Are you sure you can't drive both transistors off that same output? You're guaranteed 0.44 mA, which would be easily be enough to drive two typical low power NPNs into saturation with 15mA loads.
Sylvia.
Yes, I could drive both transistors from the same CMOS output. Yes, I could also piggyback on the LED. Yes, I could also use two transistors (which I stated in my initial post).
I had already considered those solutions, but I decided against them for a number of reasons before I made my initial post. While I did mention the last one, I apologize for not listing the first two in my initial post.
What I am attempting to do is to make use of the inverted output coming from the CMOS IC, using just one transistor (and associated circuitry to run that additional transistor). So far, Pimpom's suggestion of using a zener on the base of a PNP seems to be the only solution proffered.
Thanks,
Jon
Thanks Jamie, that would work, but it seems a bit overkill (to me) for driving a little piezo beeper at 15mA. I'd also have to order it, as there is no NTE sub for that part, and I don't have that in my parts bin.
Jon
That would throw out the idea of using logic 0 to activate it. I still like my idea of using his existing circuit to drive that nice opto. Either way, 2 active components are going to be needed here, unless he elects to use a zener of ~ 15 volts in line with Rb.
Jamie
You know, that's a good point... :)
I don't think so.
Assume the piezo requires Ip=15mA and Vp=? volts, for now. When the piezo is sounding at 15mA, your R2 must be
I can't find any statement that the IC will sink current when it's output is high, much less how much current it will sink. Maybe it's obvious from the FETs in the output stage that it will sink current (I'm not at all familiar with FETs) but there's still the question of how much, and in particular, how much it can sink and remain operational over the long term.
Sylvia.
No, just drive the emitter or source. But logic inversion is easy anyhow, often free.
John
--
Indeed; that's why I'll stick with my original solution of an N MOSFET
(with, conceivably, no gate resistor) being driven by the 4013's Q
output or an NPN with a gate resistor being driven from the same
point.
That is, if Q isn't being loaded so heavily by what's already there
that my scheme won't work.
_
If it won't, and he's forced to use Q, then I'd bite the real-estate
bullet and use two NPN's, like this:
+18
+18 |
| [BUZZER]
[10k] |
| C
+--[10k]--B 2N3904
| E
_ C |
Q>--[10k]--B 2N3904 GND
E
|
GND
No Zeners, no tricks, no marginal operation, just a rock-solid switch.
It'll (theoretically) sink as much as it can source. Think about it... in the "high" state the P-channel MOSFET is on... it's just a resistor until IDSS is exceeded.
I've actually used this to actively snub relatively low current inductive loads. ...Jim Thompson
--
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| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
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| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
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I love to cook with wine. Sometimes I even put it in the food.
About that last statement, just for information: The output FETs in 4xxx ICs can drive several mAs reliably. The output current is partly limited by the FETs' output resistance. The original datasheets by RCA- also adopted by TI, etc - show output curves for typical and worst-case devices.
For example, those curves show that the short-circuit output current for a typical device with a 5V supply is about 4.2mA. That's 21mW dissipation per logic unit. With a 10V supply, this goes up to 20mA and 200mW.
You're right. I missed that post. Sorry, no plagiarism intended.
-- Excellent reply, John! v8k5j6pi5emb2nqfq3f74j0v6jefsm7khu@4ax.com
The specific chip is a TC4013BP, datasheet here:
My chip is running off of a 5V supply.
I tested the transistor (KN2907) I will be using, and I was able to successfully run the 15mA beeper with a base current of 0.8mA, using a base resistance of 22k. I could probably go lower, given that the Hfe of the transistor is measured at 170.
Which brings up another question: what percentage of Hfe is usually considered adequate to drive the transistor into saturation (how low of a base current could I practically go to reliably drive it into saturation)?
Jon
I discarded that route initially, as it would require too much reworking of the original board (more reworking than I wish to do). There already exists an empty solder pad coming off of the inverted output, hence my desire to make use of it.
Jon
You don't need the 10K between C and B of the 2N3904s.
He could also use one of these, plus one resistor:
Where it is specified is more than adequate, usually Ic/Ib = 10. I often use 20, unless operation at temperature extremes is required (and even then, self heating may help the cause, but it requires analysis).
-- Right. When the bottom transistor's off, the top one would get Ib through the 10k to 18V, and when the bottom one's on, the base of the top one gets pulled to GND. Good catch; thanks. :-)
No. The OP wants to turn the beeper on when the FF output is low.
There can be no hard and fast rule because the measured Hfe depends on the levels of current and Vce at which it was measured (even if we disregard the effects of temperature). Hfe decreases with Vce, rapidly as saturation is approached. Ic/Ib = 10 is usually used for specifying Vce(sat) in datasheets. My own rule of thumb is to use that one-tenth rule (Ib = Ic/10) for low-beta devices and Ib = Ic/20 or Ic/30 for higher betas. These are for absolute peace of mind. I sometimes use Ic/40 or Ic/50 in a pinch for types like the BC547B I mentioned earlier (PNP version = BC557B).
The KN2907 is obviously a clone of the popular PN2907 which is itself a plastic version of the venerable 2N2907. 0.8mA of base current should be quite adequate. You probably won't notice any difference with 0.5mA. Even if the transistor is not driven into hard saturation, that kind of load will not burn it up at any level of base drive.
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