# A Simple Circuit concept to study biasing transistors

#### fredric58

Joined Nov 28, 2014
203
Hello everyone. I would like to accomplish (2) things. The first of course is building a project. However my knowledge in electronics is some what limited. The second is to learn how to bias transistors to use them a switches. I understand the basic principles behind NPN and PNP's, Beta, saturation, cutoff, what the base is for and how it affects the emitter and collector etc. Did I mention FRYING a few? I have designed and attached what I think is a relatively simple circuit containing an explanation of what I am trying to accomplish at the bottom. Before I start the project I would like to ask you to take a look and let me know if my "concept / idea" will actually WORK. I learn faster and it's more rewarding to learn on my own projects than simply copying someone else's. thank you

all guidance appreciated!

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#### MrChips

Joined Oct 2, 2009
22,578
I would simply say that you are off base.

You will not gain any timing information with the setup you have shown. Transistors switch in nanoseconds. You need to use a high speed oscilloscope in order to examine switching speeds of transistors.

Transistors blow up for any or all of three reasons:

1. Max voltages are exceeded
2. Max currents are exceeded
3. Max temperatures are exceeded

Solution to avoid destroying transistors:
1. Do not exceed max operating voltages.
2. Limit the currents with series resistors.
3. Limit the power dissipation or mount transistor on appropriately sized heat sink.

To bias a transistor as a switch, all you need to control is the base current. The rule of thumb is the ON base current must be 1/10 of the collector current. This guarantees that the transistor is in saturation mode. Use Ohms Law, I = V/R to determine the appropriate turn-on voltage and current limiting resistor.

#### fredric58

Joined Nov 28, 2014
203
I would simply say that you are off base.

You will not gain any timing information with the setup you have shown. Transistors switch in nanoseconds. You need to use a high speed oscilloscope in order to examine switching speeds of transistors.

Transistors blow up for any or all of three reasons:

1. Max voltages are exceeded
2. Max currents are exceeded
3. Max temperatures are exceeded

Solution to avoid destroying transistors:
1. Do not exceed max operating voltages.
2. Limit the currents with series resistors.
3. Limit the power dissipation or mount transistor on appropriately sized heat sink.

To bias a transistor as a switch, all you need to control is the base current. The rule of thumb is the ON base current must be 1/10 of the collector current. This guarantees that the transistor is in saturation mode. Use Ohms Law, I = V/R to determine the appropriate turn-on voltage and current limiting resistor.

#### dl324

Joined Mar 30, 2015
12,271
Circuit in question:

To calculate base resistors, we need to know the load that the transistor will be switching.

In the first case, you don't give the switched load or the driving voltage.

For the last switch, we don't know what the transistor is switching. You suggest a PNP transistor which would imply high side switching, but you drew it as low side.

Since you don't seem to know what you're doing, perhaps you should describe what you're trying to accomplish. One switch at a time.

#### fredric58

Joined Nov 28, 2014
203
hi, thank you....this has nothing to do with timing. I am using a digital "timer" but that is simply to turn the project on at whatever hour I want. what I am interested in knowing is if these proposed switches (transistors) with the "correct" base resistors will turn on the components as described

a digital timer starts a signal at say 8:00, that signal is (x) volts and is made to conform (with a base resistor) to the requirements of T1 to saturate and "open" it, therefore sending 5v to the PIR.

step (2) the PIR is now "on". movement activates it and it sends a signal of (3.3V) that is made (with a base resistor) to conform to the requirements of T2

which in turn, sends 5V to the sound card/board.

sorry for the confusion.

#### dl324

Joined Mar 30, 2015
12,271
a digital timer starts a signal at say 8:00, that signal is (x) volts and is made to conform (with a base resistor) to the requirements of T1 to saturate and "open" it, therefore sending 5v to the PIR.
That isn't the normal design process. We typically make the transistor switch conform to the signal that's driving it.

You have already been told that to calculate the base current required to guarantee a transistor will saturate, we need to know the load it will be switching. That, combined with the control signal voltage, can then be used to calculate an appropriate base resistance.

