Equation for BJT voltage controlled amplifier

In summary, even though the control voltage envelope (V_b) and input signal (V_c) are derived from ideal voltage sources, the resulting voltage (V_e) is not. To model the behaviour of a transistor in an analog device, you would need to use another resistor (load) to bridge the emitter potential (Ve) to ground.
  • #1
kvtb
11
0
Hi,
as mentioned in my previous thread, I'm currently busy with a 'small' project of translating an analog device (synthesizer, based on schematics) into a digital (software) version.

I know 'nothing' about EE, I'm a Computer Science guy, so hopefully this question is not too easy :-)

This question is about how to model the behaviour of a transistor. In the analog device, a single transistor is used as a voltage controlled amplifier.
- The control voltage envelope [tex]V_b(t)[/tex] enters via the base of the transistor,
- the voltage to be shaped according to the envelope [tex]V_c(t)[/tex], enters via the collector
- the output is a voltage [tex]V_e(t)[/tex] of which was until now modeled as simply [tex]V_e(t) = V_b(t) \cdot V_c(t)[/tex]However... later I noticed that every website about transistors say that [tex]I_e = I_b + I_c[/tex], so that would imply that it is impossible to get [tex]V_e(t) = V_b(t) \cdot V_c(t)[/tex]So, my question is simply,
- given a control voltage as function of time [tex]V_b(t)[/tex]
- given an input voltage [tex]V_c(t)[/tex]

attachment.php?attachmentid=23532&stc=1&d=1265308751.png
How do I calculate [tex]V_e(t)[/tex] in terms of [tex]V_b(t), V_c(t), R1, R2, R3[/tex] and any transistor dependent constants?

Thanks in advance!
 

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  • #2
Even if Vb and Vc are derived from ideal voltage sources, you cannot just dangle Ve like that. (Well, you "can", but I don't think that the transistor would do anything. I don't know what would happen, actually.) For instance, perhaps there is another resistor (i.e. a load) connected from Ve to ground?

BTW, what kind of transistor is it? The symbol that you have used suggests npn, perhaps 2222.
 
  • #3
kvtb said:
Hi,
as mentioned in my previous thread, I'm currently busy with a 'small' project of translating an analog device (synthesizer, based on schematics) into a digital (software) version.

I know 'nothing' about EE, I'm a Computer Science guy, so hopefully this question is not too easy :-)

This question is about how to model the behaviour of a transistor. In the analog device, a single transistor is used as a voltage controlled amplifier.
- The control voltage envelope [tex]V_b(t)[/tex] enters via the base of the transistor,
- the voltage to be shaped according to the envelope [tex]V_c(t)[/tex], enters via the collector
- the output is a voltage [tex]V_e(t)[/tex] of which was until now modeled as simply [tex]V_e(t) = V_b(t) \cdot V_c(t)[/tex]

However... later I noticed that every website about transistors say that [tex]I_e = I_b + I_c[/tex], so that would imply that it is impossible to get [tex]V_e(t) = V_b(t) \cdot V_c(t)[/tex]


So, my question is simply,
- given a control voltage as function of time [tex]V_b(t)[/tex]
- given an input voltage [tex]V_c(t)[/tex]

attachment.php?attachmentid=23532&stc=1&d=1265308751.png



How do I calculate [tex]V_e(t)[/tex] in terms of [tex]V_b(t), V_c(t), R1, R2, R3[/tex] and any transistor dependent constants?

Thanks in advance!

The 1st equation highlighted in bold is dimensionally inconsistent. You cannot multiply 2 voltages and get a voltage. Are you trying to perform an analog multiplication of 2 signals? There are circuits that do that. I would search "analog multiplier" or "Gilbert multiplier".

The 2nd bold highlight can be answered as follows. By "Ve", I presume you mean the emitter potential with respect to ground. Look up the "hybrid pi equivalent" model of the bjt operating in small signal mode. The bjt parameters are "r_pi, hfe, & gm or re". Also, r_pi = hfe/gm, where gm = Ic/Vt, Vt = kT/q. You can then compute the ac gain, i.e., Ve will be a function of Vb, the resistor values, and bjt properties. Selecting the right values for the R's minimizes the bjt parameters' influence on circuit behavior.

Did I help? A good reference text will answer your questions in detail. Horowitz is a good text.

Claude
 
  • #4
turin, cabraham, thanks for your replies.

I think my question was perhaps not clear or incomplete:

turin: The schematic is just a small part of a larger schematic to illustrate my question, that's why there's no ground.
If you interested, please take a look at transistor Q65 in cell J8 of this schematic:
http://www.kolumbus.fi/janne.husu/specs/909vb.gif
The transistor in question is a 2SC2603-F (no longer in production - but the datasheets are available)

cabraham: I do not try to do (analog) multiplication of two voltages. In the software model I'm building, I incorrectly assumed that the behaviour of a transistor VCA could be approached by multiplying the envelope signal (at base) with the input signal (at collector).

I will look into the hybrid pi equivalent model to see if I can use it to model the behaviour of a single-transistor VCA.

Thanks again for your help.
 
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  • #5
Looks like Q65 is just gating through noise from the diodes D91 and D92 with a decay envelope. Hence the rim-shot. I may be wrong however.
 
  • #6
kvtb said:
... I do not try to do (analog) multiplication of two voltages. In the software model I'm building, I incorrectly assumed that the behaviour of a transistor VCA could be approached by multiplying the envelope signal (at base) with the input signal (at collector).
What's the difference? What's VCA? Anyway, that is certainly not the behavior of Ve given Vb and Vc.

