# 555 PNP Driver Problem

Discussion in 'The Projects Forum' started by dude_500, Nov 9, 2009.

1. ### dude_500 Thread Starter New Member

Oct 22, 2009
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0
I'm working on building a square wave generator from a 555 and since I need higher voltage and current (negative polarity) to drive some power triodes, I have a PNP transistor between the 555 and the triode grid.

I think I hooked it up right, however I'm fairly new to transistor design so I may have done something obviously wrong. I'm wondering if the atypical polarity that I'm running the 555 at is an issue and how to get around this since a PNP needs a negative drive signal.

When I put a scope at test point B with ground reference, I get no signal, however if I cut the circuit at either cut A, C, or both, then I get a clean -12V square wave.

I imagine there is a better way to drive this transistor, any ideas to improve the circuit so it works?

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2. ### hgmjr Moderator

Jan 28, 2005
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What is the value of your base current limiting resistor R3?

hgmjr

Oct 22, 2009
16
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R3 = 640 ohm

4. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
The problem is that a bjt 555 can't get closer than about 1.3v less than it's Vcc, which you've placed at ground potential. So, when the output of the 555 timer is high, it's actually at about -1.3v.

The cutoff voltage of your PNP transistor will be around -0.5v; it will be well into it's linear conduction at around -0.65v.

You may have to resort to a 390 Ohm pull-up resistor from the base of the transistor to ground, to get it into cutoff.

5. ### dude_500 Thread Starter New Member

Oct 22, 2009
16
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Ok, that makes sense. Why is it that there is no signal at the test point though when the transistor is connected? Shouldn't the signal still be there without making either of the disconnects at A or C?

6. ### SgtWookie Expert

Jul 17, 2007
22,183
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OK, what do you mean by "no signal"? Is it flat-lined at something less than -12v?

When A and C are connected, what voltage do you measure at point B?

What transistor are you using?

If you're measuring close to 0v when the emitter and base are connected, the transistor may be shorted due to overvoltage.

You will need to use a high-voltage transistor in your application.

7. ### dude_500 Thread Starter New Member

Oct 22, 2009
16
0
With A and C connected it measures a flat line on the scope just a hair under zero volt line, maybe -0.5V or -1V. Disconnecting either A or C brings it back to the -12V square wave

8. ### SgtWookie Expert

Jul 17, 2007
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What transistor are you using?

What is the part number of the 555 timer?

9. ### dude_500 Thread Starter New Member

Oct 22, 2009
16
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555 is a KA555, and I've tried two transistors each with similar results:
2N6520 and MJE15031G

10. ### SgtWookie Expert

Jul 17, 2007
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OK, what's the value of the collector resistor?

11. ### dude_500 Thread Starter New Member

Oct 22, 2009
16
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It's a 400 ohm, but it behaves this way even if the collector pin isn't connected to anything. Applying power to the collector or wiring it in has no impact and the voltage output out of the final output shows no modulation at all

12. ### SgtWookie Expert

Jul 17, 2007
22,183
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Strange.

You know that the 400 Ohm collector resistor should have a 50W rating, right?

Unless you're running a really short duty cycle.

13. ### dude_500 Thread Starter New Member

Oct 22, 2009
16
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It's a 20 watt wirewound, and usually it's only set to -65 or -70V, 100 is just the maximum. It is short duty cycle of only a minute or two at a time so I'm not too concerned.

14. ### SgtWookie Expert

Jul 17, 2007
22,183
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OK, the fact that it's a wirewound resistor could be causing a big problem.

When the PNP transistor turns off, the current will still try to keep flowing through the resistor; even though it's technically a resistor, wirewounds have a good deal of inductance. That inductance will result in a large reverse EMF spike, which could easily exceed the transistor's ratings.

Try connecting a 200PIV diode or higher across that resistor; cathode to ground, with a new transistor.

15. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Here is a circuit I designed a while back for another OP. The zeners try to control shoot through, where both transistors are on at the same time, and will blow the finals PDQ.

I have never built this, but I think it will work. A fuse is a must.

16. ### dude_500 Thread Starter New Member

Oct 22, 2009
16
0
SgtWookie, even if the resistor and -high voltage are removed it still behaves this way so that has no effect on the more pressing issue.

Bill_Marsden, thanks for the schematic however it is a positive polarity design and I need to be switching negative polarity to drive the grids of triodes

Last edited: Nov 9, 2009
17. ### CDRIVE Senior Member

Jul 1, 2008
2,223
99
What's a Triode?? .... Just kidding!
Since grid bias does not need to supply power I think you would be far better off with an arrangement like this. It's advantages are that Rc and all the base fwd bias resistors can be made very large which will reduce your current and pwr dissipation considerably.

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18. ### Wendy Moderator

Mar 24, 2008
20,772
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Just a suggestion, but pay attention to how I drove the PNP in my design, it will apply.