#### fredric58

Joined Nov 28, 2014
203
thank you, let me be more specific. this isn't a test and I'm not asking anyone to solve equations. i see people on you tube switch on LED's using transistors. in its simplest form my question is with the correct components does anyone think this would work?

the last switch should be really simple to understand, pins 0-9 are different tracts. to trigger it you send the pin to ground which is why there is a PNP, of course i don't know the resistor values yet. but my understanding is that the current from these pins will travel thru the base to ground and if the base is lower than the voltage it will open and the pin will go to ground

right?

#### dl324

Joined Mar 30, 2015
12,271
thank you, let me be more specific. this isn't a test and I'm not asking anyone to solve equations. i see people on you tube switch on LED's using transistors. in its simplest form my question is with the correct components does anyone think this would work?
You've already been given sufficient information to calculate the base resistor valuess for the two NPN switches.
the last switch should be really simple to understand, pins 0-9 are different tracts.
It isn't simple to understand because you haven't indicated what the switch is switching.
to trigger it you send the pin to ground which is why there is a PNP, of course i don't know the resistor values yet. but my understanding is that the current from these pins will travel thru the base to ground and if the base is lower than the voltage it will open and the pin will go to ground

right?
If I understand what you're trying to do; no.

You've mentioned transistors opening a couple times now and your usage is backwards.

Trying to use a PNP transistor to do low side switching is problematic. To turn the transistor on, you need to forward bias the emitter-base junction. Depending on the load being switched, the emitter voltage may drop too low to keep the transistor saturated.

#### crutschow

Joined Mar 14, 2008
26,071
Relays open and close.
Transistors turn on and off.

#### fredric58

Joined Nov 28, 2014
203
problematic? i want 1 of the pins to go to ground. i choose with the rotary switch

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#### dl324

Joined Mar 30, 2015
12,271
problematic?
Yes. Using a PNP transistor as a low side switch is problematic for the reasons I mentioned earlier.
i want 1 of the pins to go to ground. i choose with the rotary switch
Why don't you just use the common pole of the, presumably single pole 10 position, switch to provide ground?

#### fredric58

Joined Nov 28, 2014
203
"Why don't you just use the common pole of the, presumably single pole 10 position, switch to provide ground?"

that would work exceptionally well

and be of zero value toward knowledge gained.

isn't voltage relative in PNP? no matter how low the emitter voltage goes, the base voltage will still be proportionately lower?
in order for it to not work, wouldn't it have to go to zero V?

#### fredric58

Joined Nov 28, 2014
203
Circuit in question:
View attachment 138589

To calculate base resistors, we need to know the load that the transistor will be switching.

In the first case, you don't give the switched load or the driving voltage.

For the last switch, we don't know what the transistor is switching. You suggest a PNP transistor which would imply high side switching, but you drew it as low side.

Since you don't seem to know what you're doing, perhaps you should describe what you're trying to accomplish. One switch at a time.
I thought that might be quite obvious. I have a 5 V system. A digital timer. At a given time it sends out a signal, or goes high, or say it sends out a current. That current goes through a resistor to a transistor and turns on a PIR.

The PIR does the same thing, when actuated it sends out a signal. That signal goes through a resistor to the base of a transistor and turns on a sound module.

Immediately, which ever pin I have chosen, zero through nine will go to ground that's the trigger for the playback.

#### hobbyist

Joined Aug 10, 2008
889
However my knowledge in electronics is some what limited. The second is to learn how to bias transistors to use them a switches.
Ok so you are using transistors as switches to sink current from the supply to ground through each load.

You are asking if this circuit would work, yes after proper resistor values are chosen to drive each transitor into saturation, while not overcurrenting the base terminals.

You could just throw any small size resistor into the circuit and see if it would work, but then you risk popping transistors, as you have already experienced.

So now after all that your ready to design a switching transistor circuit using calculations and empiricle methods.

In order to know what resistors you need for current limiting, as someone mentioned you need to know your collector current then divide that by 10 to supply base current.

