Jump to content
Electronics-Lab.com Community

BJT biasing formulas


Recommended Posts

  • Replies 82
  • Created
  • Last Reply

Top Posters In This Topic

Top Posters In This Topic

Posted Images

Hi Autir,
I don't have formulas; just Ohm's Law, the transistor's datasheet and experience.
Your 2nd circuit won't work because the current gain and base-emitter voltage of any certain transistor is different for each one, and change with temperature changes.

For the 1st transistor circuit, I selected a few things:
1) A common and cheap 2N3904 transistor.
2) A reasonable 10k output impedance and fairly low supply current.
3) An easy to calculate 10V power supply.
4) A voltage gain of only 10 for low distortion.

A transistor's collector is a high impedance current sink, so the output impedance is its collector load resistor.
5.5V was chosen as its idle collector voltage so it can swing equally up and down.
The transistor's typical current gain of 230 and base-emitter voltage of 0.65V are in the datasheet.
Ohm's Law calculated the resistors.
Any questions?

post-1706-14279142469884_thumb.png

Link to post
Share on other sites
Any questions?


As always, many  ;D

The circuit without the emitter resistor is indeed a bad design, as it is dependent upon the hfe of each transistor and the shifts of the hfe value caused by temperature etc. change. I included it just to see how the formulas change with the transition from the simple (no Re) type to the other.

Using Thevenin's theorem in your circuit dictates that the input impedance of the circuit will be Zin=R1//R2//(re+R4), which is roughly equal to R1//R2//R4.
So Zin=976 Ohm and not 34900 Ohm. Or not?  ???

How did you choose the Ib (2.4uA) to I(R1R2) (27uA) ratio?

It is indeed a very stable circuit which, when being simulated in Multisim, provided exactly the same output with a variety of transistors and hfe's. When I added a bypass capacitor paralleled with R4 the above characteristic dissapeared.

The Ib/Ic ratio in your circuit is proof of the hfe value of the transistor you used. Why, then, does it work the same with different hfe's? Does this mean that the Ib/Ic ratio can be decided by me without any consenquences regarding the circuit's function?

The gain factor is calculated by the R3/R4 ratio, if I'm not mistaken?



@Alun:

Your link is invaluable. However, there are certain "shady" parts in the text.

In the absence of any good reason for making some other choice we might just as well assume that the available voltage should be shared equally between Re, Rc, and the transistor. We therefore want about 5 volts across Re, 5 volts across Rc, and 5 volts between the collector and emitter. This means that the amplifier should have Vc=10 V,  Ve=5V, and  Vce=5V.


Abscence of any good reason? What happened with the Q point, isn't it good enough a reason?!?!?!?

In practice the simplest convenient choice is to pick something like Io=25*Ib so I suggest you choose that. Note, however, that you could choose almost anything from Io=Ib up to Io=100*Ib and it would still probably be possible to make the amplifier work despite having chosen very different currents and resistor values!


Isn't the Io/Ib ratio GIVEN AND DEFINED by the usage of Thevenin's theorem?

To change the voltage across Ce we have to move charge in or out of the capacitor. This takes time. So if we keep changing our mind and waggling the input voltage up and down quickly we don’t give this a chance to happen.


"This takes time"... why?  ??? ??? ??? ??? ???

Another question: Why do we use two distinct resistances in the Emmiter area?

Could someone please explain the Re - Rg part a little better to me?

Thanks.
Link to post
Share on other sites

I think you're jumping in at the deep end here a bit, at this stage all you need is simple ohm's law and nothing more.

I know the following is a bad design but it's a model good for understanding things.

In the following simulation, Tr1 has a gain of exactly 100 and a Vb of exactly 664mV (the default transistor model)

We want the voltage across Rc to be half the supply voltage for maximum swing so:
Vc = VRc = 1/2 +V = 10 * 1/2 = 5V.

In order for Rc to have a voltage drop of 5V the current though it needs to be:
V/R = 5 / 1k = 5mA.

Tr1 has a gain of 100 so the current going into the bas needs to be 100th of the current though the collector so Rb needs to limit Ib to:
5mA /100 = 50uA.

