is similar to what an op amp or digital drive circuit does. So those pull down buffer the signal. since s1 and s2 are push buttons you would want that to avoid debounce. Howver if driven with a microcontroller then you need not worry about that.
it would be a different reason if mosfets were being used as the gate capacitance could cause a delay in turn off of the LED
It looks like no one has explained to you what impedance is in general.
Here is the gist about voltage, current, and impedance. Every circuit has impedance. In this case let's simplify it to resistance. A circuit with lower resistance has greater ability drive circuit with higher impedance. It is like arguing. Whoever screams louder wins the argument.
The base has pretty high impedance, in other words, it doesn't take much to drive it. If you have a low impedance circuit like you finger touching the base, you can drive the base and turn on the transistor.
A 4.7k resistor is pretty low so it would be somewhat difficult to accidentally drive the base.
Finally, the switch obviously has much lower impedance than the 4.7k, so the base is easily driven by pressing the button.
They are there so people in all other reddit subs stop complaining about weird behavior when they move their hand close to the circuit when it's supposed to be off
Obviously pull-downs like people are saying, but the values are pretty odd, especially given the 10k resistors on the gate. The majority of the input signal is getting shunted to ground, which is not something most people would do on purpose unless they had to. It suggests that they're there to deal with something weird about the context of the circuit, like the input voltage being too high or something.
Normally I'd expect pull-downs to have a resistance high enough that you can almost ignore them when the circuit is active. That's definitely not true here.
They are probably not needed. BJT transistors are controlled by base current... when the switch is open the base current stops and the transistor turns off.
The one thing that comes to mind - when the switches are open the line from the switch to the transistor is basically an antenna. If you use high frequency, high gain, high base impedance, low base capacitance transistors it might be possible to pick up RF. These resistors eliminate any risk of that. In practice you are probably using the cheapest transistors you could find and it won't be an issue.
BTW - BJT transistors are mostly only seen in textbooks these days - everything is done with MOSFETs. If you redesign this circuit to use MOSFETs then you will need pull-down resistors.
Tip - if someone comes to you and asks what type of transistor they should use... listen carefully to all their requirements, consider it for a moment, then say MOSFET :-)
Thanks for your quick responses. So it’s there to prevent the transistors from being triggered by stray currents induced by magnetic fields and static. Got it.
I believe professionals call it a pull-down resistor, as was mentioned, to prevent floating potentials and return the base voltage to a known default state (0V) in the absence of an input signal. If you want to use the bjt as a switch, then this is how you connect it. As an exercise, try to do the same but with PNPs.
It’s part of a kit: Electronics Learning Lab by Forrest M. Mims. It was sold at RadioShack but you can still find them second hand. It’s a great learning tool for newbies.
I have that whole series of “engeneer’s mini handbook” s. And the big paperback. Written so well, and the animations are funny (ie overheated led)
How i miss RS days, and the old Frys Electronics stores..R.I.P rest in the past
they're pulldown resistors. they're just to make sure the voltage is never floating, so the transistor base is either on or off.
without a pulldown resistor things like stray capacitance and electromagnetic interference could much more easily cause issues.
They ensure that the base of the two transistors are pulled down all the way to ground when the switch contacts are open.
When designing logic, you never allow your inputs to float - they all must be either of the two states, 1 or 0, and nothing in between.
Pull down resistors for when switches are open. That is, set a voltage level (ground counts as well)
This. If you leave it open you can sometimes experience glitches due to 'undefined' input noise.
Yes, elecrronic noise, antenna effects, among others
Bjts will inherently have a diode between the base and emitter, so their inputs are physically not floating.
That is a gross oversimplification, you should still use a pull down resistor
Yeah, I'd still call those floating even with the base-emitter and base-collector diodes. You need those pull downs or this circuit won't work great.
Think of the tenth of a cent per unit you could be saving without those though!
And the power savings too! Assuming always on, those resistors use about 3/4 of a cent of electricity per year!
