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# Goofingn off with another Transistor problem. Slightly diffe

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#### expelleior

##### Member level 1
I grabbed this picture, of some site and thought it would be simple to work out, however I can't make sense. Does the second 1k resistor to 0V decrease the base current ? If it was meerly a pulldown i would have thought a 10K resistor would do the job.

I calculated the base current v/1k = 9mA ,

As the base current is supposed to be 1/10 of collector i know this must be wrong as there is only a LED with a 680 Resistor drawing collector current.

Sorry i uploaded same picture twice but at least i know how to make the correct size GIF's now!
Shoudnt the resistor being connected to the te V0 be more like a 10K resistor ?

I calculated the base current of the transistor and it was 9v/1000 = 9ma

dont want motr than that flowing through the collrcter.

1.What LEDS capable of drwainmg curren t. I know ther arre A few sizeds.
2. As there asre two 1k restistlors in the circuit ; is this taking solme current fvrolm the Base of the transistor:idea:.

Thaks` Mkike
Also

Mike.

daftar base transistor

The vertical 1K resistor does absolutely nothing but draw 9mA from the supply. It plays no part in the operation of the circuit whatsoever.

The horizontal one passes ((9 - Vbe)/1000) or about 8.4mA.

The transistor would almost certainly be in saturation so the LED current would be set by the 680 Ohm resistor so its current would be ((9 - Vled) / 680) or about 9mA.

Note to the purists: I have not run a simulation and both LED and transistor specifications are missing so the figures are not exact!

Brian.

### expelleior

Points: 2
silicon transistor leakage versus temperature

betwixt said:
The vertical 1K resistor does absolutely nothing but draw 9mA from the supply. It plays no part in the operation of the circuit whatsoever.

The horizontal one passes ((9 - Vbe)/1000) or about 8.4mA.

The transistor would almost certainly be in saturation so the LED current would be set by the 680 Ohm resistor so its current would be ((9 - Vled) / 680) or about 9mA.

Note to the purists: I have not run a simulation and both LED and transistor specifications are missing so the figures are not exact!

Brian.
Daer Brian, are you aware of a freebie simulation softwaere,? Or tireial versionds ?

one transistor off turning another transitor on

There are many simulators, a search on Google will find them.

A good place to start:

Brian.

### expelleior

Points: 2
680 divided by 1000

There is no real need for the two each 1k resistors and the transistor. Just the 680 resistor and the LED in series with the switched 9 V would do the same thing.

### expelleior

Points: 2
typical transistor problem base collector current

flatulent said:
There is no real need for the two each 1k resistors and the transistor. Just the 680 resistor and the LED in series with the switched 9 V would do the same thing.
WOW,you learn something everyday, so long as its not about transistors.

betwixt said:
The vertical 1K resistor does absolutely nothing but draw 9mA from the supply. It plays no part in the operation of the circuit whatsoever.

The horizontal one passes ((9 - Vbe)/1000) or about 8.4mA.

The transistor would almost certainly be in saturation so the LED current would be set by the 680 Ohm resistor so its current would be ((9 - Vled) / 680) or about 9mA.

Note to the purists: I have not run a simulation and both LED and transistor specifications are missing so the figures are not exact!

Brian.

Thats what I calculated Brian, but i tought i must be making a blunder of somesort ar the base current is no-where near 1/10 of collector current ?

Betwit, being a hypothetical and all, still can you explain your previous statement that the BASE is 1/10 running through collector anmd emmitter ? Is this 1/10 the minimum needed ?

base current calculator transistor thumbs rule

That was your statement not mine!

The base current is nearer Ic/Hfe so it is less if the transistor has higher gain. I think the myth of 1/10 comes from a rule of thumb that (rightly or wrongly) says you should have about 10 times more current flowing through your bias network than is actually carried by the base connection. If you use that rule, it suggests that if your base current is say 100uA you should allow for 1 mA down the bias network with 90% of it passing straight through, 10% being conducted through the base. It doesn't apply unless you have a resistor to supply and a resistor to ground with the base connected to their junction.

In your schematic, the overriding factor is Vbe, the voltage between the base and emitter. As this is a PN junction, it will hold a fairly constant 0.6V (for a silicon transistor). As all the base current flows through the 1K resistor and the current in a series circuit is the same throughout, it can be calculated by using Ohms law, the voltage is 9V minus the 0.6V in the B-E junction divided by 1000 Ohms which gives 8.4mA.

Assuming your transistor is not selected for dismally low current gain, 8.4mA is enough to make the transistor fully conduct so the voltage between its collector and emitter will be very low ("Vce sat" from its data sheet) and can be ignored. Its as though the transistor was a switch in the 'on' setting so the current through the LED can also be calculated directly by Ohms law.

