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Old 21st March 2012, 12:55 PM   #1
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Default Question about Vceo

I was wondering if anyone could tell me how much margin should be given to Vceo in practical circuits. I have some boards with MPSA18s on them. My power supply runs a little high, around 43V. Vce on the input pair would be just a couple volts lower than that. Since Vceo for these devices is 45V, I'm concerned they will either eventually fry or they will have higher distortion on the brink of breakdown. Any thoughts would be appreciated.

Thanks.

Henry
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Old 21st March 2012, 02:42 PM   #2
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As a rule of the thumb, usually the operating peak voltage would be about 75% of the maximum recommended by the manufacturer. If transistor in question is in class A, the peak will be when Ic goes to the minimum, then it approach Vcc, so, a VCC of 43V for a device rated to 45 is too high.

Good luck.
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Old 21st March 2012, 03:32 PM   #3
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Quote:
Originally Posted by Osvaldo de Banfield View Post
As a rule of the thumb, usually the operating peak voltage would be about 75% of the maximum recommended by the manufacturer. If transistor in question is in class A, the peak will be when Ic goes to the minimum, then it approach Vcc, so, a VCC of 43V for a device rated to 45 is too high.

Good luck.
Thanks for the advice. Yeah, I agree. I'll put in a simple front end regulator to drop the voltage to a safe level.

The power transformers I got were too good a deal to pass up, but about four volts too high...

-Henry
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Old 21st March 2012, 03:55 PM   #4
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Originally Posted by hpasternack View Post
Thanks for the advice. Yeah, I agree. I'll put in a simple front end regulator to drop the voltage to a safe level....

-Henry
...or try a higher Vceo transistor, there must be other low noise transistor compatible with your system.
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Old 7th April 2012, 06:28 PM   #5
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Hi,
This is my first post - had to jump-in here before introducing myself properly.

The parameter that matters in this case is Vcbo. Never exceed this, but as long as there is a bias on the base then Vceo is irrelevant. The base current 'shields' the emitter junction, but there is no way to protect the base-collector junction.
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Old 8th April 2012, 03:55 PM   #6
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Originally Posted by valgamaa View Post
The parameter that matters in this case is Vcbo. Never exceed this, but as long as there is a bias on the base then Vceo is irrelevant. The base current 'shields' the emitter junction, but there is no way to protect the base-collector junction.
Interesting proposition, and quite logical too, even if it goes against some conventional wisdom.
I had to put it to the test.
Result: it does actually work, but the resistance seen by the base has to be extremely low, lower than 1 ohm in the case of the BD131 I tested.

This somewhat limits the practical usefulness of this trick in existing circuits, but it good to know nevertheless.
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Old 8th April 2012, 05:48 PM   #7
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Vce0 is the limiting voltage when zero current flows. It is not a reference to whether the base is connected or not.
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Old 8th April 2012, 07:12 PM   #8
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Originally Posted by AndrewT View Post
Vce0 is the limiting voltage when zero current flows. It is not a reference to whether the base is connected or not.
The "o" from Vceo means open. It would be possible to mimic that condition by connecting the base to a circuit that synthetizes an infinite impedance, but it looks like a self-defeating proposition.
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Old 8th April 2012, 07:16 PM   #9
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AndrewT, Vceo is measured with a base open-circuit, and is the voltage at which the base region is effectively removed by sweeping the doping from the base region, leaving a resistor. This is why the current rises sharply. If the electrons/holes can be replaced by a current flowing into/from the base then the junctions remain correctly biased and no harm is done. As Elvee correctly states, this can require a low impedance drive to the base, but is a reasonable proposition in most circumstances.
It is very common for RF amplifiers (particularly class-C) to operate with collector voltages 2-3 times higher than Vceo, but as long as there is sufficient base drive no harm is done.
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Old 8th April 2012, 07:25 PM   #10
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Originally Posted by Elvee View Post
The "o" from Vceo means open. It would be possible to mimic that condition by connecting the base to a circuit that synthetizes an infinite impedance, but it looks like a self-defeating proposition.
Agree fully - it is a figure of merit, independent of how the device might be used.
If the base of a transistor was connected to the emitter, and a potential applied between the collector and emitter, which junction will fail first? The B-E one can't as it is shorted (the low impedance in Elvee's example), but the C-B one sees the full potential.
Fortunately both breakdown mechanisms are fully reversible as long as there has been no thermal damage.
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