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help calculating current for transistor

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patbb

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Hi

Here is a very basic question for you,

I have a project and i want to make a circuit that uses the audio signal from the pc and display it.

I got an amplifier and bandpass filters etc... figured out but i need to use the output of this to drive a high voltage transistor (MJE340 NPN transistor).

My display must range between 0 and 20 mA.

So the base of the transistor is connected with the output of the bandpass filters, but how do i modulate the current for the display?
I know the more voltage at the base, the more current that is going to flow through the transistor but what are the formulas and how to calculate this?


Thank you so very much!
 
What kind of display uses a high voltage? Many LEDs in series?
The collector current of a transistor is the base current times hFE. But hFE varies a lot and also varies when the temperature varies.

So add an series emitter resistor to ground. If it is 500 ohms then if the base voltage is 10.7V then the emitter voltage will be +10V and the emitter and collector currents are 20mA.
If the base voltage is +5.7V then the collector current is 10mA. If the base voltage is a little less than 0.7V then the collector current is almost 0mA.
 
Did you read this link I gave you in your last thread? Did you get a HV power supply working?


Below is a sim of the drive method shown in the link above:

First, I don't have a model of the MJE340, so I used a similar hv transistor. The IN-9 bar graph tube is modelled as 105V fixed voltage drop plus the 400 Ohm dynamic resitance mentioned in the link.

The independent variable of the simulation is the Base drive voltage (X-axis). The dependent variables plotted are the current through the IN9 (light blue trace), the voltage at the collector (dark blue), the base current (green trace) and the power dissipated in the transistor (red trace).

Note that nothing happens until the base drive gets to ~0.6V, but from that point on, the increase in current through the IN9 increases more-or-less linearly with increasing base voltage. It takes only a fraction of 1mA to drive the base hard enough to drive ~35mA thru the IN9.
 

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