That literally is not a gross oversimplification if you are at all familiar with semiconductor physics. > BJTs can be thought of as two diodes (P–N junctions) sharing a common region that minority carriers can move through. A PNP BJT will function like two diodes that share an N-type cathode region, and the NPN like two diodes sharing a P-type anode region > In the discussion below, focus is on the NPN BJT. In what is called active mode, the base–emitter voltage VBE and collector–base voltage VCB are positive, forward biasing the emitter–base junction and reverse-biasing the collector–base junction. In this mode, electrons are injected from the forward biased n-type emitter region into the p-type base where they diffuse as minority carriers to the reverse-biased n-type collector and are swept away by the electric field in the reverse-biased collector–base junction. https://en.m.wikipedia.org/wiki/Bipolar_junction_transistor#
In the real world there is stray capacitance and inductance on components that can give undesired results. The physics is based on best case theoretical scenarios.
Where's the part that says "if you leave the gate floating, it will always behave exactly as you expect"?
There is no gate in a bjt. You can in many cases leave the **base** floating without issues. Adding a pull down resistor can speed up transistor switching, as when VBE approaches 0, the diode will sink exponentially less current, as opposed to a linear sink with a resistor. Another commenter mentioned that the collector base leakage current sometimes may be large enough to allow for a significant about of collector emitter current, however this often is not an issue.
“Most of the time it’s ok so we can just not add this part that will make sure no undesirable effects happen” I’m sure nothing bad will happen and our product will definitely be safe and won’t be recalled
Lol there are so many commercial circuits which leave the base of npn transistors floating. Here are the circuit diagrams of the lm324 and lm358, arguable some of the most mass produced ic’s ever. https://components101.com/article/difference-between-lm358-and-lm324 Notice how there aren’t pull up/down resistors at the input of the bjts? Also notice Q2 and Q3, along with Q1 and Q4, when Q1 and Q4 have the no voltage applied between their base and emitter, they inherently leave the base of Q2 and Q3 floating, as there is no path to the supply voltage. According to you this will make the circuit unreliable. But the circuit is not unreliable. Why? Because you don’t know what you are talking about.
I hate that I have to say this, but an OpAmp isn't a digital logic gate. If my AND gate produces an analog signal, its unusable. Likewise my OpAmp if it won't. Also, that's not a Darlington pair as they don't have a common collector.
Right off the bat you’re insulting my knowledge of semiconductors without even asking how much I know which tells me you only have a basic understanding of BJTs and are overcompensating. Yes calling a BJT a diode is a vast oversimplification, they are LIKE a diode because they have a p-n junction but they are vastly different
It wasn’t an insult, but you are certainly not helping your case. I didn’t say a bjt was a diode, I said the base emitter junction forms a diode. Did you completely gloss over the wiki article I linked? Please explain how the pn junction formed between the base and emitter junction in an NPN bjt is vastly different to a p-n diode. If you draw the band diagram of the pn junction it will be exactly that of a pn diode. > An ordinary BJT is a low input impedance current driven device. If there's no connection to the base, then base-emitter diode forward conduction/leakage will quickly take the base potential down to the emitter potential. The transistor will be off. https://electronics.stackexchange.com/questions/527665/transistor-with-floating-base
Then we get Neo, and Agent Smith… Neo’s aight, but Agents are no good.
During instances like this, when you use a resistor for a particular application other than purely resisting the flow of current(pull down, pull up, current sensing, etc) , what goes into determining what the actual resistance value needs to be? Is there a standard?
It depends on your application. If you're using an FET, the gate (acts as/is) a capacitor, so the pull-up or pull-down resistor value would be chosen based upon how fast you need the FET to shut off. For a BJT, you're generally using a pull-up or pull-down to keep stray (antenna or otherwise induced) charges from making your circuit act weird...5k to 100k, depending on how much power is available (battery life, etc). Or... The intent may be to use the BJT in the non-saturated region, where it acts like a current amplifier, and what looks like pull-up and pull-down resistors are actually called bias resistors, to set the Base voltage at a known value (0.7v-1.0v typically), but that's a whole different conversation. Current sensing resistors are generally really small, <1ohm, because it's in series with the load, and you don't want to effect the voltage delivered to the load, and you don't want to dissipate power (heat) at that resistor. Edited for clarity.
Thanks for the thorough explanation. That makes total sense. I think this is a start for my better understanding, because sometimes when I read schematics with different sensors and opamps, I see resistors sprinkled everywhere as if they were something people just throw in "for good measure" or without a rhyme or reason. And they so common that it seems like I'm just expected to just know why.
I was going to say, I eyeball it, plus check my drawer for what I have. You gave a much better answer.
This
I think “if a line is expected to be low, pull it down” is a fine rule of thumb for non-engineers.
