Hi all,
I am trying to learn small signals amplifying theory and I am having trouble measuring simple characteritics curves for a NPN BJT BC504 transistor.
My setup is as simple as it can be: a single transistor and a standard dual power supply.
Firstly, I give a set current to the base (making sure it is in "ON state) and then I measure the collector current at different Vce voltages.
The problem I am having is that Ibe gets out of wack as soon as I apply Vce... as if Vce is directly linked to Ibe. This dependency problem gets fixed as soon as I place a resistor is series between the supply and the collector and the changes in Vce doesn't alter the Ibe current.
My question is why is this happening but more importantly how are characteristics curves measured if this is happening?
Thanks!
Charles
I am trying to learn small signals amplifying theory and I am having trouble measuring simple characteritics curves for a NPN BJT BC504 transistor.
My setup is as simple as it can be: a single transistor and a standard dual power supply.
Firstly, I give a set current to the base (making sure it is in "ON state) and then I measure the collector current at different Vce voltages.
The problem I am having is that Ibe gets out of wack as soon as I apply Vce... as if Vce is directly linked to Ibe. This dependency problem gets fixed as soon as I place a resistor is series between the supply and the collector and the changes in Vce doesn't alter the Ibe current.
My question is why is this happening but more importantly how are characteristics curves measured if this is happening?
Thanks!
Charles
Your 'IBE' measuring device is picking up Ic plus Ib (=Ie) actually.
Put an accurate 1k00 /1% resistor is series with your VBE supply.
Measure both the VBE supply and the real Vbe at the base of the transistor and divide by 1000, that will give you the input current Ib.
Then sweep VCE with different settings of Ib for measuring the corresponding Ic.
That's the output (Ic is a function of Vce for a set of Ib's) characteristic your aiming for.
It's a very flammable debate if the BJT (Bipolar Junction Transistor) is current driven (me) or voltage driven (me also).
Ic = Icob * { e^(Vbe/Vt - 1) where Vt = (k * T / q) {this being approx 26mV @ 300K}, indicates voltage driven.
However...
Ic = β * Ib indicates current driven.
Both are valid.
I split my brain and load them in the halved walnuts separately for the desired superposition view.
Put an accurate 1k00 /1% resistor is series with your VBE supply.
Measure both the VBE supply and the real Vbe at the base of the transistor and divide by 1000, that will give you the input current Ib.
Then sweep VCE with different settings of Ib for measuring the corresponding Ic.
That's the output (Ic is a function of Vce for a set of Ib's) characteristic your aiming for.
It's a very flammable debate if the BJT (Bipolar Junction Transistor) is current driven (me) or voltage driven (me also).
Ic = Icob * { e^(Vbe/Vt - 1) where Vt = (k * T / q) {this being approx 26mV @ 300K}, indicates voltage driven.
However...
Ic = β * Ib indicates current driven.
Both are valid.
I split my brain and load them in the halved walnuts separately for the desired superposition view.
Totally right, I wasn't reading Ibe correctly.
Is the resistor really that necessary? Couldn't I just control the Ibe current with the powe supply?
The Ib/Vbe slope is very steep. You need a very steady hand to keep it in control.
Start with Vbe = 0.65V and raise it to 0.66V. The increase of Ib or Ic will be 1.5x.
From 0.65 to 0.67 it is 2.2x.
From 0.65 to 0.7 it is 7.4x.
You've now 0.05V in control on your dual power supply...
It is a very steep curve.
Start with Vbe = 0.65V and raise it to 0.66V. The increase of Ib or Ic will be 1.5x.
From 0.65 to 0.67 it is 2.2x.
From 0.65 to 0.7 it is 7.4x.
You've now 0.05V in control on your dual power supply...
It is a very steep curve.
How long would you like the transistor to survive?
There's usually an absolute maximum rating for Ib in the data sheet for the transistor. Most power supplies don't have that fine control over their output current so you risk blowing the transistor during your measurement. A better tool would be a source meter such as the Keithley 236, but in lieu of that a resistor in series with the base does the job just fine.
I'd limit the collector current as well.
Tom
This makes total sense... What I have come to realise is that the diode is acting more as a short circuit -current wise- above ~0.6V (in terms of voltage, it will always "eat up" 0.7V).
In a short circuit, current is immense and uncontrolable for any voltage set by the power supply so we conviently use a resistor to "limit the current", which will control the curent to voltage relationship in a linear fashion. This finally makes me understand the deeper meaning of "current limiting" by a resistor! Thank you Citizen124032
This leads me to another question about class A signal amplifier designs (see schematic below). I have learned that R1 and R2 are used to forward bias the base voltage to turn the base to emitter diode "ON". As soon as the emitter diode turns "ON" shouldn't the voltage divider ratio (R1 and R2) be affected? The new current path to ground, from base to emiter and through resistor Re, should alter the ratio to give a lower voltage bias at the base. Is this taken to account for when choosing R1 and R2 by calcuating a higher voltage offset? Couldn't you just omit R2 and make the bias voltage dependent on the ratio set by R1 and RE?
