| Electronics > Beginners |
| Voltage Divider Power Question |
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| I2C:
Voltage dividers can have many combinations to achieve the same output voltage. A simple example is if your input is 10V and you want a 5V output (half) then a 1k & 1k will work, or a 10k & 10k (plus many other combinations); the important thing to know is that the maximum current you can draw will decrease as the resistance increases. As rstofer said, if it's an op-amp that's on the output then the current-draw is almost nothing so you're fine, but if the output current is higher then you can risk a significant voltage-drop, or even burning-out the resistors. It's best to check and see what the maximum current draw is first. I hope this helps! |
| bryancostanich:
I think we need to step back a bit and resolve some more fundamental questions I have. If i have two divider circuits of the following configuration: Of the following values: ``` Voltage Divider 1: R1 = 8ohms, R2 = 12ohms Total R = 20ohms Total I = 5V / 20 = 0.25A = 250mA I @ R1 = 5V / 8ohms = 0.625A Voltage Divider 2: R1 = 80ohms, R2 = 120ohms; Total R = 200ohms; Total I = 5V / 200 = 0.025A I @ R1 = 5V / 80ohms = 0.0625A ``` So first, if I understand correctly, then the max I can draw on Vout is 625mA in circuit 1, and 63mA in circuit 2. Since that's the max amount of current that can get through R1 at that voltage. Is that correct? Second, what about power loss? For this question, let's forget it's a voltage divider network and just assume it's a serial resistance. Let's say I'm running this off an 5V battery (assuming it's an idea voltage source) that has 1000mAh. In circuit 1, I can expect to have 1000mAh/250mA = 4hrs. In circuit 2, I would have 40 hrs of charge (1000/25). Yeah? Is that right? Or am I misunderstanding how resistors work, because I've read that the excess power is dissipated. but i'm not sure what that _really_ means. |
| suicidaleggroll:
I think it's important to note that a voltage divider is NOT a voltage regulator. The equation Vout=Vs*R2/(R1+R2) only applies if there's no load. As soon as you start drawing ANY current out of a voltage divider, you change the voltage. This is because you're effectively adding a load resistor in parallel with R2. Then your divider is no longer R2/(R1+R2), it's RX/(R1+RX), where RX is the parallel combination of R2 and Rload. --- Quote from: bryancostanich on September 14, 2017, 08:54:14 pm ---I think we need to step back a bit and resolve some more fundamental questions I have. If i have two divider circuits of the following configuration: Of the following values: ``` Voltage Divider 1: R1 = 8ohms, R2 = 12ohms Total R = 20ohms Total I = 5V / 20 = 0.25A = 250mA I @ R1 = 5V / 8ohms = 0.625A Voltage Divider 2: R1 = 80ohms, R2 = 120ohms; Total R = 200ohms; Total I = 5V / 200 = 0.025A I @ R1 = 5V / 80ohms = 0.0625A ``` So first, if I understand correctly, then the max I can draw on Vout is 625mA in circuit 1, and 63mA in circuit 2. Since that's the max amount of current that can get through R1 at that voltage. Is that correct? --- End quote --- Well, kinda, but in order to pull that much current out of the divider you'd have to short the output to ground (Rload=0), which means your "voltage divider" is now putting out 0V and isn't "dividing" anything. What you need to focus on is how much current you can pull out of the divider without changing the voltage more than X%, where "X" depends on your application. This will be likely be several orders of magnitude lower than the numbers you calculated above. --- Quote from: bryancostanich on September 14, 2017, 08:54:14 pm ---Second, what about power loss? For this question, let's forget it's a voltage divider network and just assume it's a serial resistance. Let's say I'm running this off an 5V battery (assuming it's an idea voltage source) that has 1000mAh. In circuit 1, I can expect to have 1000mAh/250mA = 4hrs. In circuit 2, I would have 40 hrs of charge (1000/25). Yeah? --- End quote --- Yes, if this voltage divider is your entire circuit, those calculations are correct (ignoring ESR in the voltage source). If your circuit consists of other components you'll need to add in their current draw as well. |
| Brumby:
--- Quote from: bryancostanich on September 14, 2017, 08:54:14 pm ---So first, if I understand correctly, then the max I can draw on Vout is 625mA in circuit 1, and 63mA in circuit 2. Since that's the max amount of current that can get through R1 --- End quote --- Yes, that IS the maximum current you can get through R1 - but that can only happen when R2 or the network associated with R2 is zero. You have implicitly stated this in the formula you used: --- Quote ---I @ R1 = 5V / 8ohms = 0.625A --- End quote --- By having no other resistances in the formula, you are stating every other part of the circuit is a simple wire. The 8 ohms is the total load seen by the source. --- Quote --- at that voltage. Is that correct? --- End quote --- If by "voltage" you mean Vs, then Yes. If you mean Vout - then NO. To get that maximum current, the output voltage will be Zero. This is the issue with simple resistor dividers - any additional load you put on them will affect the voltages at all points of the network outside of the source. |
| rstofer:
The resistors dissipate power no matter what. The power P in Watts is given by: P = I * E P = I^2 * R P = E^2 / R If you remember one formula, you can easily derive the others by using Ohm's Law. For example: Look at P = I^2 * R. Expand that to P = I * (I * R) Well, we know that E = I * R so the equation falls back to P = I * E. P in Watts I in Amps E in Volts R in Ohms |
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