I think they are to avoid potential noise triggering when the switch is open and the wire is not otherwise terminated. Every floating wire is a radio antenna.
Fair, but these are BJTs, not FETs.
BJTs are not immune to noise. Yes they are better against noise than FETs, but it's still best to pull the control line to a known low state when the switch is not engaged.
Is it common practice to use pull down resistors with bjts? Unless the common emitter current gain (Hfe) is super high, I would be surprised if noise would actually be large enough to be able to deliver enough current through the base emitter diode junction to allow any non negligible currents to flow through the collector.
Even if noise isn't a problem, a secondary effect of pull-downs may be needed to sink Icb reverse bias leakage current. If this is not done, then it will be carried by the base-emitter-junction and can cause device turn-on. This is a genuine real world effect which is well known and well documented but not always well-taught in courses.
Hm, that’s very interesting I never knew that. I still imagine it isn’t usually a practical consideration, unless hfe is very large. Usually leakage current in silicon rectifier diodes (like the np junction formed at the collector base) are very small, so even after the common emitter current gain, it should still be on the order of micro amps at max right? Unless I’m overlooking something.
They're there to prevent the input to the transistor from floating up and triggering the gate. Yes, it will work without them, but with any static or magnetic fields present it can rise to the level of vcc.
The purpose of the pull down resistors is to ensure that the transistor remains in the "off" state, or a logical "0," when there is no input signal. By permanently pulling the base to ground in the absence of a logic high, they limiting unwanted current flow which helps to which helps to reduce power loss and stabilize the transistor's state.
Yes. People sometimes forget what happens outside of nominal conditions. At high temperatures, beta gets much higher, Vbe gets much lower, and collector to base leakage current becomes significant. At high temperatures quality bipolars can indeed generate self-sustaining collector current with an open base terminal. In my experience (IC analog) most circuit design work is spent keeping the corner conditions functional.
what is the booktitle you use to learn?
It’s a kit; Electronics Learning Lab, by Forrest M. Mims. It’s out of production, but you can find it used.
>Electronics Learning Lab, by Forrest M. Mims I was just thinking, damn, that looks like Mims, but I've never seen that one. Was just about to ask what the title was. I've got several of his. Going to have to track this one down.
Love me some Forrest M.Mims. Good shit
I thought it was an XKCD thing at first with the handwriting
And by adding two small capacitors in parallel with these two resistors, you can implement an anti-key bounce, as well.
Eh, switch bounce is only really well mitigated with hysteresis. Capacitance helps, but only so much.
And what would that be for?
Mechanical Make and break switches (on the microsecond time scale) “bounce” or “ring” when the contactors make contact due to mechanical momentum slamming together until the mechanical position settles (can be seen if hook an o-scope up to location of R3&R4 if resistors are removed). Because electrons move Faster (speed of light) than mechanics parts the electrons carry current on the atomic scale which can be seen as “noisy” transients. Use a capacitor to offer a low impedance path (to create a transient filter) to “suck” these transient currents to ground preventing a voltage from building across R3 or R4 and falsely turning on Q1 & Q2 as the currents are generated. Originally, Adding R3 and R4 in the first place improves the noise immunity of Q1 and Q2 because with the resistors present at the gates of the transistors NOW real (not environmental transients like AM radio) powerful DC voltages supplied by your switches turn on transistors as intended (ie: are not turned on by some random un- engineered chance or event).
Yes, this. There's two common ways of thinking this through. First, as above, the noise is considered as AC and conducted to ground through the low capacitive reactance (impedance, resistance in an AC circuit). Second, the capacitors can be thought of as DC filters, in conjunction with the resistors, as RC circuits, being able to calculate the time constant or period that the signal will be smoothed over. As with most sciences when there are two schools of thought, both are correct.