Well yes, and this is why assuming a transistor Beta of about 100, you set R1 roughly equal to 20 times RL (VR1 ~=2x VRL) so that IR1 is about 10% of IRL. That is the base current (1%) has minimal effect on the divider voltage. I suggest you download LTC spice and simulate this circuit and others. If you chose the resistor values poorly you will see poor results. Note that RE is typically much smaller than RL, just enough to make Ic predictable and not sensitive to temperature. Without R2, the circuit becomes unpredictable and very dependent on the transistor Beta, which is unpredictable. Typical BJT transistor Beta specs have a 10:1 range and is very temperature dependent. An "acceptable" alternative is to connect R1 to the collector instead of Vcc, and then R2 and RE can be omitted.
I had a hard time beleiving this at first because I didn't quite think about voltage dividers fully.
I realise that Vbe is set by the voltage division of R1, R2 and Re (R2 and Re being in parallel):
With the help of my mathematician friend this has been broken down to:
View attachment 1411857
Here we can see that R2 and Re occupies the same role in the equation. In order for Re to really make a difference in Vbe, it need to be small relative to R2. So for a proper bias voltage idependent of Re, Re >> R2.
Upon testing the circuit I soon realise that this is not the full picture of what is happening at Vbe because of the 0.7V bias needed at the diode as well as the voltage buildup at the collector created by Ic. I am not so sure how the voltage at the junction of the voltage divider could be expressed or even if it is relevant in the equation. I am getting lost...
Good night.
Crashcourse Calculation:
Nodes
I1, I2 > Ib: I1, I2 ≈ √β * Ib ≡ I1, I2 ≈ Ic/√β
eg Vcc = 24Vdc, Ve^= 3Vdc, Vce˅ = 1Vdc : ∆V(Rc)^ = 24 - (3+1) = 20V
Vdc(Rc) =∆V(Rc)^ / 2 = 10V: Vc {'Voutdc') = 14Vdc
Vdc(Re) =∆V(Re)^ / 2 = 1,5V: Ve = 1.5Vdc
Vb ≈ 1.5 + 0.7 ≈ 2.2Vdc
Branches
TO92: BC547 unspecified, β ≈ [100 -- 400], median = 200, rb ≈ 100Ω
set Ic = 2mA, re = Vt/Ic = 25mV / 2mA = 12.5Ω
Rc = (24 - 14) / 2mA = 5.0kΩ -> 4k7
Re = 1.5 / 2mA = 750Ω -> 680E
I1, I2 ≈ Ic/√β ≈ 2mA / √200 =0.14mA
R2 = 2.2 / 0.14mA = 15.7kΩ -> 15k
R1 = (24 - 2.2) / 0.14mA = 155.7kΩ -> 150k
Performances
Ao dc = Rc / (re + Re) = 5.0k / (12.5 + 680) = 7.22x (17dB)
∆V(Rc)tt / ∆V(Re)tt = 20 / 3 = 6.7x (deviation is 13% {good enough} )
Zin = R1 // R2 // (rb + β * (re + Re) ) = 12.4kΩ
+ C2 (Fo = 10Hz, C = 23.4μF -> 22μ/25V/105°)
Ao ac = Rc / re = 5.0k / 12.5 = 400x (52dB)
Vin~^= Vout~^ / 400 = 20 / 400 = 50mV~ -> Vintt = +/-75mV
∆Vbe = 0.7V +/- 75mV: 0.775V -- 0.625V
Zin = R1 // R2 // (rb + β * re) = 2.2kΩ
If a higher impedance is required, then bootstrapping is necessary.
Kirchhoff is King!