LEDs typically drop 1.5V across them although it varies with the color and exact manufacturing technique. So the LED current is set only by the resistor, 9V minus 1.5V divided by 680 which gives 8.99mA.

I think you may be looking at transistors 'backwards' if that makes sense. Concentrate on the collector current and then work out the base current needed to achieve it. It looks like you are setting a base current first then wondering why the collector current isn't ten times more. Other factors, such as the LED and 680 Ohm resistor will determine the collector current, not the Ib calculation.

Brian.

### expelleior

Points: 2
transistor base current calculation

OK, and thankyou again. I have an off topic question. I have made a coulple of Keypads,connected to microcontrollers and UHF chips. These 20 or 30 keypads will be placesc on/beside machines. One computer will recieve all the keydata. Would you go UHF, Bluethooth or Another wireless protocol ?

use transistor backwards low gain

That's an enormous jump in technology from your earlier questions!

The answer is it depends on many factors, primarily the range they have to cover and if local regulations allow those frequencies to be used for that purpose.

1. Bluetooth - should be fairly easy to implement but short range and possibly unreliable. Imagine if the person using the keypad had a Bluetooth enabled mobile phone and headset, that's two other transmitters operating within arms length of yours.

2. UHF, you can probably use 433MHz or 868MHz without a license but check first, regulations vary from place to place. Modules are inexpensive and very easy to use. I use 5 networked 433MHz modules at home to control equipment in outbuildings over a 50 metre range without problems. You have to share the frequency with wireless weather stations, car remote locking and other remote controls so some sort of error protection as absolutely necessary.

I think a lot more information, particularly about the geographic layout and expected data traffic is needed before making decisions.

Brian.

### expelleior

Points: 2
transistor leakage iceo switch

expelleior,
The vertical Resistor is there to keep the transistor from turning itself on due to leakage current when the switch is off. Its value (1K) is much lower than it needs to be. At room temperature, the leakage current of a typical planar transistor is usually low enough that it is not a problem. Consult the data sheet for the leakage current (Icbo) specification, remembering to use the value at the highest operating temperature. You may have to estimate this value, or get it off the Icbo vs temperature curve (if one is supplied on the data sheet).
.
Regarding the Ic/10 rule, this is probably overkill for a transistor with a relatively high beta. The idea is to provide enough base current to guarantee that the transistor is saturated. The rule that I use is Ib = 10 * Ic/Hfe. This gives a conservative (high) value of base current. Datra sheet Hfe values are given at a relatively high Vce value e.g. 10V. The Hfe value at or near saturation is always lower than the data sheet value, but it is almost never specified. This is why a conservative value of Ib should be used. Remember that you must use worst case (low) Hfe at the coldest operating temperature for this calculation. If the Hfe = 100, then a base current of Ic/10 would be appropriate.
Regards,
Kral

In the first paragraph of my reply, I meant to say Iceo, not Icbo. Icbo is usually not given on a data sheet.

### expelleior

Points: 2
override factor + transistor

So where does the base current (and collector current) come from when the switch is off ???

I know where you are coming from, a 'leak' or 'bleed' resistor is to discharge internal leakage current in the transistor base junction so it doesn't turn on by itself if it isn't driven. In this case the whole supply is turned off when the switch s opened.

Brian.

can a transistor run on 1.5v

If, as betwixt states, the whole supply is turned off when the switch is opened, then you don't have a problem with leakage, and you can leave the resistor out of the circuit. If, as betwixt states, the whole supply is turned off when the switch is open, why don't you drive the LED directly from the supply thru the current limiting resistor, and eliminate the transistor circuit?
Regards,
Kral

Re: Goofingn off with another Transistor problem. Slightly d

betwixt said:
I think the myth of 1/10 comes from a rule of thumb that (rightly or wrongly) says you should have about 10 times more current flowing through your bias network than is actually carried by the base connection.
No.
We are not talking about a voltage divider with 10 times the base current.
We are talking about the base current should be 1/10th the collector current for most transistors to saturate well, as shown on datasheets.

hFE is used for a linear amplifier transistor that is not saturated so it has plenty of collector to emitter voltage.

It seems to be foolish for this transistor to have as much base current as its collector current.

Bluetooth - should be fairly easy to implement but short range and possibly unreliable. Imagine if the person using the keypad had a Bluetooth enabled mobile phone and headset, that's two other transmitters operating within arms length of yours.

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### expelleior

Points: 2
Re: Goofingn off with another Transistor problem. Slightly d

betwixt said:
There are many simulators, a search on Google will find them.

A good place to start:

Brian.

Thanks for the link to SPICE, its alittle more complex for me at this time. I am enrolled in Electronic Engineering next year with some emphsis on Microcontroller programming . 2 year course.