So if you keep that in mind throughout this design you should be able to make this work.

First determine where to begin a design. Every circuit has different starting points for design work, so you need to familiar yourself with where is the best approach for this circuit.

Lets start from the input side and see if it will work for the starting point.

You have a current that needs to feed into the base of T1 base from the timer to switch on T1, then you want at least 10 times more current flowing through T1 to drive its load the PIR, then you need current from the PIR to supply base current into base of T2 to turn it on, then 10 times more current flows through T2 to drive the soundboard. Then the soundboard supplys a current through the rotary switch and the PNP is sinking this current to ground. Since there is no signal current at the base of the PNP, (fixed resistor bias) then we can disregard it as any usefull output at this time.

So it would be that the soundboard is the final output of this circuit function.

As shown, you need to know collector current before base current because collector current is the driving currents to power on each usefull load.

So demonstrated here, it is obvious that starting from the input side of this design is not feesable, because collector current needs to be determined first before base current.

So lets start from the usefull load output side, the soundboard.
Here is where empericle measurements is needed.

First try to find any specs on this soundboard, to see what voltage and current is needed to make it work. If you cannot find any documentation on this, then experimenting is needed.

If you need to experiment to find these values, then do this.
Connecting your soundboard across the supply to determine how well it works, now on the negative side start putting in low value resistors 1-10 ohms for starters, and keep checking its performance, find out at what value of resistance where the board is no longer able to work properly, and that will give you an idea of how much current loss it can take before its performance is affected.

Next using a 1 ohm resistor hook it to the soundboard and check the voltage across this resistor, calculate the current through it and write it down. Then take your last resistor that still kept the sound board working properly and repeat the same procedure.
Now you have a low and high value of current that will drive your board.

Now you have a range of currents that T2 collector current can be made to work at for the board.
Now you need to see if the input source to T2 can supply the base current needed.

You'r PIR shows 3.3 volts that is probably open circuit voltage (unloaded), you need to determine what load can this handle, so again take your PIR and turn it on across the supply as ususal, now begin putting a resistor at its output to ground and check for performance, find the smallest resistor value that affects its performance, that would be the limit of load it can handle.
Measure voltage across and calculate current flow through it.

Now take that value and multiply it by 10 to see if it is in the range of the current values for the soundboard. Now you don't want to load this PIR to its maximum, so go somewhere in the middle value of current calculated and multiply by 10 to see if its in the range of your soundboard calcualted currents range.

This procedure allows you to determine usefull collector and base currents so you'r driving source (PIR) can turn on your transistor switch, and the switch can sink the needed current for the oputput load(soundboard), for proper performance.

Now you can calculate a approximate value for R2.
Choose a resistor to load the PIR, check output voltage from ground, calculate current through this resistor, and if this is the current you want then this will be used in the calcualtion of R2.

Take the voltage measurement at the load, and subrract Vbe (0.7v) then divide this result by the current you calculated through this load resistor, and that will give you a range value of base resistor R2.

Now build a circuit consisting only of the soundboard connected to collector of T2, and connect the output of the PIR to the base of T2 through R2, and see if the soundboard turns on when the PIR outputs a signal. If it doesn't take measurements, and determine how much to drop the R2 value until T2 turns on the soundboard.

Once that's working properly, then repeat all the steps above for the next input source for T1 and so on, each input source and output load need to be matched along the signal path so you may have to change values back and forth to make each load and input source work with each stage.

In other words, your timer may have a very high output impedance that may require a very light load on it, and the base current it takes from it may not be enough to drive the T1 transistor into saturation with required load current for the PIR input.
So thats where you have to go back in forth between stages changing values or adding different stages to make everything run in sync, from input to output.

Now when you get everything running from the timer to the soundboard, you have one more switching stage you chose to incorporate, as your the designer so you can add anything you desire.

You chose to use a PNP to sink current from a load (soundboard) through a rotary switch to ground.

The reason people on the board are rejecting this PNP design, is because the base emitter and base collector voltage determines the turn on voltage and current to drive the transistor into saturation as a switch.