Nodes
I1, I2 > Ib: I1, I2 ≈ √β * Ib ≡ I1, I2 ≈ Ic/√β
eg Vcc = 24Vdc, Ve^= 3Vdc, Vce˅ = 1Vdc : ∆V(Rc)^ = 24 - (3+1) = 20V
Vdc(Rc) =∆V(Rc)^ / 2 = 10V: Vc {'Voutdc') = 14Vdc
Vdc(Re) =∆V(Re)^ / 2 = 1,5V: Ve = 1.5Vdc
Vb ≈ 1.5 + 0.7 ≈ 2.2Vdc
Branches
TO92: BC547 unspecified, β ≈ [100 -- 400], median = 200, rb ≈ 100Ω
set Ic = 2mA, re = Vt/Ic = 25mV / 2mA = 12.5Ω
Rc = (24 - 14) / 2mA = 5.0kΩ -> 4k7
Re = 1.5 / 2mA = 750Ω -> 680E
I1, I2 ≈ Ic/√β ≈ 2mA / √200 =0.14mA
R2 = 2.2 / 0.14mA = 15.7kΩ -> 15k
R1 = (24 - 2.2) / 0.14mA = 155.7kΩ -> 150k
Performances
Ao dc = Rc / (re + Re) = 5.0k / (12.5 + 680) = 7.22x (17dB)
∆V(Rc)tt / ∆V(Re)tt = 20 / 3 = 6.7x (deviation is 13% {good enough} )
Zin = R1 // R2 // (rb + β * (re + Re) ) = 12.4kΩ
+ C2 (Fo = 10Hz, C = 23.4μF -> 22μ/25V/105°)
Ao ac = Rc / re = 5.0k / 12.5 = 400x (52dB)
Vin~^= Vout~^ / 400 = 20 / 400 = 50mV~ -> Vintt = +/-75mV
∆Vbe = 0.7V +/- 75mV: 0.775V -- 0.625V
Zin = R1 // R2 // (rb + β * re) = 2.2kΩ
If a higher impedance is required, then bootstrapping is necessary.
Kirchhoff is King!
I feel like there is alot of wisdom in all these calculations but I am getting quickly tangled in the meaning of the variables you are setting. Is there a way you could give more context to your equations? I am really starting out on learning more theorized knowledge of electronics, already know about kirchhoff stuff tho!
Charles
Attachments
A very good resource is a book titled Small Signal Audio Design. Google that phrase and you will find a copy.
There is a more-controllable way to perform some of the measurements you want to do. You need to connect your transistor in a common-base mode. For an NPN, this means that the base is grounded. If you want to measure the base current, OK, insert a current meter between the base and ground.
Apply some voltage, say 5V, to the collector. Extract a known current out of the emitter -- ideally, using a current source. But you can use a second power supply and a resistor instead. Pulling a known current out of the emitter, measure the voltage on the emitter with respect to ground. That is the Vbe of the transistor for that emitter current.
The collector current should be equal to the emitter current minus the base current.
The advantage of this approach is that you are not tasked with fiddling with millivolt changes on the base to get the desired currents flowing through your transistor. You also can get a pretty good feel for just HOW sensitive the base-emitter junction is to voltage, by changing the emitter current and seeing how that affects the emitter voltage. Not much, huh!
The old Tek curve tracer avoided some of the problems you are seeing by forcing CURRENT steps into the base and then sweeping the collector voltage, while grounding the emitter. In this mode, again, you are not tasked with forcing very small and precisely-regulated voltages on the base. Transistors are a LOT more linear w/respect to currents than voltages. The curve tracer is a bit more complicated due to that current generator. But not a big deal if you know how to make current sources....
Apply some voltage, say 5V, to the collector. Extract a known current out of the emitter -- ideally, using a current source. But you can use a second power supply and a resistor instead. Pulling a known current out of the emitter, measure the voltage on the emitter with respect to ground. That is the Vbe of the transistor for that emitter current.
The collector current should be equal to the emitter current minus the base current.
The advantage of this approach is that you are not tasked with fiddling with millivolt changes on the base to get the desired currents flowing through your transistor. You also can get a pretty good feel for just HOW sensitive the base-emitter junction is to voltage, by changing the emitter current and seeing how that affects the emitter voltage. Not much, huh!
The old Tek curve tracer avoided some of the problems you are seeing by forcing CURRENT steps into the base and then sweeping the collector voltage, while grounding the emitter. In this mode, again, you are not tasked with forcing very small and precisely-regulated voltages on the base. Transistors are a LOT more linear w/respect to currents than voltages. The curve tracer is a bit more complicated due to that current generator. But not a big deal if you know how to make current sources....
Decomposing the solidified crashcourse feels somewhat counterintuitive - the steep curve is the mastering way... or not
^is top value, ∆ is the allowable 'spread' within the given dc settings, // is 'in parallel', '--' is 'range'.
Counting the iterations up to solve such a nice & humble electronic circuit is up to one to infinte - better take the shortcut!
Take the values as posted and see how the circuit performs.
I'd say Ic is better up to 2.2mA...
^is top value, ∆ is the allowable 'spread' within the given dc settings, // is 'in parallel', '--' is 'range'.
Counting the iterations up to solve such a nice & humble electronic circuit is up to one to infinte - better take the shortcut!
Take the values as posted and see how the circuit performs.
I'd say Ic is better up to 2.2mA...