Voltage Divider Verification

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
I just want to thank everyone for the support! You've all helped me out a LOT and I hope that in the future I can achieve my goal with transistors! That is, look at any transistors specs and be able to saturate it and make a class-A amp out of it.
 

Jony130

Joined Feb 17, 2009
5,487
You've said that with the 2N3904, the base current should be 1/10th that of the collector current in other posts as well. But, I thought that it was for saturation. I recall reading in your other posts that you needed this to saturate the transistor, is it supposed to help it be a Class-A amp?
Well, you misunderstand Audioguru.
Its base current is about 1/10th the voltage divider's current so the different hFE of each transistor has a small effect on the divider's voltage.
And this mean, that current that is flow through voltage divider should be at least ten times larger the the base current.


Also, you said "The graph" and I would like to know which one you're referring to. The current gain over collector current one?


What should Vce be then if It shouldn't be 1V What should Vce be then if It shouldn't be 1V
Typical Vce(sat) is equal 0.2V so 1V is very close to saturation.
When you set the bias point at Vce=1V then for positive swing you get form Vce=1V to Vcc. And for negative swing of a output voltage, you get from Vce=1V to 0V (Vce(sat)).
And if you set bias point at Vcc/2 then positive swing is equal negative swing.
Positive swing form Vce=4.5V to 9V and negative swing from Vce=4.5V to 0V

Also, you mentioned that hFE isn't important in saturation, where I think that it could be. Since transistors act like a switch in saturation, couldn't you use hFE to calculate how much current gain you get for your transistor switch?
No, because in saturation Ib=Hfe*Ic don't hold any more.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
Well, you misunderstand Audioguru.
I thought I did. ;)

I thought that was the graph as well. ;)

Typical Vce(sat) is equal 0.2V so 1V is very close to saturation.
When you set the bias point at Vce=1V then for positive swing you get form Vce=1V to Vcc. And for negative swing of a output voltage, you get from Vce=1V to 0V (Vce(sat)).
And if you set bias point at Vcc/2 then positive swing is equal negative swing.
Positive swing form Vce=4.5V to 9V and negative swing from Vce=4.5V to 0V
Interesting, you cleared that up nicely. So actually the "On Characteristics" represent saturation of the transistor? According to the specs, how would I find a good value for Vce if I want it as a Class-A amp? In addition, what would be a good typical value, aside from that of the specs Vce?

No, because in saturation Ib=Hfe*Ic don't hold any more.
Ahh...ok. Thanks for clearing that up too.

When you say swing, I'm not quite sure what you mean. I understand that Vc=Vcc/2 so in this case it would be 4.5V, which gives it a nice "swing" (as you might say) that can vary the voltage. Supposing you didn't have this, what would happen?
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
When you say that Vce = 0.5 x Vcc, is that typically what one should do or is that in the specs? I'm assuming that it's what one would usually do?
 

Audioguru

Joined Dec 20, 2007
11,248
The spec's for a 2N3904 transistor show the hFE when the Vce is 1V and 5V so the transistor is not saturated.
They show the max saturation voltage drop when there is plenty of base current that is 1/10th the collector current.

You want the Vce of a class-A amplifier transistor to be half the supply voltage so that the collector can swing equally in a positive direction and in a negative direction then its output is max with low distortion. You must allow for the DC voltage across the emitter resistor that effectively reduces the output swing.

hFE is beta which is current gain, not voltage gain.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
The spec's for a 2N3904 transistor show the hFE when the Vce is 1V and 5V so the transistor is not saturated.
They show the max saturation voltage drop when there is plenty of base current that is 1/10th the collector current.

You want the Vce of a class-A amplifier transistor to be half the supply voltage so that the collector can swing equally in a positive direction and in a negative direction then its output is max with low distortion. You must allow for the DC voltage across the emitter resistor that effectively reduces the output swing.

hFE is beta which is current gain, not voltage gain.
Ok, I'll keep that about Vce in mind when I do more designs.

So with the voltage divider, The Electrician said that I don't need to calculate the current output of the voltage divider. How am I supposed to know if the base current is 1/10th of the collector current? I suppose it's just like power supplies, where a power supply can supply 1A, but the load only needs 500mA (example). Being the case, the voltage divider can supply so much current but the base will take what it needs. Am I right?
 

Jony130

Joined Feb 17, 2009
5,487
I recommend you read again the post written by The Electrician.
But this time with more care.
current that is flow through voltage divider should be at least ten times larger the the base current
 

hobbyist

Joined Aug 10, 2008
892
electronerd

also look at my post #24, the bottom picture shows a base current of around 1.3uA.

I calculated the divider current to be around 45uA. not even knowing what the base current would be, just by making R2 10 times bigger than R4. That was done for starters, if the bias voltage was way out then I would adjust, even right now the bias at the collector is around 3V. not 1/2VCC yet it works within a small enough signal voltage input, it could be adjusted to 1/2VCC by readjusting the divider res.


