# Voltage regulation/change between two ICs

Discussion in 'General Electronics Chat' started by SPQR, Nov 29, 2012.

1. ### SPQR Thread Starter Member

Nov 4, 2011
379
49
Hello all, another simple question for you.

My question is practical (I'm building something now), and theoretical (I'd like to learn about a new IC or learn a bit more about transistors).

I'm building what I call a "Poor Man's Oscilloscope" (I'll post it for fun when I finish it, you'll all have a good laugh).

I would like to take the output of a 555 (9V) and drive the clock terminal of a 74HC4020 (binary counter). The 4020 will only take a maximum of 6V.

So my question is, what is the best way to lower a voltage down to a level to put on an IC pin, but be able to maintain frequency response? (1Hz to 100kHz).
I thought of a 7805, but it rejects ripple above 120Hz, so I can't use that.

So I was thinking of the circuit below.
(Is this an emitter follower circuit?)
1. The output from any circuit (3-18V) connected to the base of the transistor.
2. A "pull-down" resistor between the emmiter and ground, with the pin of the following IC connected to the emmiter/resistor connection.

1. Is this circuit reasonable and possible?
2. I have some ifrz46n and TIP125 hanging around, can I use those transistors?
3. Is there a better way to do this?

The "theoretical" question.
Can you drive a transistor's base at a higher voltage than the voltage connected to the emitter and collector?
Is this commonly done to change voltages between two parts of a circuit?

I thank you all in advance.

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2. ### GopherT AAC Fanatic!

Nov 23, 2012
8,025
6,788
Put the resistor on the Collector side of the transistor and the output to your logic chip between the transistor and the resistor.

You can bring down the input level from your 555 by simpling using two resistors connected to ground. The node between the two resistors connect to the base of your transistor.

I'll let you calculate the values but make sure total load on the 555 is not excessive (that is, not too many amps). Also make sure there is enough current going through those resistors to drive your transistor. Also make sure the resistors are the right ratios to insure you can send (1) a logic low and (2) a logic high. Check the datasheets on your logic to make sure you are in the safe range. Remember, the 555 is normally outputting 1/3 Vcc for Low state and 2/3Vcc for high state. Depending on your logic device's datasheet, you might be able to avoid the transistor all together.

Good luck.

3. ### kubeek Expert

Sep 20, 2005
5,345
1,009
So basically you want to make a frequency counter of some sorts? Because if you simply convert the input signal to digital it has nothing on similar with an oscilloscope.
But if you want to use it like this, then you should use a 74hc4040 which has ouputs from all the stages of the counter. Also instead of a simple transistor you should use a comparator, this will allow you to set the "trigger" level of the counter.

4. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,958
1,827
Do you realize you can run the 555 off the same 5V as the 4020 and not have level shifting problems?

Keep it simple. Just sayin'.

5. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
+1 Just lower the voltage into the 555 and then it's output will match the needs of the 4020.

6. ### SPQR Thread Starter Member

Nov 4, 2011
379
49
Excellent!

@GopherT - very nice recommendation on the transistor. I'll play a bit tonight and I've got a small shipment of many different types of transistors coming soon, so I can experiment with your ideas.

@kubeek - Aha! A chip I haven't used before. I've just ordered some comparators, and I think they'll make a nice addition to my future projects. Thats a very slick way of converting X signal input volts to the supply voltage of the comparator.

@ErnieM and bountyhunter - Yes, I did that, and everything seemed to work fine. What I was hoping was to have a more "general" solution. For example, if I had a circuit that was running off of a 9V battery, I'd have to figure out a way not to fry the 4042. Further, this gives me the opportunity to learn a bit about transistors, and about a new chip - the comparator. I was playing with the 4020 the other night, and accidently inverted the VCC and Ground - boy did it get HOOOOOT!! - but after I switched the wires around, it cooled nicely and even worked!!

7. ### MKCheruvu Member

Nov 20, 2012
30
6
With reference to your question :"Can you drive a transistor's base at a higher voltage than the voltage connected to the emitter and collector?",I would like to add the following-
In this case the driven Transistor base gets clamped to C-B Forward Bias(say5.6V) due to collector supply 5v.Hence ,Under normal working conditions the emitter also gets clamped. However when the 555 output is higher than (say 5.6v) it will drive transistor C-B gets into forward conduction mode driving into 5v supply and unless limited this current is drawn from 555 output/and its supply.

8. ### SPQR Thread Starter Member

Nov 4, 2011
379
49
Ok, excellent.
So the answer is, you can drive the base-emitter with a certain voltage/current combination (as long as you don't fry the transistor), but the collector-emitter voltage "clamps" or "limits" the maximum voltage on base-emitter.

So I might have a collector-emitter voltage of say 5V, and the base emitter could be driven by a voltage source of 30V, but the internal workings of the transistor would only allow 5V to be seen at the base terminal of the transistor.

Am I getting that right?

9. ### ramancini8 Active Member

Jul 18, 2012
473
145
A CD4020B has a 20 volt rating, so you can run the 555 and 4020 off the same 9V. If required, clamp the 4020 output to the logic supply to limit the input voltage to the next stage.

10. ### crutschow Expert

Mar 14, 2008
20,514
5,810
Yes. You can view the base-emitter and base-collector junctions as two diodes with the anodes tied to the base (for an NPN). Normally the base-emitter junction is forward biased and the base-collector junction is reverse biased, but when the base voltage becomes higher than the collector voltage then both junctions become forward biased and the base voltage can go no higher than the collector voltage (plus the junction voltage drop of course).

SPQR likes this.
11. ### SPQR Thread Starter Member

Nov 4, 2011
379
49
Oh...OK. But I was looking at THIS datasheet, and it lists the maximum Vcc as 6V.
That was my hesitation for going any higher.
Thanks very much!

12. ### SPQR Thread Starter Member

Nov 4, 2011
379
49
Excellent! And I thank you.
This gives me greater insight into the "workings" of the transistor, and a different perspective on how to think about transistors - A series of two diodes connected together.

Superb, as usual. Thanks!

13. ### crutschow Expert

Mar 14, 2008
20,514
5,810
The problem is there are two different devices with 4020 designation. The old slower CD4020 device can tolerate higher voltage than the newer and faster SN74HC4020.

14. ### atferrari AAC Fanatic!

Jan 6, 2004
3,071
1,149
I was used to consider 15V the highest usable Vdd for the CD40XX series.

At least, when I learnt that, reading an RCA databook 35++ years ago.

15. ### SPQR Thread Starter Member

Nov 4, 2011
379
49

Ah yes, the subtlties of expertise! I thought that there was only ONE of everything!
There are good things you can learn on this forum!
Thanks again.

16. ### atferrari AAC Fanatic!

Jan 6, 2004
3,071
1,149
Subtleties? Oh no, just in case check the range of acceptable Vdd for an HC chip. Definitely different from an CD40XX CMOS.