BTW, the model of the transistor can be very dependent on 1) the type that you are using, and 2) the surrounding circuit connections to the three electrodes. For example, hybrid-π model may or may not be applicable. (Hybrid-π model is a small-signal model, correct?) I haven't looked at your schematic yet. I'll let you know if I can figure out what the transistor is supposed to do.

dlgoff said:
Looks like Q65 is just gating through noise from the diodes D91 and D92 with a decay envelope. Hence the rim-shot.
What's a rim-shot? I've never heard of such a circuit. Is this some sort of musical circuit? (wikipedia says that a rimshot is a drumming technique.)

EDIT:
Upon cursory inspection, here's how I visualize your transistor circuit:

Firstly, it can be reduced no further than:
Q65, D91, D92, R408, R415, R417, R418, R423

It has three input voltages (ultimately for collector control) and one input current (base), and it produces one output current (emitter).

The three input voltages come from three op-amps? and are presented at the three resistors: R408, R415, and R417. The three currents feed into the voltage divider circuit of R418 and R423, but the maximum input voltage to this divider is clamped by the diode circuit of D91 and D92. (Is that the right term - "clamped"?). The output of the voltage divider provides a Vc at your transistor.

Since the emitter is connected directly to a negative op-amp? input, the positive op-amp? input is grounded, and the op-amp? has negative feedback, the emitter voltage is held at 0 V.

So, the three input voltages basically control Vce (subject to the diode clamp constraint), without negative emitter feedback. The output of the transistor is a current output that flows through the op-amp? feedback RC circuit.

With the given Vce, the output current is basically determined by the input current at the base. This current comes from Q64 through two resistors: R401 and R403. I think that these resistors are mainly for filtering and feedback to Q64.

So, this looks like a base-to-emitter current amplifier circuit that is somehow controlled by three voltages (which somehow work together to determine the current amplification by controlling Vce). I don't know if the diode circuit is just for protection, or if it has a functional purpose.
 
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  • #7
turin said:
What's a rim-shot? I've never heard of such a circuit. Is this some sort of musical circuit? (wikipedia says that a rimshot is a drumming technique.)
The circuit looks to be an electronic drum (percussion) set. It looked like the diodes were involved in making the noise (used to simulate snare durms) but I'm not sure since there is also a "noise section" in the circuit. Here is a way to generate and amplify thermal noise from a diode:
broadband-noise-generator-circuit.gif

D1 generates noise,...
http://www.simplecircuitdiagram.com/2009/12/25/broadband-noise-generator/"

The three wave forms, F1, F2 and F3 are the envelopes used to modulate the noise. Which makes the sound of a rim-shot. i.e. the stick strikes the head and rim of the snare drum at the same time.
 
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  • #8
dlgoff, turin, thanks for your replies.

let me try to explain the circuit in the link I've given, using another link;
http://mitglied.multimania.de/raf909/rimshot.htm

the three op-amps and surrounding capacitors and resistors are sine wave generators, these generate a sine wave (at three different frequencies) with different damping.

the three sines are then mixed together, and 'wave shaped': the two diodes are 'limiting' the amplitude of the mixed signal.
(In other parts of this schematic, you'll also see this construction with these two diodes. For example, to 'translate' a triangle wave into a 'sine' wave. So these diodes are not related to noise generation).

For those who have LTspice installed, and are interested in the output of the Rimshot circuit, please load and simulate the attached file. In the LTSpice file, you'll see transistor Q3, which is the transistor that does the VCA.

You may want to 'probe' the following nets:
F1, F2, F2 (to see the sine waves)
shaped (effect of the two diodes, and input to the collector of the transistor)
ENV ( the envelope applied to the base of the transistor)
 

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  • #9
So these diodes are not related to noise generation
I stand corrected.
 
  • #10
Can you describe the "trigger"? I'm most curious about the amplitude of the trigger (above ground). I imagine that it is just a quick square voltage pulse of several volts amplitude (above ground) that is produced by a key press or something. I don't want to know where it comes from, just what are its expected characteristics.
 
  • #11
Hi, the trig signal is a +5V pulse, duration is 2 msBTW, in mean while, I've tried to 'measure' a transfer function (as function of Vb and Vc) and tried to fit an equation on top of it. I'll post results later.
 
  • #12
OK. That's what I expected. I don't think I can offer any further input.
 

1. What is a BJT voltage controlled amplifier?

A BJT voltage controlled amplifier is a type of electronic circuit that uses a bipolar junction transistor (BJT) to amplify a voltage signal. It is commonly used in audio and radio frequency applications.

2. How does a BJT voltage controlled amplifier work?

A BJT voltage controlled amplifier works by using the BJT's ability to amplify current to control the voltage output. The voltage input signal is applied to the base of the BJT, and the amplified output is taken from the collector. The amount of amplification is determined by the characteristics of the BJT and the circuit design.

3. What is the equation for calculating the gain of a BJT voltage controlled amplifier?

The equation for calculating the gain of a BJT voltage controlled amplifier is Av = -gm * RL, where Av is the voltage gain, gm is the transconductance of the BJT (measured in siemens), and RL is the load resistance. The transconductance can be calculated using the BJT's current gain (β) and the bias current (IB): gm = β * IB.

4. How is the input impedance of a BJT voltage controlled amplifier determined?

The input impedance of a BJT voltage controlled amplifier is determined by the resistance of the biasing network and the input resistance of the BJT. The biasing network is typically designed to provide a high input resistance, while the input resistance of the BJT can be calculated using the BJT's transconductance (gm) and the bias current (IB): Rin = gm^-1 + (1 + β) * (1 / IB).

5. Are there any limitations or drawbacks to using a BJT voltage controlled amplifier?

One limitation of using a BJT voltage controlled amplifier is that it is sensitive to changes in temperature, which can affect the performance and accuracy of the circuit. Additionally, BJTs have a limited bandwidth and may introduce distortion to the amplified signal. Careful design and proper biasing can help mitigate these limitations.

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