I didn't even know what base current I was working with, this base current ended up being this by choosing a base ground resistor (R2) in the voltage divider to be 10 times RE (R4).

A rule of thumb is make the base to ground resistor around 5 to 20 times the emitter resistor, to get the proper bias on the voltage divider.

If you get your collector resistor and emitter resistor established then you can experimentally choose a value for the base to ground resistor, to get the proper voltage at the collector that your aiming for, if the base to ground resistor is to high then the base current will begin to load the voltage divider thereby lowering the base voltage and ultimately lowering the voltage at the cocllector.

A good way to tell if you have a proper size value base resistor is to build the complete amp (or draw in sim.) then test measure the VC. voltage at the collector with respect to ground, and check the voltage VB., base voltage with respect to ground, and if the VC is too low, and VB is very low, remove the transistor, and check VB, if VB goes back up to your calculated value, then you definately have a base current loading, so if RE is to remain the same, then you need to lower base resistor to ground and recalculate the base resistor to supply to complete the voltage divider bias, THEN RETEST, and do this until the transistor base current has very LITTLE effect on the voltage divider VB.

It usually will have some effect but with the minimum effect then your VE, and VC, should come very close to your calculated values, rhen you will have your transistor in it's linear region.

With your base voltage established, then your signal input PK. should not go above that VB value, or the transistor, will lose it's bias when the signal goes neg.

The input signal adds too and subtracts from the bias voltage, that's why the bias voltage is used, so the input voltage can fluctuate this DC value (bias) in a AC (signal) shape and value of voltage waveform.

Just my way of trying toexplain it, not the bbest but hope it helps some.

And YES I KNOW, this is NOT a proffessional way to look at this, but if it gets him in the ballpark than someone else can give him the technical side of it to get him into the game...
 
Last edited:

hobbyist

Joined Aug 10, 2008
892
electronerd

See if this helps,

Here are 6 transistors picked out at random on multisim simulator.

I designed it to run around 1/2VCC with a RC of 1K ohms.
I chose a Av. of 10.
calculated the IC and VE, VB, chose a base to ground resistor(R3) 10 x RE (R2). Then calculated the current flow through R3 and solved for R4.

I kept the same bias resistors.

I used a 10mV.pk input signal so as to use this as a small signal amp.

here are the results, notice how the VC (collector voltage) changes a little bit, BUT with a small enough input signal it keeps each transistor running in it's linear region.

2n2222A.jpg


2n3390.jpg


2n3904.jpg


2n4123.jpg


2n5088.jpg


2n6715.jpg

Hope this explains better of what were trying to convey to you about biasing the collector at close to 1/2 of VCC.
 
So with the voltage divider, The Electrician said that I don't need to calculate the current output of the voltage divider. How am I supposed to know if the base current is 1/10th of the collector current? I suppose it's just like power supplies, where a power supply can supply 1A, but the load only needs 500mA (example). Being the case, the voltage divider can supply so much current but the base will take what it needs. Am I right?
I think that there is some mixing up of saturated with non-saturated designs in this thread.

For a non-saturated design, the base current isn't going to be just 1/10 of the collector current.

The base current is related to the collector current by the β of the transistor.

First determine the collector current, Ic, you need based on the impedance of the load you need to drive. Then the base current is Ib = Ic/β; use the minimum β at your operating point. Then select the bias divider resistors so that the current in the divider is 10 times Ib.

How do we set the current in the divider?

If Vcc is known, the the current in the divider is Vcc/(R1+R2), where R1 and R2 are the divider resistors. That is, the total resistance of the divider is R1+R2, so (R1+R2) = Vcc/(10*Ib). Then you have to select R1 and R2 individually such that their ratio (while keeping their sum constant) gives the desired output voltage, ignoring the effect of Ib; in other words, assume Ib = 0 at this point. The error due to that assumption will be small enough to ignore.

It's just a little algebra.
 

Thread Starter

ELECTRONERD

Joined May 26, 2009
1,147
I think that there is some mixing up of saturated with non-saturated designs in this thread.

For a non-saturated design, the base current isn't going to be just 1/10 of the collector current.

The base current is related to the collector current by the β of the transistor.

First determine the collector current, Ic, you need based on the impedance of the load you need to drive. Then the base current is Ib = Ic/β; use the minimum β at your operating point. Then select the bias divider resistors so that the current in the divider is 10 times Ib.

How do we set the current in the divider?

If Vcc is known, the the current in the divider is Vcc/(R1+R2), where R1 and R2 are the divider resistors. That is, the total resistance of the divider is R1+R2, so (R1+R2) = Vcc/(10*Ib). Then you have to select R1 and R2 individually such that their ratio (while keeping their sum constant) gives the desired output voltage, ignoring the effect of Ib; in other words, assume Ib = 0 at this point. The error due to that assumption will be small enough to ignore.

It's just a little algebra.
Thanks, The Electrician.

It's just a little algebra.
I guess I'm not doing so bad for a kid my age...I'm pretty young.
 
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