>Because electrons move Faster (speed of light) than mechanics parts the electrons carry current on the atomic scale which can be seen as “noisy” transients Electric fields move at the speed of lights. Electrons move at like an inch per minute.
https://en.wikipedia.org/wiki/Switch#Contact_bounce
there are a few reasons you may want to do this for high voltage circuits. 1. Reducing base current 2. Reducing power dissipated Rb (R1 or R2) the idea of this circuit is similar to what an op amp or digital drive circuit does. So those pull down buffer the signal. since s1 and s2 are push buttons you would want that to avoid debounce. Howver if driven with a microcontroller then you need not worry about that. it would be a different reason if mosfets were being used as the gate capacitance could cause a delay in turn off of the LED
It looks like no one has explained to you what impedance is in general. Here is the gist about voltage, current, and impedance. Every circuit has impedance. In this case let's simplify it to resistance. A circuit with lower resistance has greater ability drive circuit with higher impedance. It is like arguing. Whoever screams louder wins the argument. The base has pretty high impedance, in other words, it doesn't take much to drive it. If you have a low impedance circuit like you finger touching the base, you can drive the base and turn on the transistor. A 4.7k resistor is pretty low so it would be somewhat difficult to accidentally drive the base. Finally, the switch obviously has much lower impedance than the 4.7k, so the base is easily driven by pressing the button.
Tie down res...to avoid false highs in noisey circuits due to high input impedance of the next input
They are there so people in all other reddit subs stop complaining about weird behavior when they move their hand close to the circuit when it's supposed to be off
Obviously pull-downs like people are saying, but the values are pretty odd, especially given the 10k resistors on the gate. The majority of the input signal is getting shunted to ground, which is not something most people would do on purpose unless they had to. It suggests that they're there to deal with something weird about the context of the circuit, like the input voltage being too high or something. Normally I'd expect pull-downs to have a resistance high enough that you can almost ignore them when the circuit is active. That's definitely not true here.
They are probably not needed. BJT transistors are controlled by base current... when the switch is open the base current stops and the transistor turns off. The one thing that comes to mind - when the switches are open the line from the switch to the transistor is basically an antenna. If you use high frequency, high gain, high base impedance, low base capacitance transistors it might be possible to pick up RF. These resistors eliminate any risk of that. In practice you are probably using the cheapest transistors you could find and it won't be an issue. BTW - BJT transistors are mostly only seen in textbooks these days - everything is done with MOSFETs. If you redesign this circuit to use MOSFETs then you will need pull-down resistors. Tip - if someone comes to you and asks what type of transistor they should use... listen carefully to all their requirements, consider it for a moment, then say MOSFET :-)
How should I set up a current mirror for an audio amp?
Yeah there's a lot of things BJT's are better at than MOSFETs.
RF comes to mind - I mean, FETs are my favorite - but sometimes you can't beat that ft.
I would start with a MOSFET and an opamp. Admittedly I haven't had to design a current mirror so I would have to think through the opamp circuit.
an opamp is built out of bjt's and literally already contains a current mirror internally. So yeah...
Lets not get carried away. BJT hasn't been put out to pasture yet... I have several low power things that I prefer a BJT on.
Thanks for your quick responses. So it’s there to prevent the transistors from being triggered by stray currents induced by magnetic fields and static. Got it.
I believe professionals call it a pull-down resistor, as was mentioned, to prevent floating potentials and return the base voltage to a known default state (0V) in the absence of an input signal. If you want to use the bjt as a switch, then this is how you connect it. As an exercise, try to do the same but with PNPs.
Curious, what would be the book that you’re reading?
It’s part of a kit: Electronics Learning Lab by Forrest M. Mims. It was sold at RadioShack but you can still find them second hand. It’s a great learning tool for newbies.
I have that whole series of “engeneer’s mini handbook” s. And the big paperback. Written so well, and the animations are funny (ie overheated led) How i miss RS days, and the old Frys Electronics stores..R.I.P rest in the past
Ah, thank you for the quick reply! I’m trying to compile a list for electronic books to get/study.
What book is this?
It looks like a Forrest Mims book, possibly one of the ones that came with the RadioShack electronics learning lab.
Floating digital inputs = bad.
Is it working exactly like the lesson describes?
they're pulldown resistors. they're just to make sure the voltage is never floating, so the transistor base is either on or off. without a pulldown resistor things like stray capacitance and electromagnetic interference could much more easily cause issues.
Pull down resistors.
Pull down resistors for low-noise applications
They ensure that the base of the two transistors are pulled down all the way to ground when the switch contacts are open. When designing logic, you never allow your inputs to float - they all must be either of the two states, 1 or 0, and nothing in between.
Thanks for sharing, I learned a lot from everyone's comments on here!
What book is that?
What book are you using?
I believe those resistors extend the life of the transistors that they are powering on and off.
What book is that?