Vb = 664mV so the potential differance across it is:
Vb = 10 - 0.664 = 9.336V

At 9.336V the resitor value to pass 50uA = V/I = 9.336 / 50uA = 186.72k, which is rounded to 187k on the schematic.

You'll see that if you use these formulae with componants having exactly the right characteristics then the voltages all round the circuit can be calculated exactly but as you know in practise this isn't true.

Do you understand all of the above calculations? Because you can't move on to the other circuits until you can grasp the fundermentals.

post-0-14279142471674_thumb.gif

Link to post
Share on other sites

You haven't read my post properly have you audioguru?


I know the following is a bad design but it's a model good for understanding things.


I know all of this, the purpose of this circuit is not to be useful in the practical sense but as a teaching aid. If/when autir returns I will explain to him the principle behind an emitter follower, and why it has a high input impedance, then I'll return to the common emitter amplifier and add Re then finally the resitor below the base. It's far easier to break a circuit up into smaller componants and understanding each, rather than attempting to grasp the whole thing in one go.
Link to post
Share on other sites

Hi autir
Audioguru's design not a holy, there are many schools in designing a transistor amplifier. One of the simplest ways you can found in this site:
http://www.electronics-tutorials.com/amplifiers/small-signal-amplifiers.htm
it is very good for you to understand what you want, read it carefully then ask.
90% of Audioguru answers are obscure at least for me. Take a look at these examples:
(1) autir ask: How did you choose the Ib (2.4uA) to I(R1R2) (27uA) ratio?
Audioguru's answer: A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2.
I try to understand it fore an hour but could not.
If i want to answer this i say: assume that Ib = 0, that is IR1 = IR2 =I
this I must be 1/5 t0 1/10 the IE
I = 10V/(330K+47K)=26.5 uA close to 27uA
If we choose I = 1/5 IE ==> IE = 0.27 ma, Audioguru IE = 0.55mA the double, that is Audioguru choose I =1/20 IE , many schools
Why Ib =0 because it is a very small value and we can negect it in our calculations.
In the link above you can read the full story it is direct and simple.
(2)  autir ask: The Ib/Ic ratio in your circuit is proof of the hfe value of the transistor you used. Why, then, does it work the same with different hfe's? Does this mean that the Ib/Ic ratio can be decided by me without any consenquences regarding the circuit's function?
Audioguru's answer: I selected a transistor whose hFE is never less than half its typical, nor more than double. The circuit would need more current in the R1/R2 divider if a transistor with a lower minimum hFE was used.
walid: hfe in Audioguru's design = IC/IB = 0.55mA/27uA = 20.37 and this is less than half its typical.
(3) autir ask: The gain factor is calculated by the R3/R4 ratio, if I'm not mistaken?
Audioguru's answer: it is long and far from a direct answer
Walid: this is the way audioguru answer 90% of the questions, indirect and obscure. His answer must be either yes or no with explenation to why yes and why no. there are many questions must answered by yes or no at the beginning.
If I try to answer i say: NO NO NO this formula you use suitable for emitter degeneration (a sublink in the same site above) look at that site and compare.
(4) In his reply to Alun, Audioguru surprised that ther is no RE and no R2, why i don't know.
I tell him that this circuit called a self biased and is found as the first circuit in a small sig amp in any electronic book and this circuit is used in realty, i see it many times in areal designs.
Finally I hope from all my heart that Audioguru accept this criticism and still my big friend.
yours
walid

Link to post
Share on other sites

No Walid, this is self bias and the formulea isn't too differant from before:

Full full swing we need:
Vc = 5V

So we want Irc to equal 5mA and to do this the base curent needs to be:
IRf = Irc / 100 = 5mA / 100 = 50uA.

The voltage across Rf is:
VRf = Vc - Vb = 5 - 664mV = 4.336V.

So for 50uA to flow is value should be:
IRf = VRf / Ib = 4.336 / 50uA = 86.72k.

Now you'll see that there's some error here, this is because Ic isn't exactly IRc as some of the current goes though to the base. We could calculate the exact value for Rf using simultanious equations but why should we bother as it's only an error of 2mV and the transistor's gain will vary a lot and introduce a greater error than this?

