Potential divider leaves me with no current! - Electronic Circuits Projects Diagrams Free

chris414 · 24th May 2009, 08:56 PM

Hey there. Okay so im new to electronics and am having a problem... I need to turn on a transistor when a certain variable resistor is at 50k ohms. By my understanding this would require me to make a potential divider with R1 = 300k ohms and R2 = variable resistor (50k) since I am running the circuit from a 5V power supply and my transistor's threshold is at 0.7 ohms. The only reason I can think of for why it's not working is that my 300k resistance means there's not enough current left to turn on the transistor... Is there another way to do this? Or should i just look for a transistor that takes a super low current?

dknguyen · 24th May 2009, 08:58 PM

Is the transistor a BJT or FET? And are you trying to just use it as a switch (full on/off) or are you using it for some analog purpose?

Assuming you are trying to make it full on/off:

For a BJT, this would be your problem (since it is acting like a current amplifier). Yes, that's why low power (ie. high resistance) dividers cannot be used to power things or provide any real current. If you lower the resistance so you can power things you suddenly have a lot of power loss from current flowing through the resistive divider doing nothing. THe way to use a low power resistive divider as a reference to provide current at that voltage, is to buffer it with an op-amp.

However, if this is a FET it will turn on, just very slowly since current is very low (the gate capacitor needs to charges up to voltage). BUT it will not turn on because the "voltage threshold" of a transistor is where it BARELY starts to turn on. You want the transistor to saturate. Which means a higher voltage (10V for most FETs, 3.3V or 5V for more logic-level FETs). THe info should be in the datasheet and is normally the gate voltage they give that has a on-resistance (RdsOn). Anything between this voltage and the maximum gate voltage should work.

chris414 · 24th May 2009, 09:15 PM

Thanks for the speedy response. I'm not sure whether the transistor is FET or BJT coz i dont know the difference (*goes to look it up*) but this is the datasheet: http://www.datasheetcatalog.org/data...hild/BC337.pdf. It's not chosen for any specific reason other than that it was lying on my desk

Oh and i'm using it as a switch for now.

dknguyen · 24th May 2009, 09:43 PM

NPN and PNP are two complimentary types of BJTs. NMOS and PMOS are the two complimentary types of MOSFETs. THat is a BJT.

So, yes, current is your problem. For it to be used as a switch, you must be above the base threshold voltage (0.7V in this case), and be able to supply enough current so that the BJT can do it's job as a switch- amplifying the base current across the collect-emitter until it can do no more, at which point the current levels off at a maximum.

You don't need a resistive divider though. Think of the base-emitter pins as a diode connecting them. What happens when you just connect a voltage straight across a diode? Short-circuit (in reality, the resistance in the divider will limit it, but still not needed). Instead, just take a voltage higher than the BJT's saturation threshold voltage and connect it to the base through a resistor Choose this resistor so that a current large so that

(worst case current gain)*(base current) > (maximum collector-emitter current).

so the BJT can act as much like a switch as possible.

The current across this base resistor can be found with V=IR:
Ibase = (VDC - Vb_sat)/Rbase

This assumes emitter and VDC have the same reference (ground) or else you must take that difference into account as well. In this case, Vb_sat is 1.2V (base emitter voltage required for saturation). The 0.7V is for the collector-emitter voltage when the BJT is fully on (saturated).

Do you see how this works? It's exactly the same as a current limiting resistor used to drive an LED.

chris414 · 25th May 2009, 08:17 AM

I'm not sure exactly what you're suggesting? I'm using the potential divider because i need to trigger the transistor when the variable resistor (the x-axis in a joystick) goes above 50k...

dknguyen · 25th May 2009, 08:41 AM

Then you should be using a comparator driving the base through a resistor. And the potentiometer is connected to one input of the comparator and the other comparator is your trigger voltage. The BJT is not going to suddenly turn on- it's going to gradually turn on with what you have in mind.

chris414 · 25th May 2009, 09:05 AM

Thanks for the help

Hero999 · 30th May 2009, 12:17 AM

What load are you driving with the transistor?

If needs to be small >33k.

