My CD4050 is leaking voltage. User error?

Papabravo

Joined Feb 24, 2006
17,616
Are you perhaps confusing the behavior of the CD4050 and the CD4050B. The "B" parts were introduced to solve some of the deficiencies of the original parts.
 

ronv

Joined Nov 12, 2008
3,770
Here is what I think is happening, though I must admit it is not clear from the data sheets. :(
1- There is a clamp diode to Vcc. On which side of the resistor, I'm not sure - perhaps both.
2- If you read the data sheet in says maximum input current is 10 ma. So don't just tie an input to +5 (although if the clamp is on the other side of the unknown resistor it may be ok.)
3- " it can function as a "CMOS to DTL/TTL Hex Converter". CMOS parts do not output much current in the high state. Maybe 1 or 2 ma.. So much less than tying the input to +5.
4- If the 3.3 volts does not have much of a load it's voltage will rise thru the clamp diode connected to +5 thru the input. Your symptom.
So, what can we do to confirm it?
If you have a variable supply we can place a nice load on the 3.3 volts. Say a 33 ohm 1/2 watt resistor.
Put a 330 ohm resistor in series with the input to limit the current to under 10 ma.
Slowly raise the output to the 330 ohm resistor until you see it draw some current. Probably around 3.9 volts (one diode drop above the 3.3 volt supply). This should confirm that it is a clamp diode.
Raise the input voltage to 5 volts and measure the voltage at the input pin. This will give us an idea of if or how large the resistor in the input of the IC is and which side of the resistor.
Calculate the current thru the input (from the drop across the 330 ohm)
Measure the 3.3 volt supply to see how much the voltage went up.
 

Thread Starter

colinta

Joined Jun 25, 2014
11
Here is what I think is happening, though I must admit it is not clear from the data sheets. :(
1- There is a clamp diode to Vcc. On which side of the resistor, I'm not sure - perhaps both.
2- If you read the data sheet in says maximum input current is 10 ma. So don't just tie an input to +5 (although if the clamp is on the other side of the unknown resistor it may be ok.)
3- " it can function as a "CMOS to DTL/TTL Hex Converter". CMOS parts do not output much current in the high state. Maybe 1 or 2 ma.. So much less than tying the input to +5.
4- If the 3.3 volts does not have much of a load it's voltage will rise thru the clamp diode connected to +5 thru the input. Your symptom.
So, what can we do to confirm it?
If you have a variable supply we can place a nice load on the 3.3 volts. Say a 33 ohm 1/2 watt resistor.
Put a 330 ohm resistor in series with the input to limit the current to under 10 ma.
Slowly raise the output to the 330 ohm resistor until you see it draw some current. Probably around 3.9 volts (one diode drop above the 3.3 volt supply). This should confirm that it is a clamp diode.
Raise the input voltage to 5 volts and measure the voltage at the input pin. This will give us an idea of if or how large the resistor in the input of the IC is.
Calculate the current thru the input (from the drop across the 330 ohm)
Measure the 3.3 volt supply to see how much the voltage went up.
This makes a ton of sense, thank you! I will try this experiment tonight and report. I do have a small variable supply handy (just arrived yesterday in fact - and a benchtop supply is wrapped up under the tree right now!)
 

Thread Starter

colinta

Joined Jun 25, 2014
11
Are you perhaps confusing the behavior of the CD4050 and the CD4050B. The "B" parts were introduced to solve some of the deficiencies of the original parts.
The parts are listed as CD4050B and CD4050BE, so if they are labeled correctly then I am using the updated parts.
 

Papabravo

Joined Feb 24, 2006
17,616
It is the case that the "B" and "non-B" parts have different protection networks. The "B" parts should be immune to overvoltage on the inputs unless those protection diodes are in reverse breakdown.
 
Top