Potential divider - Electronic Circuits Projects Diagrams Free

prasannan82 · 17th April 2007, 03:51 PM

Hi all,
A potential divider can be made by using the ratios of two resistor values.but with the same ratio many other different resistors can be chosen.like 1k,2.2k and 10k,22k and 100k,220k.But which resistor values to be used?What does that depend on and how?

**Nigel Goodwin** · 17th April 2007, 03:54 PM

Like everything else, it depends on ohms law - which you can use to calculate them, instead of a simple ratio. The actual values are dependent on the other values they connect to (input and ouput impedances) and what you are using them for.

gramo · 17th April 2007, 03:58 PM

Say you had a 10V power supply and two 10K resistors in series to earth, the total resistance is 20K, therefore the total current going through the circuit is I = V / R

10V / 20000ohm = 0.0005 (500uA)

If two 1K resistors are used, then the total resistance = 2K, so the total current would be

10V / 2000ohm = 0.005 (5mA)

prasannan82 · 17th April 2007, 04:00 PM

My application is like this.
I have to build an voltage attenuator ckt to attenuate input to 1.2v(using potential divider technique) whose input is an audio from milli volts to max 4 volts.But the output of my amplifier should clip only at 1.2v input to it.How much value resistor should i use?

**Nigel Goodwin** · 17th April 2007, 04:17 PM

You don't mention the input impedance the potential divider has to feed, nor the source impedance that feeds the divider - both of these are crucial.

prasannan82 · 17th April 2007, 04:20 PM

the input impedanc for the divider is 300 ohm and the output of divider should go to the amplifier whose input impedance is from 60k to 150k.

**Nigel Goodwin** · 17th April 2007, 04:24 PM

OK then, and you want to go from 4V down to 1.2V - so the ratio of resistors needs to be 2.8 to 1.2 - so a 28K and a 12K should be fine, 28K may need to be made by paralleling resistors, or just use a 27K instead.

prasannan82 · 17th April 2007, 04:30 PM

But why not 2.8k and 1.2k resistors.

prasannan82 · 17th April 2007, 04:31 PM

What is the proper rule for that?

**Nigel Goodwin** · 17th April 2007, 05:09 PM

It depends on the input and output impedances like I said, in this case the source impedance was very low (300 ohms) so can be ignored - otherwise it goes in series with the top resistor of the potential divider. The other consideration is the load impedance, which was 60K minimum - this is in parallel with the bottom resistor.

As a 'rule of thumb' the bottom resistor needs to be no higher than 1/5 of the load - which is why I choose 12K (the 12 actually just came from the 1.2 value for the divider output).

So the values I gave were probably as high as you should go - you can safely reduce them, but shouldn't really increase them.

prasannan82 · 17th April 2007, 05:11 PM

OK.Thank you very much.

huttojb · 21st April 2007, 10:29 AM

Wouldn't you use a T type Attenuator or a Pi type Attenuator for this sort of application??

Jason

**Nigel Goodwin** · 21st April 2007, 11:00 AM

Quote:

Originally Posted by huttojb

Wouldn't you use a T type Attenuator or a Pi type Attenuator for this sort of application??

Well he doesn't mention the application?, but you would only do that where you're wanting impedance matching, which is much less common (such as RF circuits).

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17th April 2007, 03:51 PM	(permalink)
prasannan82	Potential divider Hi all, A potential divider can be made by using the ratios of two resistor values.but with the same ratio many other different resistors can be chosen.like 1k,2.2k and 10k,22k and 100k,220k.But which resistor values to be used?What does that depend on and how?

17th April 2007, 03:54 PM	(permalink)
Nigel Goodwin	Like everything else, it depends on ohms law - which you can use to calculate them, instead of a simple ratio. The actual values are dependent on the other values they connect to (input and ouput impedances) and what you are using them for. __________________ PIC programmer software, and PIC Tutorials at: http://www.winpicprog.co.uk

17th April 2007, 03:58 PM	(permalink)
gramo	Say you had a 10V power supply and two 10K resistors in series to earth, the total resistance is 20K, therefore the total current going through the circuit is I = V / R 10V / 20000ohm = 0.0005 (500uA) If two 1K resistors are used, then the total resistance = 2K, so the total current would be 10V / 2000ohm = 0.005 (5mA) __________________ Spency. PIC Micro's - Your mind is the limit PIC's and interfacing with other devices - a PIC Basic Guide @ digital-diy.net

17th April 2007, 04:00 PM	(permalink)
prasannan82	My application is like this. I have to build an voltage attenuator ckt to attenuate input to 1.2v(using potential divider technique) whose input is an audio from milli volts to max 4 volts.But the output of my amplifier should clip only at 1.2v input to it.How much value resistor should i use?

17th April 2007, 04:17 PM	(permalink)
Nigel Goodwin	You don't mention the input impedance the potential divider has to feed, nor the source impedance that feeds the divider - both of these are crucial. __________________ PIC programmer software, and PIC Tutorials at: http://www.winpicprog.co.uk

17th April 2007, 04:20 PM	(permalink)
prasannan82	the input impedanc for the divider is 300 ohm and the output of divider should go to the amplifier whose input impedance is from 60k to 150k.

17th April 2007, 04:24 PM	(permalink)
Nigel Goodwin	OK then, and you want to go from 4V down to 1.2V - so the ratio of resistors needs to be 2.8 to 1.2 - so a 28K and a 12K should be fine, 28K may need to be made by paralleling resistors, or just use a 27K instead. __________________ PIC programmer software, and PIC Tutorials at: http://www.winpicprog.co.uk

17th April 2007, 04:30 PM	(permalink)
prasannan82	But why not 2.8k and 1.2k resistors.

17th April 2007, 04:31 PM	(permalink)
prasannan82	What is the proper rule for that?

17th April 2007, 05:09 PM	(permalink)
Nigel Goodwin	It depends on the input and output impedances like I said, in this case the source impedance was very low (300 ohms) so can be ignored - otherwise it goes in series with the top resistor of the potential divider. The other consideration is the load impedance, which was 60K minimum - this is in parallel with the bottom resistor. As a 'rule of thumb' the bottom resistor needs to be no higher than 1/5 of the load - which is why I choose 12K (the 12 actually just came from the 1.2 value for the divider output). So the values I gave were probably as high as you should go - you can safely reduce them, but shouldn't really increase them. __________________ PIC programmer software, and PIC Tutorials at: http://www.winpicprog.co.uk

17th April 2007, 05:11 PM	(permalink)
prasannan82	OK.Thank you very much.

21st April 2007, 10:29 AM	(permalink)
huttojb	Wouldn't you use a T type Attenuator or a Pi type Attenuator for this sort of application?? Jason