5
\$\begingroup\$

I have build a circuit as follow. From 5V a 220 Ohm resistor to a LED to a 2N3904 transistor to 0V. The base of the transistor is connected to a 10K Ohm resistor to a switch to 5V.

When I measure the voltage across the transistor from collector to emitter with the switch open I measure 3.43V and with the switch closed 73.4mV.

From what I have read I expected a somewhat higher voltage drop with the switch closed. Is this low voltage to be expected with a circuit like this?
And the part I don't get is why I measure 3.43V with the switch open. Also with the switch open I measure 3mV over the LED and 0V over the resistor. That doesn't match the 4.98V measured from ground to +5V. Is my voltmeter affecting what I am reading?

circuit

\$\endgroup\$
  • 3
    \$\begingroup\$ What are you using for a voltmeter? If it's an old fashioned, unpowered, moving-needle type, it could well affect your circuit. But if it's any kind of DVM, don't worry about it. \$\endgroup\$ – JustJeff Feb 6 '12 at 12:32
  • 1
    \$\begingroup\$ How is the base connected? \$\endgroup\$ – Count Zero Feb 6 '12 at 12:47
  • \$\begingroup\$ The full circuit \$\endgroup\$ – jfasoc Feb 6 '12 at 14:10
  • \$\begingroup\$ The volt meter is a digital AC clamp meter which also measure DC. \$\endgroup\$ – jfasoc Feb 6 '12 at 14:12
  • \$\begingroup\$ If I connect the base to 0V. If the base is floating of connected directly to 0V it shows the same voltage as switch open. \$\endgroup\$ – jfasoc Feb 6 '12 at 14:13
6
\$\begingroup\$

I see you have several answers, but they are mostly off the mark. You are NOT seeing a drop on the LED due to leakage of the transistor. Adding a resistor to ground on the base won't fix anything since you essentially already have that between R1 and R2. These transistors do have a small amount of leakage with the base held at 0V, but again, that's now what you are seeing.

What you are seeing is the voltmeter acting like a resistor, which pulls down on the LED cathode enough to get the little bit of current it requires. LEDs are diodes, so the current at a function of voltage is quite nonlinear. With the voltmeter providing a small current path to ground, the LED forward voltage is apparently 1.55V. It is probably at 2V or so when really on. 1.55V is plausible to support the few µA or even nA that the voltmeter draws.

To prove this point, use the same voltmeter to measure between the LED cathode and the 5V supply. If the circuit causing the apparent voltage drop on the LED, you should now read about 1.5V. I predict you will read essentially 0V. That is because now there is no current path thru the LED to ground.

The LED cathode is a very high impedance node with the switch is open. So high that the voltmeter significantly effects the circuit.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Olin, the scheme was added by AndrejaKo based on our answers, and I don't think that the OP had the pull down resistor; anyway, what you've said about the voltmeter pulling down is possible. \$\endgroup\$ – clabacchio Feb 6 '12 at 16:45
  • \$\begingroup\$ @Olin - I think you are probably right. Thinking about it the LED should only need a tiny amount of current to drop some voltage. \$\endgroup\$ – Oli Glaser Feb 6 '12 at 19:08
  • \$\begingroup\$ @Oli: If the OP would measure accross the LED and R3 with the same voltmeter, we would know for sure whether it's the voltmeter or not. \$\endgroup\$ – Olin Lathrop Feb 6 '12 at 19:24
  • \$\begingroup\$ @Olin - I just did a quick test and my multimeter causes a drop. +1 \$\endgroup\$ – Oli Glaser Feb 6 '12 at 21:30
  • \$\begingroup\$ Kind of obvious, once you think about it; at low current, the LED looks like a Very Large Resistor, so providing a return path, even through 10M of DVM, would show a drop. I.e., you can assume your DVM doesn't affect the circuit only as long as its impedance is much much greater than the items being measured - and in this case, the assumption fails. +1 on you. \$\endgroup\$ – JustJeff Feb 6 '12 at 23:32
3
\$\begingroup\$

The 3.4V indicates that there's a tiny leakage current into the collector when the transistor is off. The 1.6V drop is across the LED. If there's no collector current at all the collector voltage should be close to 5V.
One way to accomplish this is to use a pull-down resistor on the base, like clabaccico suggests. The more common placement of the 1M\$\Omega\$ resistor is directly on the base, however. The resistor has a rather high value here, but the way clabacchio placed it it's parallel to the 100\$\Omega\$ resistor and is an extra load when the switch is closed. Placing it after the 100\$\Omega\$ resistor it's in series. I would pick 1k\$\Omega\$ for the series resistor and 10k\$\Omega\$ for the one to ground. That will give you a base current around 4mA, which is high enough for about any low-signal transistor; for a 20mA LED current you would only need an \$H_{FE}\$ of 5 (even a 2n3055 can do that!). Anything higher is OK, and most small-signal transistors will have an \$H_{FE}\$ of at least 100.
If the switch is off the 10k\$\Omega\$ resistor will make sure the base is at 0V and there's no collector current.

The 73mV saturation voltage is actually very good, but not impossible.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ When I connect the base to 0V there is still the same 3.5V drop across the transistor. Could one answer be that the transistor is somewhat faulty so it always leak a bit of current and that that also would make the saturation voltage lower. \$\endgroup\$ – jfasoc Feb 6 '12 at 15:03
  • \$\begingroup\$ Sorry @stevenvh, but I disagree with your explanation (a very rare occurrance). I think the LED is being biased on due to current thru the voltmeter. Also, there already is 20 kOhms pulling down the base when the switch is off, so it should be at 0V or close to it already. \$\endgroup\$ – Olin Lathrop Feb 6 '12 at 16:36
  • \$\begingroup\$ @Olin - Hadn't thought of it, but you're probably right: the voltmeter may be the culprit. \$\endgroup\$ – stevenvh Feb 6 '12 at 17:54
  • \$\begingroup\$ Sorry Steven, I had seen the scheme and I thought it was related to your answer; I didn't see that it was posted by the OP in the comment, so I'm reverting all changes I've made, and if you want you can put again the picure. Sorry again \$\endgroup\$ – clabacchio Mar 1 '12 at 12:03
0
\$\begingroup\$

If you are reading 0V across the resistor and 3mV across the LED with the switch open, where is the ~1.6V drop coming from?
EDIT - I think Olin is right about your multimeter causing the LED drop. Since an LED is very high impedance below it's Vf it will only take a tiny bit of current to cause a drop.

To test this I just set my bench supply to 5V, grabbed a standard 5mm red LED and connected the positive lead to anode, then put my digital multimeter in series between the cathode and the negative lead.
The reading was 3.76V, which proves it to be the multimeter (well in my case at least) causing enough current to flow to drop voltage across the LED.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I measured again. From 0V to the collector and to the LED cathode I measure now 3.49V and from 0V to the LED anode I measure now 5.04V. But measuring across the LED I measure now 1.6mV. I guess the volt meter is creating a path across the LED? \$\endgroup\$ – jfasoc Feb 6 '12 at 12:06
  • 1
    \$\begingroup\$ Or I guess the other way around. When I measure across the transistor that is creating a path that open up the LED and creates a 1.6V drop across it, but not enough current to make it light up? \$\endgroup\$ – jfasoc Feb 6 '12 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.