In your design, you are driving a changing voltage load through your emitter, meaning that your load is not a stable resistance, so as the load changes so does the Vbe of the transistor, and this voltage determines the function of the transistor VCE due to collector current.
Also in saturation the base collector junction needs to be forward biased as well as the Vbe, in your design the PNP will not reach saturation, however it may still turn on enough to sink current for the load.
That's something you could experiment with to see how it works out.

However if you choose to use a PNP in the design then putting it on top of the soundboard to supply current to it, would allow for the transistor to act as a switch.

I think this may help you get started in playing with transistors at a basic level, as you stated you would like to learn this through hands on approach.

#### fredric58

Joined Nov 28, 2014
203
Ok so you are using transistors as switches to sink current from the supply to ground through each load.

You are asking if this circuit would work, yes after proper resistor values are chosen to drive each transitor into saturation, while not overcurrenting the base terminals.

You could just throw any small size resistor into the circuit and see if it would work, but then you risk popping transistors, as you have already experienced.

So now after all that your ready to design a switching transistor circuit using calculations and empiricle methods.

In order to know what resistors you need for current limiting, as someone mentioned you need to know your collector current then divide that by 10 to supply base current.

So if you keep that in mind throughout this design you should be able to make this work.

First determine where to begin a design. Every circuit has different starting points for design work, so you need to familiar yourself with where is the best approach for this circuit.

Lets start from the input side and see if it will work for the starting point.

You have a current that needs to feed into the base of T1 base from the timer to switch on T1, then you want at least 10 times more current flowing through T1 to drive its load the PIR, then you need current from the PIR to supply base current into base of T2 to turn it on, then 10 times more current flows through T2 to drive the soundboard. Then the soundboard supplys a current through the rotary switch and the PNP is sinking this current to ground. Since there is no signal current at the base of the PNP, (fixed resistor bias) then we can disregard it as any usefull output at this time.

So it would be that the soundboard is the final output of this circuit function.

As shown, you need to know collector current before base current because collector current is the driving currents to power on each usefull load.

So demonstrated here, it is obvious that starting from the input side of this design is not feesable, because collector current needs to be determined first before base current.

So lets start from the usefull load output side, the soundboard.
Here is where empericle measurements is needed.

First try to find any specs on this soundboard, to see what voltage and current is needed to make it work. If you cannot find any documentation on this, then experimenting is needed.

If you need to experiment to find these values, then do this.
Connecting your soundboard across the supply to determine how well it works, now on the negative side start putting in low value resistors 1-10 ohms for starters, and keep checking its performance, find out at what value of resistance where the board is no longer able to work properly, and that will give you an idea of how much current loss it can take before its performance is affected.

Next using a 1 ohm resistor hook it to the soundboard and check the voltage across this resistor, calculate the current through it and write it down. Then take your last resistor that still kept the sound board working properly and repeat the same procedure.
Now you have a low and high value of current that will drive your board.

Now you have a range of currents that T2 collector current can be made to work at for the board.
Now you need to see if the input source to T2 can supply the base current needed.

You'r PIR shows 3.3 volts that is probably open circuit voltage (unloaded), you need to determine what load can this handle, so again take your PIR and turn it on across the supply as ususal, now begin putting a resistor at its output to ground and check for performance, find the smallest resistor value that affects its performance, that would be the limit of load it can handle.
Measure voltage across and calculate current flow through it.

Now take that value and multiply it by 10 to see if it is in the range of the current values for the soundboard. Now you don't want to load this PIR to its maximum, so go somewhere in the middle value of current calculated and multiply by 10 to see if its in the range of your soundboard calcualted currents range.

This procedure allows you to determine usefull collector and base currents so you'r driving source (PIR) can turn on your transistor switch, and the switch can sink the needed current for the oputput load(soundboard), for proper performance.

Now you can calculate a approximate value for R2.
Choose a resistor to load the PIR, check output voltage from ground, calculate current through this resistor, and if this is the current you want then this will be used in the calcualtion of R2.

