Common emitter as level switcher

First, I'm a transistor idiot.

I wanted to change a 5V 1 MHz clock signal to 12V.

I put this together,

+12V--10kohm----+----Vout | |/ in--5V 1MHz-1kohm-+---| 2N3904 | |\>

| | +--1k-+ Gnd

Worked, but it took forever (~100 nS) to turn off. The base was driven to near 800 mV. When the square wave clock shut off the base just drifted down slowly for 100ns and then dropped. The base to ground resistor was added to try and turn the base off quicker. It barely helped.

If I don't drive it as hard will it turn off faster? Make the input to base resistor ~5k to give 0.8V at the base with 5V in?

Thanks George H.

Oh feel free to suggest a better circuit too!

Reply to
George Herold
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(oops wrong title the first time.)

Reply to
George Herold

Look for "Bakers Clamp", that is a diode feed back from the collector to the base circuit to prevent the long term cap at the base.

Jamie

Reply to
Jamie

Excellent, Thanks Jamie.

George H.

Reply to
George Herold

Tie a Schottky from the base (anode) to the collector (cathode). This will keep the transistor out of saturation and hopefully buy you the speed you need. Another alternative would be to use an N-FET there instead of the NPN.

Reply to
krw

will

e NPN.- Hide quoted text -

Great, I was wondering if a Fet would work better. The only charge I have to worry about is whatever's on the gate capacitance?

I'll try the Bakers clamp/ Schottky first since there's already an npn 'lashed' in place. using TTL circuitry seems like a nice retro touch.

George H.

Reply to
George Herold

10K is kinda big for the collector pullup. 10K * 10 pF = 100 ns, so even the scope probe will slow it down. Try 1K. A series inductor might be fun, too. Try L = 0.5 * R^2 * C or thereabouts.

A small cap across the base drive 1K is a classic speedup. And make the Thevenin voltage drive into the base more like 0,1.2 than the existing 0,2.5, namely reduce the lower base resistor, 330 ohms maybe.

The 3904 will saturate and turn off slowly. A small mosfet, like a

2N7000, with a 1K pullup would be faster.

John

Reply to
John Larkin

NPN.- Hide quoted text -

And drain. The 10K pullup resistor has to charge whatever capacitance is on the drain (+ load). You can reduce this, obviously at the cost of power.

Also, you have a 1K base resistor, which is going to give you about 3mA of base current ((5-.7)/1K). Your collector current is only ~1mA (10V/12K). This is going to saturate the hell out of the transistor. You might reduce the base drive an order of magnitude (or two).

The problem is the base charge. All that charge in the base needs to flush out through the emitter, taking Beta times as much current from the collector with it. The Schottky keeps the transistor out of deep saturation (technically saturation is Vbc positive) by bleeding current from the base to the collector.

Good plan. But also try increasing the base resistor to at least 10K, maybe

50K.

That's not a bad idea either, though I'm not sure there are 1G types with 15V or 30V outputs.

BTW, if the positive transition needs to be cleaner than the negative, turn the whole thing upside down and change the sex of the transistor.

Reply to
krw

First thing that crosses my mind is the roughly 4mA of base drive and the roughly 1mA of collector current. That is hard saturation and will place a lot of electrons in the collector drift region (I think.) You might want to run it closer to a beta of 20 or 30, not 0.25. Also, the other thing I note is that you are getting 100ns turn off (no surprise) and running at 1MHz. So 10% turnoff isn't good for you. You want less? From your numbers, the two 1k resistors (moving to off state) are in parallel at provide 500 ohms to ground for that charge to leave. Have you tried a 'speed up' capacitor across your base drive resistor? I've used them with some success. And they go back a LONG TIME. I think before Baker clamps.

I'm not that much of a designer. But this is basics. So that's me. Anyway, here is a shot at it.

Take your above circuit. Get rid of the two 1k resistors and replace them with something that delivers on the order of a

20 beta. (It is a 2N3904, after all. It can do well.) So 1/20th of 1mA or 20,000 times the 4V or so differential you will have, less the Vbe of the BJT. That would get near 80k ohm, but drop it to 56k or thereabouts. At 1mA Ic, you need about Vbe=0.65V (or less, really.) So 0.65/5*56k is around 7k pull down. But that is R_tot and at 0.65V and 1/20mA base, this is about R_base=13k for the BJT so the 7k is after taking that into account, too. This means the pull down is about the same -- 13k-ish. I'd use 15k. Then put a speed up cap across the 56k. I think most BJTs have small pF to worry about. So make it a 12-15pF. Something about that size.

Then see how it goes. You won't be saturating the hell out of it, anyway. But I've no idea what you are driving, either.

Just a hobbyist thought.

Jon

Reply to
Jon Kirwan

Great that's much nicer! The base to collector zener got rid of the

100ns delay, but it was till taking another 100nS to get up to 12 Volts. (looking with a ~16pF 'scope probe)

A series inductor

Ouuu.. that's getting a little too complicated. (..tune out a bit of the C with the L.) I just wanted to add a some voltage to the clock. And now it's a growing rat's nest on the side of the prototype.

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Yeah, I got that. The Bakers clamp 'name' from Jamie was golden. There was a reference in the first editon of AoE (which lives at home.) and the 'speed up' cap was right next to the Bakers clamp. At work today I coudn't find this in the second edition.

And make

Thanks George H.

Reply to
George Herold

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OK, I can see that, Thanks. George H.

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George H.

Reply to
George Herold

I'll give it a quick try. (Reducing the base current and voltage... KRW suggested the same thing.) I've reduced the collector resistor to 1k so my collector current is a bit more now.. but I'll scale things accordingly.

George H.

Reply to
George Herold

--oops that's a base to collector Schottky diode----

Reply to
George Herold

Try 100pF, then.

Jon

Reply to
Jon Kirwan

Use a mosfet! 2N3904s belong in museums next to the 6SN7s.

John

Reply to
John Larkin

At 5-10x the cost, you just lost 1000 Joerg points.

Reply to
krw

That's only .01% of a cent.

--
You can't have a sense of humor, if you have no sense.
Reply to
Michael A. Terrell

Ok, make that 1000K Joerg points.

Reply to
krw

We pay $ 0.0208 for 2N7002s. And it doesn't need a schottly clamp diode plus two or three passive components in the base drive path.

John

Reply to
John Larkin

Since you bring it up, I never pay more than 0.6 cents for a

2N2222 or 2N3904/6 BJT. And I'm a damned hobbyist.

But the main thing is that George already said, "I'll try the Bakers clamp/ Schottky first since there's already an npn 'lashed' in place." And who knows? He might get by with a speed up, even.

Jon

Reply to
Jon Kirwan

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