This is often used with an input resistor to intorduce negitive feedback and even though the distortion is quite high compared with common emitter amplifier with large amounts of negitve feedback the output impedance is a lot lower.

About your other comments, all will be explaind later when I explain the emitter follower and then add the other resistors. I think we really need to teach people step by step as all of the articles I've read rush into things far too quickly.

I'm off to bed shortly so you'll have to wait untill tomorrow.

post-0-14279142472795_thumb.gif

Link to post
Share on other sites

Hi friend Walid,
I'm sorry you don't understand me, I'm not a teacher nor an author.


90% of Audioguru answers are obscure at least for me. Take a look at these examples:
(1) autir ask: How did you choose the Ib (2.4uA) to I(R1R2) (27uA) ratio?
Audioguru's answer: A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2.
I try to understand it fore an hour but could not.

Each part number and each sample of transistor has a different current gain. Transistors have very reduced gain at high currents. Therefore I calculate the current in the voltage divider to have about 10 times the transistor's base current, not 1/7 times the emitter current as in your tutorial, nor 1/20 times the emitter current as taught elsewhere.

The transistor I selected has a typical gain of 230 and a minimum gain of about 100 with my chosen 0.55mA emitter current. Therefore its max base current is about 5.5uA. So the minimum ratio of divider current to max base current is actually 4.82:1, not 5:1 as I said.

The divider could be 330k and 47k or 360k and 51k and the results would be the same.

You shouldn't neglect the base current because it affects the divider's voltage.
The voltage of the divider without any base current is 1.25V. With a transistor's typical base current of 2.4uA, the voltage at the base of the transistor is 1.15V. Therefore its emitter current will be 0.5V, its emitter voltage will be 0.5V and its collector voltage will be 5.0V, not 5.5V like I planned. With its max 5.5uA of base current, a low gain transistor would change the base voltage to 1.05, its emitter voltage will be 0.4v and its collector voltage will be 6V.
If you used your tutorial's formula of having a divider current equalling 1/7 of the emitter current then the divider will have a current 3 times as much as mine, and the collector voltage will change less with changes in the transistor's gain. The tutorial's author wrongly calls the divider's current the "base current".
If you used your tutorial's formula of having a divider current equalling 1/7 of the emitter current then with a power transistor having low gain at high current, the divider current won't be enough but my formula still works.   

(2)  autir ask: The Ib/Ic ratio in your circuit is proof of the hfe value of the transistor you used. Why, then, does it work the same with different hfe's? Does this mean that the Ib/Ic ratio can be decided by me without any consenquences regarding the circuit's function?
Audioguru's answer: I selected a transistor whose hFE is never less than half its typical, nor more than double. The circuit would need more current in the R1/R2 divider if a transistor with a lower minimum hFE was used.
walid: hfe in Audioguru's design = IC/IB = 0.55mA/27uA = 20.37 and this is less than half its typical.

I don't understand what you are asking because my divider's current is a multiple of the base current, not a fixed fraction of the emitter current.
I don't see where Walid said that. My transistor has IC/IB = 230.

(3) autir ask: The gain factor is calculated by the R3/R4 ratio, if I'm not mistaken?
Audioguru's answer: it is long and far from a direct answer

I said the correct and accurate formula for the gain and everything that should be considered when calculating gain. Autir's simple formula would calculate a gain of infinity if the emitter resistor was zero or bypassed which is plainly wrong. Frequently, a correct answer to a question in electronics is "long and far from being direct".

(4) In his reply to Alun, Audioguru surprised that ther is no RE and no R2, why i don't know.
I tell him that this circuit called a self biased and is found as the first circuit in a small sig amp in any electronic book and this circuit is used in realty, i see it many times in areal designs.

Alun's circuit isn't self-biased. A self-biased circuit has its base resistor connected to its collector, not its supply, for negative feedback which controls its operating point when a transistor having a different gain is used or the temperature changes.
Link to post
Share on other sites

Hi all my friends in this community especially Audioguru

I'm sorry for the many mistakes in my last reply.
Audioguru is true I'm false
(1) hfe is 230 and not 20.3 as i do, because it is 0.55m/2.4u and not 0.55m/027u
(2) I mean a fixed bias and not a self-bias sorry I'm confused
(3) All other thing I told is not correct if Audioguru commented it.
I'll back from the beginning and ask these questions, it may seem boring but I'm insistent.
AUDIOGURU: " A transistor's collector is a high impedance current sink, so the output impedance is its collector load resistor. "
Walid: I need more explanation about this.