Take the voltage measurement at the load, and subrract Vbe (0.7v) then divide this result by the current you calculated through this load resistor, and that will give you a range value of base resistor R2.

Now build a circuit consisting only of the soundboard connected to collector of T2, and connect the output of the PIR to the base of T2 through R2, and see if the soundboard turns on when the PIR outputs a signal. If it doesn't take measurements, and determine how much to drop the R2 value until T2 turns on the soundboard.

Once that's working properly, then repeat all the steps above for the next input source for T1 and so on, each input source and output load need to be matched along the signal path so you may have to change values back and forth to make each load and input source work with each stage.

In other words, your timer may have a very high output impedance that may require a very light load on it, and the base current it takes from it may not be enough to drive the T1 transistor into saturation with required load current for the PIR input.
So thats where you have to go back in forth between stages changing values or adding different stages to make everything run in sync, from input to output.

Now when you get everything running from the timer to the soundboard, you have one more switching stage you chose to incorporate, as your the designer so you can add anything you desire.

You chose to use a PNP to sink current from a load (soundboard) through a rotary switch to ground.

The reason people on the board are rejecting this PNP design, is because the base emitter and base collector voltage determines the turn on voltage and current to drive the transistor into saturation as a switch.

In your design, you are driving a changing voltage load through your emitter, meaning that your load is not a stable resistance, so as the load changes so does the Vbe of the transistor, and this voltage determines the function of the transistor VCE due to collector current.
Also in saturation the base collector junction needs to be forward biased as well as the Vbe, in your design the PNP will not reach saturation, however it may still turn on enough to sink current for the load.
That's something you could experiment with to see how it works out.

However if you choose to use a PNP in the design then putting it on top of the soundboard to supply current to it, would allow for the transistor to act as a switch.

I think this may help you get started in playing with transistors at a basic level, as you stated you would like to learn this through hands on approach.
THANK YOU! Your guidance is very much appreciated. I will begin as instructed and have already determine this is going to be a great learning experience. Many thanks

#### hobbyist

Joined Aug 10, 2008
889
If I may suggest, transistor circuit design can be very frustrating, when first starting out to learn how to use transistors.

You'r project is not a beginner project to learn efficiently circuit design, because you are using active loads.

Before building and experimenting with your project, it would be more of a training in transistor behaviour by doing some simple experiments as shown here.

#### fredric58

Joined Nov 28, 2014
203
cool. i'll give it a go. I just finished the first part suggested in the previous notes. dug up a data sheet, pulled out the resistors. connected the sound card/board to a reg 5V source and started with a 1 ohm resistor from the boards gnd, to a MM and then to ground . my findings are as follow. I didn't notice a difference in performance between 1 and 10 ohms. so I put a 330 in and noticed a huge difference. I went back to 10, 12, 15....up to 86 ohms. the current remained close to the data sheet at 29-32mA during operation through the entire pile of resistors. what I came to realize is with each additional amount of resistance, the volume was being slightly diminished. tomorrow i'll stop by the studio and pick up my decibel meter so I will be able to "see" when the actual loss begins to appear. ??? Was this executed properly and were these the right results to be expected???

on the second part you mention. I am not quite sure how to proceed.

"Next using a 1 ohm resistor hook it to the soundboard and check the voltage across this resistor, calculate the current through it and write it down. Then take your last resistor that still kept the sound board working properly and repeat the same procedure.
Now you have a low and high value of current that will drive your board."

is that 1 ohm from the + to the sound board? measure V, write down. reinstall the last resistor on the (-) side that I settled on, then add 1,2,3,4,5 ohms......to the positive side till I begin to lose amplification/results not satisfying?

#### hobbyist

Joined Aug 10, 2008
889
is that 1 ohm from the + to the sound board? measure V, write down. reinstall the last resistor on the (-) side that I settled on, then add 1,2,3,4,5 ohms......to the positive side till I begin to lose amplification/results not satisfying?
No not at all, the resistors are used to take the place of the transistors temporarily to see what minimum currents each load can handle, so you have a range of how much current it takes for each load for the transistors to sink.

