# Pullup resistor-Pls explain

Discussion in 'General Electronics Chat' started by winall, Mar 11, 2008.

1. ### winall Thread Starter New Member

Mar 11, 2008
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0
Hi,

I know abit about transistor circuit, but dont really understand how connecting a resistor can really "pullup" the output.
Can anybody explain.

Thanks.

2. ### jpanhalt Expert

Jan 18, 2008
7,533
1,808
Look at the diagram below. In "A" there is a device with an open collector. If you measured the output, it would be near zero. There is no source of current at the output. In "B" I have added a pull-up resistor. Now when the device turns on, current flows from the supply, through the resistor, collector, device, and to ground. The output will show a voltage proportional to the current.

Hope that helps. John

3. ### winall Thread Starter New Member

Mar 11, 2008
4
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What I see is that when the transistor turns on, the voltage from output to GND is near 0v (0.7VDC) because it's just the voltage drop on emitter in its open state. I dont understand the explanation of the voltage drop proportional to current flow on the resistor because it's above the output line. Can u explain more?

4. ### beenthere Retired Moderator

Apr 20, 2004
15,808
295
Ohm's law explains it. The transistor will pass current if it has proper bias and some base current flowing. Current through the transistor collector has to be the same as current in the resistor, as it is a series circuit.

5. ### jpanhalt Expert

Jan 18, 2008
7,533
1,808
An explanation must always be viewed in the eyes of the reader, which I am not very good at doing. On re-reading my explanation, I see where your confusion comes from.

In A, the voltage at the "output" will be low (it is floating) regardless of whether the transistor is turned on or not. In B, the the voltage at the "output" at the collector will be high when the transistor is off, and low when the transistor is turned on. The voltage at the "output" when the transistor is off is "pulled high" by the resistor; hence, the term "pull-up resistor."

I hope that is a little clearer. John

6. ### winall Thread Starter New Member

Mar 11, 2008
4
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Hi John,

Yr new explanation is much clearer now.
But in B, when the transistor is off, the voltage at the collector output is "floating" because its ground connection has been cut due to transistor off-state. It means that the voltage at the collector output is dependent on the other ground connected to it, if one connected.
Can you give more explanation?

7. ### jpanhalt Expert

Jan 18, 2008
7,533
1,808
The voltages are measured relative to the same ground. Thus, in B, the voltage at the collector is not floating. It is at Vcc relative to the emitter and to ground. That is, the switch is open, and you are simply measuring the voltage across the switch.

By analogy, Vcc is not normally directly connected to ground with a wire or closed switch, yet it is measured relative to its ground and is not said to be floating.
John

8. ### winall Thread Starter New Member

Mar 11, 2008
4
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If so, you can connect the output directly to Vcc, not through a pullup resistor. BTW, because the output is "open", not connected to other circuit, there's no current flowing to the output in open state, thus no voltage drop on the pullup resistor, in effect the voltage at output is almost the same as Vcc

9. ### jpanhalt Expert

Jan 18, 2008
7,533
1,808
I am not sure I understand everything you said. I would not connect an open collector directly to Vcc without knowing more about the circuit. For example, if the emitter is connected to ground, as in a comparator, then when the output turns on, you effectively would have a short to ground if the collector were tied to Vcc. The datasheet for IC devices will tell you whether you can leave the collectors floating or not.
John