AUDIOGURU: "5.5V was chosen as its idle collector voltage so it can swing equally up and down."
Walid: Why 5.5V why not 1/2 Vcc thats 5.

AUDIOGURU: "Ohm's Law calculated the resistors."
Walid: I want now try to calculate them:
IC = (10 - 5.5)/10k = 0.45 mA not 0.55mA confused at the start

AUDIOGURU:"A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2."
WALID: this is the main problem for me to understand

AUDIOGURU: Each part number and each sample of transistor has a different current gain.
WALID: Its OK
AUDIOGURU: Transistors have very reduced gain at high currents.
WALID: u mean if IC =100mA beta reduced, WHY?
AUDIOGURU: Therefore I calculate the current in the voltage divider to have about 10 times the transistor's base current..
WALID: how you calculate it and is it a rule to be this ratio? Please teach us.

AUDIOGURU: The transistor I selected has a typical gain of 230 and a minimum gain of about 100 with my chosen 0.55mA emitter current. Therefore its max base current is about 5.5uA.
WALID: OK
AUDIOGURU: So the minimum ratio of divider current to max base current is actually 4.82:1, not 5:1 as I said.
Walid: 4.82:1 ratio between what?
AUDIOGURU: The divider could be 330k and 47k or 360k and 51k and the results would be the same.
Walid: What the rule?

Finally, dear AUDIOGURU to save your efforts and to answer all question now and in the future about this subject tell us the story step by step like this:

I want Vcc = x and I choose a transistor x (you actually said this) IC = x and....
We have these rules.... Then calculate it to make Walid understand
Thank you
Yours
Walid

Link to post
Share on other sites

Hang in there walid, there is a good reason why the current in the potential divider needs to be 10 times Ib, it's all to do with the emitter follower and how the transistor's gain affects the base impedance.

So let's look at the emitter follower.

When Vb is applied to the base the transistor turns on causing a current to flow through Re.

This causes a potential difference proportional to the current to form across Re until Ve is 0.66V lower than the base because it can't rise any higher as the base is a diode. This is known negative feedback, a signal enters the base, the emitter voltage rises in turn cancelling out the change that would normally occur in the other common emitter amplifier without Re I posted previously.

Now Ib is always 1/100 of Ic as the transistor has an amplification factor of 100 and as the collector current is dependant on Re the impedance at the base is 100 times larger than the impedance of Re.

In the previous examples I did the calculations for you, I still have, you have the answers in the simulation below, but I didn't show you how I got those figures. Try doing you own calculating to see for your self, experiment with different gains and values of Re.

By no you should beable to see that the base impedance is directly proportional to the gain of the transistor. For this reason the current in the potential divider needs to be significanty higher than the base current as the latter can be highly variable and affect the voltage in the potential divider. Don't forget we want to bais the transistor so it's gain has as smaller affect on the circuits behaviour as possible.

In my next instalment I'll add Re to the previous common emitter amp and then finally the resistor below the base and I'll explain everything as I go along and I'll reninforce the point I've made above.

post-0-1427914247355_thumb.gif

Link to post
Share on other sites

(1) hfe is 230 and not 20.3 as i do, because it is 0.55m/2.4u and not 0.55m/027u

Yes.

(2) I mean a fixed bias and not a self-bias sorry I'm confused

Yes.

AUDIOGURU: " A transistor's collector is a high impedance current sink, so the output impedance is its collector load resistor. "
Walid: I need more explanation about this.

Attached is a graph comparing the impedance of a 2.5k resistor with the collector of a BC547 transistor operating over the same voltage range. The collector has an impedance that is 23 times the resistor.

AUDIOGURU: "5.5V was chosen as its idle collector voltage so it can swing equally up and down."
Walid: Why 5.5V why not 1/2 Vcc thats 5.

