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Modulating Ic on a BJT?

juansg

May 4, 2013
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Hi,

I am working in a project with a circuit that a college gave me, but he is not with me anymore, so I am trying to figure this out with no luck.

The idea is to be able to modulate the current in Ic, using the voltage source (ultimately, the Vs will actually be the pwm output of a microcontroller). I am simulating this circuit, but it's not working as I was told it should work. Anybody has any insight on how to fix it?

thanks
 

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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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In order to assist you in fixing it, we need to know what's going wrong.

Oh, and also what you think it should do when it's working correctly.

However, D1 looks to be exactly wrong as far as I can tell.
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
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Your Photo-Voltaic model delivers 100A??? In your dreams! :rolleyes:

Chris
 

duke37

Jan 9, 2011
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It goes through 100 ohms, that needs 10kV.!

I presume the current is 100mA.

Where is the load?
 

juansg

May 4, 2013
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thanks for the replies. The current source + resistors on the right side was just a poor attempt to model the photovoltaic module. It's a 285W module, with the I-V curve that I've attached. The idea is to be able to use a PWM ramp at the Op Amp's Vin and being able to regulate a current flow from 0A to about 8A.

The original circuit was actually working, and I was given a hand written version of it. I just got access to the full circuit (see it attached), but I don't get how's that possible. Please, note that the wire connected to the emisor and going upwards in the diagram, just goes to a current sensor and then to ground. Also, somebody made a mistake drawing the 500uOhm resistor, it's actually a 0.01Ohm resistor in the circuit

The main problem I see is that using such a tiny load resistor, the voltage and current at the base of the BJT go crazy high. Any comments are more than welcomed.
 

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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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The simple explanation is that the PWM input is effectively integrated by the resistors R3, R4, and the capacitor C2.

The op amp turns on Q1 sufficiently so that the voltage across R2 is equal to that across C2.

If the voltage across C2 is 1V, then the collector current of Q1 is 100 amps (assuming that the current source can provide this current).

If you have a real current source in the circuit you have shown, then the first thing that would happen is that the 1N4148 would vaporize. The next thing is that the output voltage would rise to a very high value.
 

CDRIVE

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While it will effectively regulate Watt/Hrs (average wattage/time) PWM isn't a voltage or current regulator. Excluding transition points, at any point in its cycle it's either on or off. It isn't a sawtooth waveform either.

Chris
 
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