It looks like you are getting a good handle of testing before designing process,

So now you can work with those values of current to decide on base and collector currents for T2.

As a side note to keep any frustrations from taken hold, (because transistor circuit design can be very finiky when your first starting out)

As I said different loads sometimes don't want to play together, so you may decide to use a collector current value for this T2 transistor, and find out that when you calculate the divide by ten rule for base current, that this base current loads down the input stage, (PIR) (these things are always happening in circuit design), so thats where you have to keep on changing values of current for T2, that's where it is good practice to keep a log of what currents work for each stage, the lowest value resistor on the output of a load, allows you to determine how much current you can draw from it before it loses performance.
Also be aware, you may have a input source that has such a high output impedance, that no matter what resistor you use to limit base current to the transistor it is driving, that the base current loads the input source to make it useless, thats where you learn how to design buffer circuits, with transistors, and dc amps ect....

Also remember all the calculations are just ballpark figures, to get you close to design specs, you will find that tweaking component values is a given when designing transistor circuits.

And above all Have fun.

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#### fredric58

Joined Nov 28, 2014
203
this is fun and I have never found anything about it frustrating. I was blessed with an unbelievable amount of patience. it looks like I didn't fully understand the first part. the resistor being "substituted" for the transistor. I did discover increasing the resistance between the board and gnd. is similar to closing a valve on a water supply (diminishing the volume/sound out) the water is still there but it isn't flowing. what I don't understand is why I still got a reading of 29-32mA of current flow regardless of which or how much resistance I added. but each increase, decreased the volume output from the speakers. I don't believe I measured correctly. if fact I am pretty sure I didn't. I would like to focus on this one part.

5+ to sound board // insert resistor in the gnd line // measure resistance from which point to which point?
what is known: performance will suffer with each increase of resistance. what is believed: is that i will be able to add a limited amount of resistance before i begin to lose performance in the board. that i can test with the db meter.

have faith. slow starter, but like a freight train...........

thanks again

#### hobbyist

Joined Aug 10, 2008
889
I was blessed with an unbelievable amount of patience.
A Blessing indeed.

OK, lets take it one step at a time, the resistor on the ground side of the load (soundboard), is to determine how much current loss it can take before performance is affected.

However I would like to caution, some systems may require a minimum amount of current and no lower or else risk damaging the circuit board, as well as damage by excess current, it depends on the system design.

Since the current remains stable when changng resistors, may be that there is a voltage or current regulation going on on the soundboard to keep everything happy during different supply anomolies.

Thats why it is best to find the data sheet for a system you are working with.

29-32mA for the soundboard, gives you a good springboard to work off of.

T2 needs to be biased to sink around 29-32mA of current.
To saturate T2 you would need to use a base resistor that would supply 1/10 of that, around 2.9-3.2mA. of base current.

Now you did a fine job in determining the output voltage of the driving source (PIR), so now you can take the output voltage of the PIR minus Vbe of the T2, then divide into that the base current you calculated above to give you a ballpark figure for resistor R2 that will be your base resistor for T2.
Build and tweak this resistor value until you get good performance.

Now you could just hook up the PIR at this point and see if it all works together.

However I would like to suggest for training, design a voltage divider that will give you the PIR output voltage, using a ground resistor thats 1/10th of the value of R2, then hook up T2 to it and see if the transistor saturates as a switch and turns the soudboard on.
Then raise the value of the ground resistor to the same value of R2, and readjust the resistor value at the top of the divider to give around the same voltage as before, and hook it to your T2, and see how much the input voltage drops and see if your transistor can drive the soundboard, that will give you an idea of how the transistor needs to always have a higher input impedance with respect to the source driving it, and you will see how that affects the collector current driving the load.

Or then again : you could just jump right ahead and hook up your PIR to the transistor and see if it all works together at this point of the design.

Because at the stage your are at right now, you are matching the output impedance of the PIR to your transistor switch, while at the same time providing the transistor switch with all the current it needs from the PIR to successfully drive its load (soundboard).