You want max voltage swing at the collector. When the transistor is saturated, its current is (10V/(1k+10k)= 0.91mA. Therefore its emitter voltage is 0.91mA x 1k= 0.91V. With an idle collector voltage of 5V then the collector can swing in a negative direction only 5.0V-0.91V= 4.09V. With the idle collector voltage at 5.5V then the collector can swing 5.5V-0.91V= 4.59V which is more.

AUDIOGURU: "Ohm's Law calculated the resistors."
Walid: I want now try to calculate them:
IC = (10 - 5.5)/10k = 0.45 mA not 0.55mA confused at the start

Sorry, I made a mistake. :-[

AUDIOGURU:"A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2."
WALID: this is the main problem for me to understand

A 330k and 47k divider produces 1.247V without a load. A 360k and 51k divider produces 1.241V, about the same.
I was talking about having a ratio of 10:1 of the divider's current to the base current. If the gain of the transistor is low, its base current is doubled and the ratio becomes 5:1. If the gain of the transistor is high, the base current is halved and the ratio becomes 20:1. In my previous post I described how the idle collector voltage would change by only 1V.

AUDIOGURU: Transistors have very reduced gain at high currents.
WALID: u mean if IC =100mA beta reduced, WHY?

I don't care about why, but a 2N3904 transistor is guaranteed to have more than 0.3 times the current gain at 100mA that it has at 10mA. A 2N3055 power transistor is guaranteed to have a gain of at least 20 at 4A but a gain of only 5.0 at 10A. It is rated at 15A so guess how low its minimum gain would be.

AUDIOGURU: Therefore I calculate the current in the voltage divider to have about 10 times the transistor's base current.
WALID: how you calculate it and is it a rule to be this ratio? Please teach us.

That's what I do and it always works. The guy in your tutorial uses a divider ratio that is a division of the emitter current and it fails at high current/low gain.

AUDIOGURU: So the minimum ratio of divider current to max base current is actually 4.82:1, not 5:1 as I said.
Walid: 4.82:1 ratio between what?

The divider's current is 10V/(330k+47k)= 26.5uA. With its minimum gain of 100, the base current with a 0.55mA collector current is 0.55mA/100= 5.5uA. The ratio is 26.5uA/5.5uA= 4.82. The ratio is 11 when the transistor has a typical gain of 230.

AUDIOGURU: The divider could be 330k and 47k or 360k and 51k and the results would be the same.
Walid: What the rule?

I showed they are the same earlier.

Finally, .......calculate it to make Walid understand

I showed and explained how it is designed. ;D

post-1706-14279142473742_thumb.png

Link to post
Share on other sites

Dear Audioguru
I take 4 peices of paper, write every hard aspect u tell me in a descrete papare. then I read all this document from the beginning, when reach a hard point i look to the related paper, finally i understand many points (but not all) and i descover that u are a genius, yes you are really genius and we all must profit from your experience.
At least for me it take at average an hour to understand one paragraph of your answers. You talk from a real world and this strikes the info (from theoritical textbooks) in my brain, this is why i can't understand you at once!!
Each time I read it i disvover somthing new deserve to ask about, but before this i want to redisign your V-divide Amp and ask you to correct me:
Any design needs three points:
1) assumptions
2) Rules
3) calculations and corrections

1) Assumptions:
here you must choose the components and the intial conditions of your circuit as follows:
Audioguru choose the famous 2N3904 NPN transistor whose typical hfe=230 and minimum hfe=100
Also he choose VCC=10v and IC to be 0.55mA and Vc = 5.5VDC fom max signal swing.

2) RULES:
a) Ohm's law; V=IR
b) VBE = 0.65 V approx.
c) hfe = IC/IB

3904.pdf

Link to post
Share on other sites

Walid,
Please do not ignore my posts, audioguru's are informative but please read mine as well because they explain everything from the ground upwards. Hopefully they will answer your questions and help you figure things out for yourself.

Link to post
Share on other sites

i descover that u are a genius, yes you are really genius and we all must profit from your experience.

I have experience but I am not a genius.

