Voltage Divider Problem

[Moleculo]

I need some help from all you smart circuit / electronics people. This seems like a simple problem, but apparently I don't understand it: I wanted to drop the input voltage a little for a project I'm working on (more about that later) so I made a simple voltage divider. When I looked around at my boxes of resistors, I found some that would let me drop the input voltage of 18 volts to just under 16 volts. I used a 68Kohm for R1 and two 200Kohm resistors in series for 400Kohm total for R2 leg. I measured the voltage before hooking up the circut and it was perfect...just under 16v. The problem is, when I hook it up to the circuit, the input voltage at the board was only 1v! While it was hooked up, I checked the voltage at the R1/R2 junction and it was just under 16v, but at the board it was still 1v. I bypassed the voltage divider altogether and ran the 18v straight to the board and it measured at 18v at the board. Why is my voltage divider showing proper voltage when not attached to the circuit, but only 1V when attached? What am I missing?

 

[Captain Kilowatt]

Sounds like you have R1 and R2 reversed and you won't see much (if any) voltage drop until you start to draw current. Remember that V= I²R and if I=zero then the voltage drop will be zero. Either that or the circuit draws a LOT more current than you think.

 

[Moleculo]

I already checked/recheck to make sure that I didn't have R1 & R2 reversed. The primary current consumption on this circuit is a transistor that has a maximum collect current rating of 70mA.

Is this just a case of using too much resitance and it's causing a ton of voltage drop under load as a result? Perhaps if I picked up some small value ohm resistors?

 

[Captain Kilowatt]

How did you come up with using 68K and 400K in the first place? Remember that R=V x I so with both V and I being a small value the R will also be a very small value. Why the worry about 16-18 volts anyway? A simple series resistor of 28.5 ohms ( or 27 ohms close enough standard value) should work FB.

Are you sure there is not a short somewhere like a transistor being latched on when power is applied? Still sounds like waaaay to much current certainly a lot more than 70mA.

Try this:


http://www.raltron.com/cust/tools/voltage_divider.asp

 

[Moleculo]

Voltage Divider Calculator

Haha...that's the same calculator I used...because I'm math lazy Actually, I'm not really that worried about a 2V difference, but I was doing it for posterity because the kit is only "rated for 16v". Interesting enough, I originally was using much lower value resistors rated at 1/4 watt, but they couldn't handle the current draw & started smoking. So the values I ended up using were the 1/2 watt ones I had on hand that would achieve the same voltage according to the calculator. Looking at the simple scematic (I'll post it after I scan it in the morning), it does look like the transistor is latched on while power is applied, so I think you're right about the current draw being potentially fairly high. As I'm typing this, I'm thinking, "No shit, dumbass...if the 1/4 watt resistors smoked the current draw is higher than you thought".

Anyway, I'll post the schematic...

 

[bob85]

what are you missing? load impedance,
you wont get a load stable voltage even if the divider or series resistor is fed from a regulated 18v supply, the higher the divider network impedances the more it will drop under load,
using high wattage resistors of lower impedance should make it less load sensitive but it also loads the supply far more wasting energy as heat in the divider and supply,
if your load is dynamic or anything but very low current / high load impedance you would be much better off using a 3 terminal regulator,

try a 7815 or at a push a 7812 with a resistor between the middle leg and ground to pull the voltage up to where you want it under load,

good luck.

 

[Captain Kilowatt]

Bob is right about the dynamic load part. Any change in load current will result in a change in voltage drop. If the 1/4 watt resistors were smoking there was a LOT more than 70mA flowing.

 

[Dr_DX]

Why don't you just use a few diodes? At .6v~.7v per diode, 3 in series will drop you to below 16v and still have plenty of current carrying capability. 1N4007's will give you 1 amp, 1N5408's will give you 3 amps.

 

[bob85]

diodes will drop the voltage, i have used them in low current applications where stability is not an issue with good results nicad chargers ect but you have a similar problem to a divider network with dynamic loads,

the trouble mole is seeing with his divider is that the load impedance of the circuit hes trying to power is effectively in parallel with r2 of the divider which pulls the voltage down or swings it around if the load is not a constant current draw..

 

[Dr_DX]

I guess I am missing something. What I meant was scrap the voltage divider all together and just use the 3 diodes to reduce the 18v to 16v going to the input of his "project". No R1 or R2 to deal with. Not really what he asked but it is his overall goal.

 

[bob85]

dr_dx you are not missing something
volt drop across a diode is a function of the current through the diode and diode temperature,
at low currents you have little vdrop, as current increases vdrop increases,
if you only use a string of diodes with low current dynamic loads the voltage at the load will swing up and down in an unregulated fashion, at higher currents the voltage becomes more stable

 

[linearone]

I would use 1-3 amp diodes like 1n5408's or 1n40001's and just use two of them in series. each will drop .9 volts. plenty of current handling

 

[HiDef]

Last time I looked it took a very tiny amount of current to realize the junction voltage drop of a 1N5408. DX's idea should work just fine.

 

[Moleculo]

Thanks for all the suggestions guys. I understand what's going on now and thanks for helping me come up with a different way of doing it.

BTW, here is the schematic I was feeding:

 

[linearone]

what is it? an amp of some kind? looks like a tiny amp judging by the transistor. like a receive amp?