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24th May 2009, 08:56 PM	#1
chris414	Potential divider leaves me with no current! Hey there. Okay so im new to electronics and am having a problem... I need to turn on a transistor when a certain variable resistor is at 50k ohms. By my understanding this would require me to make a potential divider with R1 = 300k ohms and R2 = variable resistor (50k) since I am running the circuit from a 5V power supply and my transistor's threshold is at 0.7 ohms. The only reason I can think of for why it's not working is that my 300k resistance means there's not enough current left to turn on the transistor... Is there another way to do this? Or should i just look for a transistor that takes a super low current?

24th May 2009, 08:58 PM	#2
dknguyen	Is the transistor a BJT or FET? And are you trying to just use it as a switch (full on/off) or are you using it for some analog purpose? Assuming you are trying to make it full on/off: For a BJT, this would be your problem (since it is acting like a current amplifier). Yes, that's why low power (ie. high resistance) dividers cannot be used to power things or provide any real current. If you lower the resistance so you can power things you suddenly have a lot of power loss from current flowing through the resistive divider doing nothing. THe way to use a low power resistive divider as a reference to provide current at that voltage, is to buffer it with an op-amp. However, if this is a FET it will turn on, just very slowly since current is very low (the gate capacitor needs to charges up to voltage). BUT it will not turn on because the "voltage threshold" of a transistor is where it BARELY starts to turn on. You want the transistor to saturate. Which means a higher voltage (10V for most FETs, 3.3V or 5V for more logic-level FETs). THe info should be in the datasheet and is normally the gate voltage they give that has a on-resistance (RdsOn). Anything between this voltage and the maximum gate voltage should work. __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^ Last edited by dknguyen; 24th May 2009 at 09:06 PM.

24th May 2009, 09:15 PM	#3
chris414	Thanks for the speedy response. I'm not sure whether the transistor is FET or BJT coz i dont know the difference (goes to look it up) but this is the datasheet: http://www.datasheetcatalog.org/data...hild/BC337.pdf. It's not chosen for any specific reason other than that it was lying on my desk Oh and i'm using it as a switch for now.

24th May 2009, 09:43 PM	#4
dknguyen	NPN and PNP are two complimentary types of BJTs. NMOS and PMOS are the two complimentary types of MOSFETs. THat is a BJT. So, yes, current is your problem. For it to be used as a switch, you must be above the base threshold voltage (0.7V in this case), and be able to supply enough current so that the BJT can do it's job as a switch- amplifying the base current across the collect-emitter until it can do no more, at which point the current levels off at a maximum. You don't need a resistive divider though. Think of the base-emitter pins as a diode connecting them. What happens when you just connect a voltage straight across a diode? Short-circuit (in reality, the resistance in the divider will limit it, but still not needed). Instead, just take a voltage higher than the BJT's saturation threshold voltage and connect it to the base through a resistor Choose this resistor so that a current large so that (worst case current gain)(base current) > (maximum collector-emitter current). so the BJT can act as much like a switch as possible. The current across this base resistor can be found with V=IR: Ibase = (VDC - Vb_sat)/Rbase This assumes emitter and VDC have the same reference (ground) or else you must take that difference into account as well. In this case, Vb_sat is 1.2V (base emitter voltage required for saturation). The 0.7V is for the collector-emitter voltage when the BJT is fully on (saturated). Do you see how this works? It's exactly the same as a current limiting resistor used to drive an LED. __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^ Last edited by dknguyen; 24th May 2009 at 09:54 PM.*

25th May 2009, 08:17 AM	#5
chris414	I'm not sure exactly what you're suggesting? I'm using the potential divider because i need to trigger the transistor when the variable resistor (the x-axis in a joystick) goes above 50k...

25th May 2009, 08:41 AM	#6
dknguyen	Then you should be using a comparator driving the base through a resistor. And the potentiometer is connected to one input of the comparator and the other comparator is your trigger voltage. The BJT is not going to suddenly turn on- it's going to gradually turn on with what you have in mind. __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^ Last edited by dknguyen; 25th May 2009 at 08:42 AM.

30th May 2009, 12:17 AM	#8
Hero999	What load are you driving with the transistor? If needs to be small >33k. __________________ I do not answer private messages asking for help because no one else can: benefit from advice I may give or correct me if I'm wrong. Please ask on the open forum if you have a question and I'll be happy to help, if I know the answer.