1) Assumptions:
here you must choose the components and the intial conditions of your circuit as follows:
Audioguru choose the famous 2N3904 NPN transistor whose typical hfe=230 and minimum hfe=100
Also he choose VCC=10v and IC to be 0.55mA and Vc = 5.5VDC fom max signal swing.

The hFE (DC) is 100, 230 and about 400. The hfe (AC) is less and is shown on Fairchild's datasheet.
The 1st thing was to choose a voltage gain of 10.

3) Calculations:
a) If vcc =10v & Vc=5.5v and IC = 0.55mA then RC = (VCC-VC)/IC = 8.18K round it to 8.2Kohm.

I made a small mistake with it so leave the collector resistor as 10k.

b) For our 3904 transistor, the max IB is when hfe is minimum (=100), so, IB(max)= 0.55mA/100 = 5.5uA
c) IB(max) = 5.5uA, so Id = 5 * 5.5u = 27.5 uA take it 27, 28 or as it is.

Correct.

d) from © ==> R1 + R2 = 10V/27uA = about 370 Kohm. Now you must know what Audioguru's VB, or exactly what the rule of choosing VB, like someone say VB = 1/3 or 1/4 VCC. Audioguru takes VB=1.2V, that is 12% of VCC. It seem that he reach this in a reverse approach, starting IE=0.55m RE=1K then VE=0.55V ==> VB=VE+0.65=1.2V.

I didn't use a reverse approach. The gain of 10 and requirement for fairly low current demanded that R3= 10k and R4= 1k. Since idle current through R4 would raise the emitter voltage, I knew that the operating point must be with the collector voltage about 5.5V for max output swing. HFE and Ohm's Law calculated the remaining parts.

post-1706-14279142474216_thumb.png

Link to post
Share on other sites

Hi my best friend Alun,
I'm never ignore your replies, I read it carefully every time and I'm waiting for your next instalment that you will add Re in it to the previous common emitter amp and then finally the resistor below the base and you will explain everything, I wait for this tell now.
The problems are that:
(1) Your answers are direct, simple and easily understood compared to Audioguru's
(2) Your answers in most are similar to what I already read in many textbooks. It is true that there are some of helpful additives.
(3) It seems to me that the more hot discussion is that with Audioguru, and you instead of helping me to stand together face to face with Audioguru, you a rise subjects (only to read) not pour into the point of the hot subject.
Please don't admonish at (or to or from) me. I looked at this as there are many points in Audioguru's replies were hard to understand and I expect from you and others to help me to over this, then if any of you have an addition welcome, but not to ignore yours. I respect you and others from all my heart. I feel you are my electronic family and if I meet any of you someday in a road, I'll be very pleasure and I'll help him as I could.
Yours, Walid

Link to post
Share on other sites

Hi AUDIOGURU

Audioguru:"The 1st thing was to choose a voltage gain of 10"
Walid: Voltage gain Av = Vo/Vin, how this affect my calculations of resistors, tell me please.

Audioguru:"so leave the collector resistor as 10k."
Walid:" Why you insistent to take it 10K, 8.2K is a close value to not to recalculate it from the beginning!"

Audioguru:"I didn't use a reverse approach. The gain of 10 and requirement for fairly low current demanded that R3= 10k and R4= 1k. Since idle current through R4 would raise the emitter voltage, I knew that the operating point must be with the collector voltage about 5.5V for max output swing. HFE and Ohm's Law calculated the remaining parts."
Walid:" Please Audioguru I don't understand this. You can calculate it as I do, save your time. The best way to make someone easily understand is to show him in form of calculated example, so why they do this in any technical book. Numbers solve the problem, thank you."

please Audioguru I want not to repeat my questions back and forth, with numbers and examples you make me understand fast and saving your time to others who need your experience.
yours, walid.

Link to post
Share on other sites
I wonder why Autir doesn't learn this basic stuff in school. ???

Because the electronics I have learnt is that of the computer scientist, that is very few and almost all of it digital.
Plus my theory books were meant for physicists and electrical engineers. They are too complicated.
Regarding electronics I am a hobbyist, not a professional. ;)

@walid:
Thank you for your reply. I have visited this link and found it to be horrible - the guy just throws in the Rb values and implies that Vce=0! No, thanks
Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...

×
  • Create New...