# How do I figure out these resistors?

Discussion in 'General Electronics Discussion' started by xgabrielx, Oct 19, 2015.

1. ### xgabrielx

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0
Jan 1, 2010
I was looking at 555 timer circuits. I can see what this does...it is supposed to turn the transistor on and off at about 10khz. But I haven't really looked into transistors yet. I'm kinda learning stuff either as I need it, or, like now, when I see something that raises a question. I guess the one closer to the IC is like the grid resistor on a vacuum tube? They haven't labelled them...what does the other one do? The one between the base and emitter? How would I calculate the values of both using this circuit as an example? Oh, and how critical is it? Surely there's a wide range of what works? I don't know how this will react as the battery voltage drops from the 9.8v or whatever of a fresh battery, but I'm just interested in the actual maths, not their specific circuit. I guess if they just use a value somewhere between a fully charged battery and a "dead" battery, everything will be within acceptable tolerances?
I haven't looked into transistors much, but I thought to use them as a switch, you just needed a current flowing from base to emitter, and that as long as it was enough current for that tranmsistor, then collecter to emitter would simply function like an open switch. (or closed..can never remember which way round, because it's a bad choice of word) Did I miss some things? I guess I did, bceause I don't see a point in those resistors based on what I've read so far. Is there some relationship between the base to emitter current and the collector to emmitter current that must be kept within a certain range? Is that the point of the resistor on the left?
Then, what's the one from base to emitter for? Is it some non-essential thing, or does it have to be a specific value?

1,114
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Aug 13, 2011
I count 10 question marks. Would you care to focus on one?

3. ### xgabrielx

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Jan 1, 2010
Well, I guess it can be narrowed down to 2 questions. What do they do, and how do you calculate what values are required?

4. ### Martaine2005

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912
May 12, 2015
Hello,
They are both voltage dividers of sorts.
The 8k2 and 68k charge the capacitor. When charged the 68k discharges it.
The component values used determine the frequency, ie flash rate. on/off rate.
The datasheet explains the math involved Here.

Martin

Last edited: Oct 19, 2015
5. ### xgabrielx

6
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Jan 1, 2010
Oh, I know ;D I meant the two that don't have values next to them. Sorry, I should have been more clear about that

6. ### Martaine2005

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May 12, 2015
No, you were quite clear.
But I cannot help with the math. I have a friend that helps me with that!!!
The resistor from the base to 0v is a pull down resistor. It is to do with leakage or noise from collector to emitter. It basically keeps the transistor turned off .
Would you believe I have edited this about 10 times!! . Because I don't wont to start a war!!
Only because I can't win it.

Martin

Last edited: Oct 19, 2015
Arouse1973 likes this.
7. ### davennModerator

13,802
1,941
Sep 5, 2009
LOL yup

PLEASE desist from any argument/discussion on that topic in this thread. ... anything further like that and I will delete those posts

cheers
Dave

8. ### Martaine2005

3,336
912
May 12, 2015
It never fails......Just different players...Lol.

Martin

9. ### hevans1944Hop - AC8NS

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Jun 21, 2012
The 555 has a "totem-pole" output. See attached PDF. The output on pin 3 is either pulled up toward Vcc or pulled down toward ground. You can use just about any value resistor in the range of 1 kΩ to 10 kΩ to provide base current for the transistor that is switching the inductor on and off. The resistor from the base to the emitter is optional. You can safely omit it. If you want to leave it in, choose any value greater than 1 kΩ.

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10. ### dorke

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665
Jun 20, 2015
heavans,
How would a beginner learn if you try to "chew " it for him...?

xgabrielx,
If you want to learn, come back to this thread .
there is much more on this issue...

11. ### davennModerator

13,802
1,941
Sep 5, 2009
OK a move of several posts to a new thread and a deletion of a bunch of others

please just discuss the purpose and values of the 2 specified resistors

Dave

12. ### GPG

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66
Sep 18, 2015
The 555 will pull down much better than up So the b-e resistor is immaterial. The resistor from pin3 should saturate the transistor. If you calculate it based on a low beta this will do. The idea is called forced beta
http://whites.sdsmt.edu/classes/ee320/notes/320Lecture24.pdf
In your case the relay is 38Ω and the supply 9V so relay current = ~240 mA
Forcing beta to say 20 = Ib~ 20 mA
Under worst case the 555 Vout will be ~ V+ -1,5 and Vbe = .6v so the voltage available to drive 20mA is~7V
So 7V/ 20mA =~350Ω Nearest E12, 380Ω And that is being conservative.

Last edited: Oct 25, 2015
13. ### eetech00

95
8
Nov 17, 2014
Basically, you need to know:
Base input voltage
Relay drive current
Gain of transistor

Hi..

Example:
Vb = 9v <--base input voltage (same as (Vcc) supply voltage)
Ic = 236mA <--Collector current (current to drive relay = 9v/38 = 236mA (estimate))
hfe = 100 <--Transistor Gain = 100
Vbe = 0.7v <-- Voltage drop across Transistor BE junction

How much base current to pass 236mA through collector?:
Ib = Ic/hfe = 0.236/100 = 0.00236
Ib = Ic/hfe = 0.300/100 = 0.003

Base Resistor:
Rb = (Vb-(0.7v))/Ib = 9v-(0.7v)/0.003 = 2767
Use nearest value = 2.7k

Estimate Resistor from base to ground
10 * Rb = 27670 use 27k

Make sure Tip41c is on heatsink.

14. ### GPG

452
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Sep 18, 2015
.Correcting error:
Forcing beta to say 20 = Ib~12 mA
Under worst case the 555 Vout will be ~ V+ -1,5 and Vbe = .6v so the voltage available to drive .12mA is~7V
So 7V/ 12mA =~583Ω Nearest E12, 560Ω.
https://www.fairchildsemi.com/datasheets/TI/TIP41C.pdf
beta may be as low as 30
Using a logic level fet makes things easier.

15. ### dorke

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Jun 20, 2015

Taking that circuit as a theoretical drill only.

Correction:
For the relay to be on, we need the Tr. in saturation.
(deep saturation that is ,not just on the verge of it)

The Beta we need is the minimum one (take into account temp,IC current, etc.)
for TIP41C,Beta would be 15.
BTW, this Tr. is an overkill for driving that relay.

VBE sat=0.8V
VCEsat=0.2V

IC=8.8/38=231ma
IB=231/15=15ma

So design for:
IB=at least 20ma.
VBEsat=0.8V

and that TR. doesn't need an heat sink at all!